Representing endothermic and exothermic processes using energy diagrams

WEBVTT
Kind: captions
Language: en

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- [Instructor] Let's
say we run an experiment

00:00:01.730 --> 00:00:05.510
to determine if a reaction
is endo or exothermic.

00:00:05.510 --> 00:00:10.510
For our hypothetical reaction,
A reaction would B to form C.

00:00:10.580 --> 00:00:13.290
And let's say this reaction
takes place in aqueous solution

00:00:13.290 --> 00:00:15.180
in a beaker.

00:00:15.180 --> 00:00:18.259
We can define our system as
the reactants and products

00:00:18.259 --> 00:00:20.850
that make up our chemical reaction,

00:00:20.850 --> 00:00:23.130
and everything else is
part of the surroundings,

00:00:23.130 --> 00:00:25.475
for example, the water and also the beaker

00:00:25.475 --> 00:00:28.630
the reaction is taking place in.

00:00:28.630 --> 00:00:30.100
Let's say we run the reaction

00:00:30.100 --> 00:00:31.667
and we put our hand on the beaker

00:00:31.667 --> 00:00:34.710
and we feel that the beaker is warm.

00:00:34.710 --> 00:00:35.860
If the beaker is warm

00:00:35.860 --> 00:00:38.700
since the beaker is part
of the surroundings,

00:00:38.700 --> 00:00:41.130
energy must've been
transferred from the system

00:00:41.130 --> 00:00:42.920
to the surrounding.

00:00:42.920 --> 00:00:45.790
So heat flowed from the
system to the surroundings.

00:00:45.790 --> 00:00:49.160
And that's an example of
an exothermic reaction.

00:00:49.160 --> 00:00:53.280
So Delta H is negative.

00:00:53.280 --> 00:00:54.938
We can determine the amount
of energy that flowed

00:00:54.938 --> 00:00:56.700
from the system to the surroundings

00:00:56.700 --> 00:00:58.177
by looking at the energy profile

00:00:58.177 --> 00:01:01.670
for our hypothetical reaction.

00:01:01.670 --> 00:01:04.680
In an energy profile, potential
energy is in the y-axis

00:01:04.680 --> 00:01:06.733
in kilojoules per mole
and reaction to progress

00:01:06.733 --> 00:01:08.980
is on the x-axis.

00:01:08.980 --> 00:01:12.230
So as we move to the right on the x-axis,

00:01:12.230 --> 00:01:14.820
the reaction is occurring.

00:01:14.820 --> 00:01:16.810
Our reactants, which are A and B

00:01:16.810 --> 00:01:18.750
have a certain amount of potential energy.

00:01:18.750 --> 00:01:21.780
So that's right here
on our energy profile.

00:01:21.780 --> 00:01:25.350
So that part represents the
energy of our reactants.

00:01:25.350 --> 00:01:28.000
Our reactants react
together to form our product

00:01:28.000 --> 00:01:30.840
which is C and that's at the very end.

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So over here, this line
represents the energy

00:01:34.480 --> 00:01:37.533
the potential energy of our products.

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Notice how the potential
energy of our reactants

00:01:41.320 --> 00:01:43.800
is higher than the potential
energy of the product.

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So if we were to find
the change in energy,

00:01:46.900 --> 00:01:49.070
that would be the final minus the initial.

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So the energy of the
products minus the energy

00:01:52.760 --> 00:01:56.120
of the reactance for this energy profile,

00:01:56.120 --> 00:02:00.100
the energy of the products is
about 50 kilojoules per mole,

00:02:00.100 --> 00:02:03.310
and the potential energy
of our reactance is at 100.

00:02:03.310 --> 00:02:06.946
So this would be 50 minus 100

00:02:06.946 --> 00:02:11.946
which is equal to negative
50 kilojoules per mole.

00:02:15.470 --> 00:02:18.470
So on our energy profile,
we could show Delta E

00:02:18.470 --> 00:02:20.940
which would be this difference right here.

00:02:20.940 --> 00:02:23.300
So that represents Delta E

00:02:23.300 --> 00:02:27.374
and the change in energy Delta
E is also equal to the change

00:02:27.374 --> 00:02:32.374
in the enthalpy Delta H for this reaction.

00:02:32.730 --> 00:02:34.530
So we know that the change in enthalpy

00:02:34.530 --> 00:02:37.900
is equal to negative 50
kilojoules per moles.

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Let's go ahead and plug that
in over here on the left.

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So by feeling the outside of the beaker,

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we knew that the reaction was exothermic,

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but the energy profile
allowed us to figure out

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how much energy was
transferred from the system

00:02:53.220 --> 00:02:55.510
to the surroundings.

00:02:55.510 --> 00:02:58.560
So for an energy profile, when
the energy of the reactants

00:02:58.560 --> 00:03:01.180
is higher than the energy of the products,

00:03:01.180 --> 00:03:05.300
this is the energy profile
for an exothermic reaction.

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Let's say we ran a similar reaction

00:03:07.460 --> 00:03:10.090
where A plus B turned into C,

00:03:10.090 --> 00:03:11.780
but this time when we felt the beaker,

00:03:11.780 --> 00:03:14.620
the beaker felt cool to the touch.

00:03:14.620 --> 00:03:16.400
If that's the case, it's because energy

00:03:16.400 --> 00:03:21.150
was being transferred from the
surroundings to the system.

00:03:21.150 --> 00:03:23.920
And since these surroundings
was losing energy,

00:03:23.920 --> 00:03:27.000
that's why the beaker felt cool.

00:03:27.000 --> 00:03:29.840
So heat flows from the
surroundings to the system,

00:03:29.840 --> 00:03:33.400
and this occurs in an endothermic reaction

00:03:33.400 --> 00:03:36.940
and the change in entropy,
Delta H is positive

00:03:36.940 --> 00:03:39.810
for an endothermic process.

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When we look at the energy profile

00:03:41.123 --> 00:03:43.740
for an endothermic reaction.

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The energy of the reactants.

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So go ahead and write reactants in here.

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The energy of the reactants is lower

00:03:50.990 --> 00:03:54.083
than the energy of the products.

00:03:56.343 --> 00:03:59.313
So this time, if we find Delta E

00:03:59.313 --> 00:04:03.000
that would be the energy of our products

00:04:03.000 --> 00:04:06.210
minus the energy of our reactants.

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And the energy of our products

00:04:07.710 --> 00:04:11.050
is about 100 kilojoules per mole

00:04:11.050 --> 00:04:13.528
and the energy of our
reactants is about 50.

00:04:13.528 --> 00:04:16.600
So let's say it's 100 minus 50,

00:04:16.600 --> 00:04:21.513
which would be positive
50 kilojoules per mole.

00:04:24.890 --> 00:04:28.840
So on our diagram if we represent Delta E,

00:04:28.840 --> 00:04:32.520
that would be this difference
here on the energy profile.

00:04:32.520 --> 00:04:36.998
And once again, Delta E
is equal to the change

00:04:36.998 --> 00:04:41.600
in the enthalpy Delta H for the reaction.

00:04:41.600 --> 00:04:44.230
So Delta H for this hypothetical reaction

00:04:44.230 --> 00:04:48.583
is positive 50 kilojoules per mole.

00:04:51.310 --> 00:04:52.369
Since Delta H is positive,

00:04:52.369 --> 00:04:54.794
we know that energy was
transferred from the surroundings

00:04:54.794 --> 00:04:56.690
to the system.

00:04:56.690 --> 00:04:58.550
And that's the reason why the products

00:04:58.550 --> 00:05:01.890
have a higher potential
energy than the reactants

00:05:01.890 --> 00:05:03.663
in our energy diagram.


Endothermic and exothermic processes

WEBVTT
Kind: captions
Language: en

00:00:01.400 --> 00:00:02.660
- [Educator] Before we get into the terms,

00:00:02.660 --> 00:00:04.090
endo and exothermic,

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We need to look at some
other thermodynamics terms

00:00:06.480 --> 00:00:07.590
that are used.

00:00:07.590 --> 00:00:09.060
For example, system.

00:00:09.060 --> 00:00:12.520
The system refers to
the part of the universe

00:00:12.520 --> 00:00:14.270
that we are studying.

00:00:14.270 --> 00:00:17.770
For our example, we're going
to consider a monatomic gas.

00:00:17.770 --> 00:00:21.460
Let's say we have some helium
particles in a container.

00:00:21.460 --> 00:00:24.970
And the helium gas represents our system.

00:00:24.970 --> 00:00:28.440
The surroundings are everything
else in the universe.

00:00:28.440 --> 00:00:31.548
So that would include this piston here

00:00:31.548 --> 00:00:33.080
and the cylinder

00:00:33.080 --> 00:00:34.580
in which the gas is in.

00:00:34.580 --> 00:00:37.720
And the universe consists
of both the system

00:00:37.720 --> 00:00:39.723
and the surroundings.

00:00:40.670 --> 00:00:43.890
Next, let's look at the
first law of thermodynamics

00:00:43.890 --> 00:00:45.080
which can be summarized

00:00:45.080 --> 00:00:47.060
by writing delta E is equal

00:00:47.060 --> 00:00:49.100
to Q plus W.

00:00:49.100 --> 00:00:52.470
Delta E is the change
in the internal energy

00:00:52.470 --> 00:00:53.620
of the system.

00:00:53.620 --> 00:00:56.460
And the internal energy refers to the sum

00:00:56.460 --> 00:00:59.510
of all the kinetic and potential energies

00:00:59.510 --> 00:01:01.750
of the components of the system.

00:01:01.750 --> 00:01:05.580
Since we have a monatomic
gas for our system,

00:01:05.580 --> 00:01:07.900
we only have kinetic energy.

00:01:07.900 --> 00:01:10.670
So if you could imagine
adding up the kinetic energy

00:01:10.670 --> 00:01:12.140
for each particle,

00:01:12.140 --> 00:01:14.910
the sum of those kinetic
energies for this example

00:01:14.910 --> 00:01:17.450
would be equal to the internal energy

00:01:17.450 --> 00:01:18.910
of the system.

00:01:18.910 --> 00:01:21.490
Q refers to the heat

00:01:21.490 --> 00:01:24.640
that's transferred from the system.

00:01:24.640 --> 00:01:28.270
So it's either transferred
to or from the system.

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And W refers to the work done on

00:01:31.730 --> 00:01:33.253
or by the system.

00:01:34.440 --> 00:01:36.080
Let's look at the sign conventions

00:01:36.080 --> 00:01:38.300
for the first law of thermodynamics.

00:01:38.300 --> 00:01:40.030
When we think about Q;

00:01:40.030 --> 00:01:43.930
when heat flows into the
system from the surroundings,

00:01:43.930 --> 00:01:47.750
we say that Q is positive.

00:01:47.750 --> 00:01:49.740
When heat flows out of the system

00:01:49.740 --> 00:01:51.520
and into the surroundings,

00:01:51.520 --> 00:01:54.943
we say that Q is negative.

00:01:56.270 --> 00:01:59.530
For work, when work is done on the system

00:01:59.530 --> 00:02:01.080
by the surroundings,

00:02:01.080 --> 00:02:03.610
the work is positive.

00:02:03.610 --> 00:02:05.070
But if work is done

00:02:05.070 --> 00:02:07.420
by the system on the surroundings,

00:02:07.420 --> 00:02:10.400
work is negative.

00:02:10.400 --> 00:02:12.840
It's very useful to think
about internal energy

00:02:12.840 --> 00:02:14.240
like a bank account.

00:02:14.240 --> 00:02:15.900
So if Q is positive

00:02:15.900 --> 00:02:17.520
and work is positive,

00:02:17.520 --> 00:02:20.950
that's like money coming
into your bank account.

00:02:20.950 --> 00:02:22.440
But if Q is negative

00:02:22.440 --> 00:02:23.800
or the work is negative,

00:02:23.800 --> 00:02:26.983
that's like money leaving
your bank account.

00:02:28.040 --> 00:02:30.610
Let's look at an example of
the first law of thermodynamics

00:02:30.610 --> 00:02:32.460
using our sample of helium

00:02:32.460 --> 00:02:33.640
that's in a cylinder

00:02:33.640 --> 00:02:35.720
with a movable piston.

00:02:35.720 --> 00:02:40.150
So let's say that 6,000 joules of energy.

00:02:40.150 --> 00:02:45.150
So 6,000 joules flows from the
surroundings into the system.

00:02:47.290 --> 00:02:49.950
And that heats up our helium particles,

00:02:49.950 --> 00:02:50.980
which then expand

00:02:50.980 --> 00:02:53.800
and push the piston up.

00:02:53.800 --> 00:02:56.580
So the piston gets pushed up,

00:02:56.580 --> 00:03:01.310
which is the system doing
work on the surroundings.

00:03:01.310 --> 00:03:03.660
And let's say that,

00:03:03.660 --> 00:03:08.300
that's 2000 joules of work done

00:03:08.300 --> 00:03:11.550
by the system on the surroundings.

00:03:11.550 --> 00:03:13.230
Using our assigned conventions,

00:03:13.230 --> 00:03:17.210
since heat flowed into the
system from the surroundings,

00:03:17.210 --> 00:03:19.520
that's a positive value for Q.

00:03:19.520 --> 00:03:22.020
And since the work was done

00:03:22.020 --> 00:03:24.380
by the system on the surroundings,

00:03:24.380 --> 00:03:26.650
that's a negative for the work done.

00:03:26.650 --> 00:03:28.200
So we can go ahead and plug in

00:03:28.200 --> 00:03:31.190
to the first law of thermodynamics.

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The heat transferred is
positive 6,000 joules

00:03:35.080 --> 00:03:40.080
and the work done as negative 2000 joules,

00:03:40.480 --> 00:03:42.100
therefore delta E

00:03:42.100 --> 00:03:45.290
or the change in the
internal energy is equal

00:03:45.290 --> 00:03:49.373
to positive 4,000 joules.

00:03:50.220 --> 00:03:53.430
Thinking about internal
energy like our bank account,

00:03:53.430 --> 00:03:56.100
we've gained 4,000 joules.

00:03:56.100 --> 00:03:59.190
So that could be like gaining $4,000.

00:03:59.190 --> 00:04:02.360
Since the system has gained 4,000 joules,

00:04:02.360 --> 00:04:06.690
that must mean the surroundings
has lost 4,000 joules.

00:04:06.690 --> 00:04:08.210
But since energy is conserved,

00:04:08.210 --> 00:04:12.180
the total energy of the
universe remains constant.

00:04:12.180 --> 00:04:14.240
Let's apply the first
law of thermodynamics

00:04:14.240 --> 00:04:16.420
to the combustion of propane

00:04:16.420 --> 00:04:19.360
and an open container
at constant pressure.

00:04:19.360 --> 00:04:20.680
For the combustion of propane,

00:04:20.680 --> 00:04:22.950
the reactants and products

00:04:22.950 --> 00:04:25.150
for the combustion reaction are considered

00:04:25.150 --> 00:04:26.280
to be the system,

00:04:26.280 --> 00:04:29.720
and everything else is the surroundings.

00:04:29.720 --> 00:04:34.400
So this combustion reaction
gives off 2,044 kilojoules

00:04:34.400 --> 00:04:35.380
of energy.

00:04:35.380 --> 00:04:38.160
So that's the heat that's
transferred from the system

00:04:38.160 --> 00:04:39.840
to the surroundings.

00:04:39.840 --> 00:04:43.020
The system also does
two kilojoules of work

00:04:43.020 --> 00:04:44.420
on the surroundings.

00:04:44.420 --> 00:04:47.140
So by convention, we make that negative.

00:04:47.140 --> 00:04:49.610
To find the change in the
internal energy of the system,

00:04:49.610 --> 00:04:51.450
we add Q plus W

00:04:51.450 --> 00:04:56.250
and get negative 2,046 kilojoules.

00:04:57.690 --> 00:04:59.220
Since this reaction was carried out

00:04:59.220 --> 00:05:01.090
under constant external pressure,

00:05:01.090 --> 00:05:03.960
we can write Q sub P here.

00:05:03.960 --> 00:05:07.330
So Q sub P is the heat

00:05:07.330 --> 00:05:10.280
that's transferred at constant pressure,

00:05:10.280 --> 00:05:14.690
and that is equal to the
change in the enthalpy,

00:05:14.690 --> 00:05:18.230
which is symbolized by delta H.

00:05:18.230 --> 00:05:20.550
So the change in enthalpy is the heat

00:05:20.550 --> 00:05:23.600
that's transferred at constant pressure.

00:05:23.600 --> 00:05:25.150
So the change in enthalpy

00:05:25.150 --> 00:05:26.920
for the combustion of propane is equal

00:05:26.920 --> 00:05:31.570
to negative 2,044 kilojoules.

00:05:31.570 --> 00:05:35.280
And notice how the change
in enthalpy is almost

00:05:35.280 --> 00:05:38.800
the same value as the change
in the internal energy.

00:05:38.800 --> 00:05:42.060
So the work done by the
system is a very small amount

00:05:42.060 --> 00:05:42.893
in this case,

00:05:42.893 --> 00:05:44.090
and that's usually the case.

00:05:44.090 --> 00:05:46.710
And since most chemical reactions are done

00:05:46.710 --> 00:05:48.700
under constant pressure,

00:05:48.700 --> 00:05:52.270
chemists care more about
the change in the enthalpy

00:05:52.270 --> 00:05:55.870
than they do about the change
in the internal energy.

00:05:55.870 --> 00:05:57.980
When delta H is negative,

00:05:57.980 --> 00:06:00.530
we call that an exothermic process.

00:06:00.530 --> 00:06:04.733
So the combustion of propane
is an exothermic reaction.

00:06:05.810 --> 00:06:09.390
An endothermic process is
where heat is transferred

00:06:09.390 --> 00:06:11.890
from the surroundings to the systems.

00:06:11.890 --> 00:06:16.310
So the system has gained
heat from the surroundings.

00:06:16.310 --> 00:06:17.520
The change in enthalpy,

00:06:17.520 --> 00:06:21.060
delta H is positive for
an endothermic process.

00:06:21.060 --> 00:06:24.880
An example could be melting an ice cube.

00:06:24.880 --> 00:06:28.230
If heat flows from the
system to the surroundings,

00:06:28.230 --> 00:06:30.890
the system has released
heat to the surroundings,

00:06:30.890 --> 00:06:32.230
and the change in enthalpy,

00:06:32.230 --> 00:06:34.190
delta H is negative.

00:06:34.190 --> 00:06:36.930
And we call this an exothermic process.

00:06:36.930 --> 00:06:38.810
And we already saw an example of that.

00:06:38.810 --> 00:06:43.240
The combustion of propane
is an exothermic reaction.

00:06:43.240 --> 00:06:44.980
So a phase change could be an endo

00:06:44.980 --> 00:06:46.800
or an exothermic process.

00:06:46.800 --> 00:06:49.010
A chemical reaction could also be an endo

00:06:49.010 --> 00:06:50.890
or an exothermic process.

00:06:50.890 --> 00:06:54.370
And even the formation of a
solution could be classified

00:06:54.370 --> 00:06:57.690
as an endo or an exothermic process.

00:06:57.690 --> 00:06:59.740
Let's think about making a solution.

00:06:59.740 --> 00:07:01.440
So let's say that we have a beaker

00:07:01.440 --> 00:07:03.040
and we have it full of water,

00:07:03.040 --> 00:07:04.770
and we take a solid

00:07:04.770 --> 00:07:07.350
and we dissolve the solid in the water

00:07:07.350 --> 00:07:10.240
to form a solution.

00:07:10.240 --> 00:07:12.853
If the dissolution process
is an exothermic process,

00:07:14.470 --> 00:07:17.510
that means the system releases heat

00:07:17.510 --> 00:07:18.640
to the surroundings.

00:07:18.640 --> 00:07:21.270
And since the beaker is
part of the surroundings,

00:07:21.270 --> 00:07:23.630
if we were to put our hand on the beaker

00:07:23.630 --> 00:07:25.250
and the beaker feels hot,

00:07:25.250 --> 00:07:26.740
we know that the disillusion

00:07:26.740 --> 00:07:30.283
of this particular solid
is an exothermic process.

00:07:31.140 --> 00:07:33.850
If we do the same thing
with another solid;

00:07:33.850 --> 00:07:36.740
so we dissolve this solid in water

00:07:36.740 --> 00:07:37.960
to form a solution.

00:07:37.960 --> 00:07:40.320
And we put our hand on the beaker,

00:07:40.320 --> 00:07:42.510
and this time the beaker feels cold.

00:07:42.510 --> 00:07:44.360
The reason the beaker feels cold is

00:07:44.360 --> 00:07:46.830
because energy was transferred
from their surroundings

00:07:46.830 --> 00:07:48.390
to the system.

00:07:48.390 --> 00:07:51.660
And so, the surroundings lost energy.

00:07:51.660 --> 00:07:53.780
And so, therefore we
can say the dissolution

00:07:53.780 --> 00:07:58.683
of this particular solid
was an endothermic process.


Representing systems of equations with matrices

WEBVTT
Kind: captions
Language: en

00:00:00.370 --> 00:00:02.450
- [Instructor] I'm a big fan
of looking at the same problem

00:00:02.450 --> 00:00:05.360
in different ways or different
ways to conceptualize them.

00:00:05.360 --> 00:00:08.210
And so, for example, if I had
a system of three equations

00:00:08.210 --> 00:00:10.290
with three unknowns,
let me just make one up,

00:00:10.290 --> 00:00:15.290
3x - 2y - z = -1,

00:00:15.620 --> 00:00:16.790
that's one equation.

00:00:16.790 --> 00:00:19.090
And in three dimensions,
this would represent a plane.

00:00:19.090 --> 00:00:20.240
And then I have another one,

00:00:20.240 --> 00:00:25.240
2x + 5y + z = 0.

00:00:25.480 --> 00:00:27.500
That would represent another plane.

00:00:27.500 --> 00:00:29.950
Now, if you had two non-parallel planes,

00:00:29.950 --> 00:00:32.850
they would intersect each
other and form a line.

00:00:32.850 --> 00:00:36.100
But then if we have a fourth
plane, so let me write that

00:00:36.100 --> 00:00:41.100
as -4x - y = 8, it's possible,

00:00:41.830 --> 00:00:43.260
not always going to be the case,

00:00:43.260 --> 00:00:45.840
but it's possible that
they will all intersect

00:00:45.840 --> 00:00:49.780
in exactly one x,y,z-coordinate.

00:00:49.780 --> 00:00:52.110
And so, in other videos, we talk about

00:00:52.110 --> 00:00:54.400
solving systems of equations like this,

00:00:54.400 --> 00:00:57.500
with three equations and three unknowns.

00:00:57.500 --> 00:00:59.950
Now, what I want to do in this
video is connect this idea

00:00:59.950 --> 00:01:03.580
to the notion of matrices
and matrix multiplication,

00:01:03.580 --> 00:01:06.330
which we've already
reviewed in other videos.

00:01:06.330 --> 00:01:09.180
So we can think of this exact same problem

00:01:09.180 --> 00:01:10.390
in the following way.

00:01:10.390 --> 00:01:12.290
If we take all of the coefficients

00:01:12.290 --> 00:01:15.400
and we create a three by
three matrix with them,

00:01:15.400 --> 00:01:16.930
So let me do that.

00:01:16.930 --> 00:01:20.020
So for example, let me get
all of the x-coefficients,

00:01:20.020 --> 00:01:22.820
a 3, a 2 and a -4,

00:01:22.820 --> 00:01:27.820
and I'll put it in this first
column here, 3, 2 and -4.

00:01:28.350 --> 00:01:30.156
Let me get all the y-coefficients,

00:01:30.156 --> 00:01:34.832
- 2, 5 and essentially a -1.

00:01:34.832 --> 00:01:37.920
- 2, 5 and a -1.

00:01:37.920 --> 00:01:39.350
And then last but not least,

00:01:39.350 --> 00:01:40.770
all the coefficients on the z's,

00:01:40.770 --> 00:01:44.850
a -1, a +1 and then there's
implicitly a 0z here.

00:01:44.850 --> 00:01:45.970
You can't see it.

00:01:45.970 --> 00:01:50.170
So it would be -1, +1 and then a 0.

00:01:50.170 --> 00:01:52.030
So these are the
coefficients on x in purple,

00:01:52.030 --> 00:01:55.300
on y in yellow and in like a salmon color

00:01:55.300 --> 00:01:57.100
for the z's right over here.

00:01:57.100 --> 00:01:59.410
If we said that that is
going to be multiplied

00:01:59.410 --> 00:02:02.833
by a three-dimensional vector,

00:02:04.320 --> 00:02:08.730
that is, I guess we could
say, unknown, x, y and z,

00:02:08.730 --> 00:02:09.700
that is going to be equal

00:02:09.700 --> 00:02:12.980
to a second three-dimensional
vector, which we do know,

00:02:12.980 --> 00:02:17.090
and that's -1, 0 and 8.

00:02:17.090 --> 00:02:18.740
And I know there's a
lot of things going on

00:02:18.740 --> 00:02:19.730
in your brain right now.

00:02:19.730 --> 00:02:21.570
You're like, Sal, this
looks somewhat magical.

00:02:21.570 --> 00:02:24.680
You just took the coefficients,
put the x,y,z's here,

00:02:24.680 --> 00:02:27.540
put the right-hand sides of
the equal sign, in this case,

00:02:27.540 --> 00:02:29.320
the sides that didn't
have the variables on it,

00:02:29.320 --> 00:02:30.240
put it over here.

00:02:30.240 --> 00:02:32.950
Does this actually make sense?
Does this actually work?

00:02:32.950 --> 00:02:36.010
And to validate that,
let's actually multiply out

00:02:36.010 --> 00:02:39.270
the left-hand side of this
equation right over here.

00:02:39.270 --> 00:02:41.130
So in other videos, we've talked about

00:02:41.130 --> 00:02:44.280
multiplying a three by three matrix

00:02:44.280 --> 00:02:48.110
times a, in this case,
a three by one matrix,

00:02:48.110 --> 00:02:51.060
and this is going to actually
give us a three by one matrix.

00:02:51.060 --> 00:02:52.980
So that's looking good already,

00:02:52.980 --> 00:02:57.200
and the reason why we know
that is these two have to match

00:02:57.200 --> 00:02:59.500
in order for the multiplication
to even be defined.

00:02:59.500 --> 00:03:01.560
And then, the dimensions of the product

00:03:01.560 --> 00:03:03.930
are going to be three by one.

00:03:03.930 --> 00:03:06.310
But let's actually multiply things out.

00:03:06.310 --> 00:03:08.550
Well, we know one way to construct this,

00:03:08.550 --> 00:03:11.300
I'm just gonna focus on
the left-hand side here,

00:03:11.300 --> 00:03:12.620
is to say, all right,

00:03:12.620 --> 00:03:17.380
let's essentially take
this row and this column

00:03:17.380 --> 00:03:19.210
and then take the sum of the product

00:03:19.210 --> 00:03:21.760
of the corresponding terms,
I guess you could say.

00:03:21.760 --> 00:03:25.880
So this is gonna be 3
times x, which is 3x,

00:03:25.880 --> 00:03:27.900
minus 2 times y, -2y,

00:03:29.376 --> 00:03:30.970
minus 1 times z, -1z, like that.

00:03:35.270 --> 00:03:38.750
And then, the next one, I'm going to take

00:03:38.750 --> 00:03:42.580
all of this business and
multiply it by this column.

00:03:42.580 --> 00:03:43.810
So it's gonna be 2 times x,

00:03:43.810 --> 00:03:46.710
and this is just a review
of multiplying matrices,

00:03:46.710 --> 00:03:49.513
plus 5 times y, plus 5y,

00:03:50.410 --> 00:03:53.643
plus 1 times z, plus 1z.

00:03:55.100 --> 00:03:58.560
And then last but not
least, if I take this

00:03:58.560 --> 00:04:00.840
and I do the same thing with that column,

00:04:00.840 --> 00:04:05.840
gonna be -4 times x, -4x, minus 1 times y

00:04:06.300 --> 00:04:09.650
minus, I can just write
-y, and then 0 times z,

00:04:09.650 --> 00:04:11.970
which I could write if
I want to or not write,

00:04:11.970 --> 00:04:12.803
but I could write it.

00:04:12.803 --> 00:04:14.280
So let me just write it
to make things clear.

00:04:14.280 --> 00:04:16.830
And so, the product of what
we have on the left-hand side

00:04:16.830 --> 00:04:18.460
is this right over here.

00:04:18.460 --> 00:04:20.620
And it might look like a three by three,

00:04:20.620 --> 00:04:22.760
but it's actually a three by one here,

00:04:22.760 --> 00:04:25.800
where this would eval..., if
you knew what x, y and z is,

00:04:25.800 --> 00:04:28.400
this is going to evaluate to some number.

00:04:28.400 --> 00:04:32.560
Likewise, this is going to
evaluate to some other number,

00:04:32.560 --> 00:04:34.730
and this is going to
evaluate to another number.

00:04:34.730 --> 00:04:37.180
And we know, from this,
I guess you could say,

00:04:37.180 --> 00:04:40.450
matrix vector or this matrix
equation that we have set up,

00:04:40.450 --> 00:04:43.720
that this, the left-hand
side, this product,

00:04:43.720 --> 00:04:47.630
needs to be equal to what we
have on the right-hand side.

00:04:47.630 --> 00:04:52.470
It needs to be equal to -1, 0 and 8,

00:04:52.470 --> 00:04:54.960
which means, and I think
things might be connecting

00:04:54.960 --> 00:04:58.950
for you now, that this
needs to be equal to that,

00:04:58.950 --> 00:05:03.160
and that this needs to be equal to that,

00:05:03.160 --> 00:05:06.370
and last but not least, -4x - y + 0z

00:05:06.370 --> 00:05:07.600
needs to be equal to eight,

00:05:07.600 --> 00:05:09.750
which is exactly what that original system

00:05:09.750 --> 00:05:11.620
of equations was telling us.

00:05:11.620 --> 00:05:14.040
Now, I know, well, there's
still probably some things

00:05:14.040 --> 00:05:16.020
that are circulating in your mind.

00:05:16.020 --> 00:05:18.860
One question is, well, that's all nice,

00:05:18.860 --> 00:05:21.460
you found a different
way of representing that,

00:05:21.460 --> 00:05:24.800
but how do, does this introduce
a new way of solving this?

00:05:24.800 --> 00:05:26.780
And the answer I'll
give you for now is yes,

00:05:26.780 --> 00:05:29.690
it will lead to a new way of solving this.

00:05:29.690 --> 00:05:31.780
Because if you think about
it, we're taking the product

00:05:31.780 --> 00:05:34.890
of a matrix and a vector
here to get another vector.

00:05:34.890 --> 00:05:36.360
If there's some way to essentially

00:05:36.360 --> 00:05:40.790
unwind this matrix multiplication,
then you might be able

00:05:40.790 --> 00:05:43.590
to do that, apply it to
this vector on the right,

00:05:43.590 --> 00:05:46.850
and then solve for this
unknown vector here.

00:05:46.850 --> 00:05:49.010
And that's the way that
things like a computer,

00:05:49.010 --> 00:05:52.360
a lot of computer algorithms,
actually try to solve problems

00:05:52.360 --> 00:05:56.450
like this, by representing
them as matrices.

00:05:56.450 --> 00:05:57.720
Now, another interesting thing

00:05:57.720 --> 00:05:59.890
just with the representation itself

00:05:59.890 --> 00:06:01.650
is it makes you think about the problem

00:06:01.650 --> 00:06:03.300
a little bit differently.

00:06:03.300 --> 00:06:07.420
You could view this as three
planes in three dimensions

00:06:07.420 --> 00:06:10.210
and what is the x,y,z-coordinates

00:06:10.210 --> 00:06:11.740
where they could intersect,

00:06:11.740 --> 00:06:14.720
or you could view this
three by three matrix here

00:06:14.720 --> 00:06:16.790
as a transformation matrix

00:06:16.790 --> 00:06:19.160
that's applying to, is being applied to

00:06:19.160 --> 00:06:21.740
some unknown three-dimensional vector,

00:06:21.740 --> 00:06:23.480
and under transformation,

00:06:23.480 --> 00:06:25.140
that unknown three-dimensional vector

00:06:25.140 --> 00:06:28.159
is equal to this known
three-dimensional vector,

00:06:28.159 --> 00:06:30.580
- 1, 0 and 8.

00:06:30.580 --> 00:06:32.810
And so the question of
solving this would say,

00:06:32.810 --> 00:06:36.580
all right, can we somehow
perform a reverse transformation,

00:06:36.580 --> 00:06:38.740
an inverse transformation, so to speak,

00:06:38.740 --> 00:06:40.860
on the right-hand side to figure out

00:06:40.860 --> 00:06:43.410
what that unknown vector really is?

00:06:43.410 --> 00:06:44.660
And I'll just leave you there,

00:06:44.660 --> 00:06:48.463
and we'll continue that line
of thought in future videos.


Invertible matrices and transformations

WEBVTT
Kind: captions
Language: en

00:00:00.320 --> 00:00:02.680
- We have two by two matrices here

00:00:02.680 --> 00:00:05.630
and in other videos we talk
about how a two by two matrix

00:00:05.630 --> 00:00:10.170
can represent a transformation
of the coordinate plane

00:00:10.170 --> 00:00:11.400
of the two dimensional plane

00:00:11.400 --> 00:00:13.100
where this of course is the x-axis

00:00:13.100 --> 00:00:16.270
and this of course is the y-axis.

00:00:16.270 --> 00:00:18.150
What we're doing in
this video is visualize

00:00:18.150 --> 00:00:20.560
these transformations and
get a visual understanding

00:00:20.560 --> 00:00:24.010
for why it's reasonable
for A to have an inverse

00:00:24.010 --> 00:00:26.000
or for why matrix A is invertible

00:00:26.000 --> 00:00:27.670
and why it's not reasonable

00:00:27.670 --> 00:00:30.280
for B to have an inverse matrix

00:00:30.280 --> 00:00:32.950
or why matrix B is not invertible.

00:00:32.950 --> 00:00:35.060
So just as a reminder transformation

00:00:35.060 --> 00:00:36.800
these transformation matrices

00:00:36.800 --> 00:00:40.020
essentially tell us what to
do with our unit vectors.

00:00:40.020 --> 00:00:43.700
For example, we have
the one zero unit vector

00:00:43.700 --> 00:00:45.840
and the first column of
each of these matrices

00:00:45.840 --> 00:00:49.000
tell us what the one zero unit vector

00:00:49.000 --> 00:00:51.560
the vector that goes one
unit in the extraction

00:00:51.560 --> 00:00:53.120
what it will get transformed

00:00:53.120 --> 00:00:55.470
to under each of these transformations

00:00:55.470 --> 00:00:59.160
and then of course you have
the unit vector in the Y

00:00:59.160 --> 00:01:02.430
direction the zero one unit
vector and the second columns

00:01:02.430 --> 00:01:05.270
here tell us how we would transform that.

00:01:05.270 --> 00:01:09.410
So let's first think about
matrix A, matrix A transforms

00:01:09.410 --> 00:01:12.910
the one zero vector
into the two, one vector

00:01:12.910 --> 00:01:16.600
so the two, one vector is going
to look something like that

00:01:16.600 --> 00:01:18.860
and it transforms the zero one vector

00:01:18.860 --> 00:01:21.740
into the two, three vector
so the two three vector

00:01:21.740 --> 00:01:23.820
is going to look something like this.

00:01:23.820 --> 00:01:27.000
So one way to think about
it instead of our grid

00:01:27.000 --> 00:01:29.020
looking like this grid
that I already had here

00:01:29.020 --> 00:01:31.220
which is just our
standard coordinate axes.

00:01:31.220 --> 00:01:36.220
This is defining a new grid
that would look like this

00:01:36.220 --> 00:01:39.120
and I define this new grid
by looking at our multiples

00:01:39.120 --> 00:01:42.830
of what the one zero vector
has been transformed into

00:01:42.830 --> 00:01:46.360
and we also look at multiples
of what the zero one vector

00:01:46.360 --> 00:01:48.390
has been transformed into.

00:01:48.390 --> 00:01:53.040
So for example, if I were to
take this point right here

00:01:53.040 --> 00:01:56.160
before it is transformed
it has one of each of these

00:01:56.160 --> 00:02:00.570
vectors, well under this
first transformation

00:02:00.570 --> 00:02:03.660
it would be one of each of these vectors.

00:02:03.660 --> 00:02:06.440
So it's going to be one two one vector

00:02:06.440 --> 00:02:11.440
plus one two three vector
so this point will be mapped

00:02:11.780 --> 00:02:15.010
to this point and by that
same logic this point

00:02:15.010 --> 00:02:19.010
right over here which is one
unit more in the extraction

00:02:19.010 --> 00:02:21.490
well now it's going to have one unit more

00:02:21.490 --> 00:02:25.640
in the direction that the X unit vector

00:02:25.640 --> 00:02:28.730
has been transformed into
this point right over here

00:02:28.730 --> 00:02:33.090
by the same logic will be one
unit more in the direction

00:02:33.090 --> 00:02:35.540
that the Y unit vector
has been transformed into

00:02:35.540 --> 00:02:36.640
so it will be there.

00:02:36.640 --> 00:02:39.490
And that this point by the
same logic will be transformed

00:02:39.490 --> 00:02:42.900
into that point and so this
region that I'm showing you

00:02:42.900 --> 00:02:46.453
in white will get
transformed to this region.

00:02:47.610 --> 00:02:50.360
Now there's some obvious
things going on here.

00:02:50.360 --> 00:02:53.100
We have a two dimensional
area that has been transformed

00:02:53.100 --> 00:02:56.090
to another two dimensional
area and in fact, it looks like

00:02:56.090 --> 00:02:59.110
it has been scaled up in
other videos we have talked

00:02:59.110 --> 00:03:02.600
about that this scale factor
is going to be the absolute

00:03:02.600 --> 00:03:05.980
value of the determinant
of A and it's clear

00:03:05.980 --> 00:03:08.010
that frankly, not only is this non-zero

00:03:08.010 --> 00:03:09.470
but it's going to be greater
than what it looks like

00:03:09.470 --> 00:03:13.320
we are scaling up our area
but the very fact that this

00:03:13.320 --> 00:03:16.950
does not equal zero tells
us that we were scaling

00:03:16.950 --> 00:03:20.900
from a two dimensional area to
another two dimensional area.

00:03:20.900 --> 00:03:23.970
And so it's completely
reasonable to be able to go back

00:03:23.970 --> 00:03:27.020
you will, for sure be able
to find a transformation

00:03:27.020 --> 00:03:30.410
that takes you from this
region to that region.

00:03:30.410 --> 00:03:35.410
So that makes us feel good
that A inverse is reasonable.

00:03:37.130 --> 00:03:40.340
Now, as a point of contrast
let's think about matrix B

00:03:40.340 --> 00:03:41.590
right over here.

00:03:41.590 --> 00:03:44.980
Matrix B transforms the
one zero unit vector

00:03:44.980 --> 00:03:46.730
into the two, one unit vector

00:03:46.730 --> 00:03:49.860
so it transforms it into this unit vector

00:03:49.860 --> 00:03:53.060
actually very similarly to how A did it

00:03:53.060 --> 00:03:55.010
but then let's see what it
does to the zero one vector

00:03:55.010 --> 00:03:58.580
transforms this vector
into the four two vector

00:03:58.580 --> 00:04:03.410
four two is this notice
the four two vector

00:04:03.410 --> 00:04:05.560
is just two of the two one vectors

00:04:05.560 --> 00:04:07.210
is going in the same direction

00:04:07.210 --> 00:04:09.800
it just has a different
size or different length

00:04:09.800 --> 00:04:11.580
or a different magnitude.

00:04:11.580 --> 00:04:13.680
So in this situation is taking things

00:04:13.680 --> 00:04:15.460
in two dimensional space
and it's turning them

00:04:15.460 --> 00:04:18.970
into combinations of things
that sit on this same direction.

00:04:18.970 --> 00:04:21.180
So everything in two dimensions
is going to be mapped

00:04:21.180 --> 00:04:24.230
to something along this line here.

00:04:24.230 --> 00:04:26.980
So this region, if I were
to apply transformation B

00:04:26.980 --> 00:04:30.660
it's going to be a region
of this line over here

00:04:30.660 --> 00:04:34.020
anything, if I were to
map it using matrix B

00:04:34.020 --> 00:04:36.900
is going to get mapped onto this line

00:04:36.900 --> 00:04:38.850
so you're going to something with area

00:04:38.850 --> 00:04:42.240
and then you're mapping it
to something that has no area

00:04:42.240 --> 00:04:45.870
so the scaling factor here must be zero.

00:04:45.870 --> 00:04:47.520
So we know since the absolute value

00:04:47.520 --> 00:04:51.810
of the determinant of B is
zero, we could also fairly say

00:04:51.810 --> 00:04:54.330
that the determinant of B is equal to zero

00:04:54.330 --> 00:04:57.600
and the how are you going
to have an inverse matrix?

00:04:57.600 --> 00:04:59.460
How are you going to have
something that can scale

00:04:59.460 --> 00:05:03.150
from zero area to something that has area?

00:05:03.150 --> 00:05:08.150
So we know that B inverse does not exist

00:05:09.440 --> 00:05:12.450
there's nothing that can transform us back

00:05:12.450 --> 00:05:14.330
and so there's a couple of things here

00:05:14.330 --> 00:05:17.660
this reinforces this idea
that if your determinant

00:05:17.660 --> 00:05:20.280
of a transformation matrix is zero

00:05:20.280 --> 00:05:23.950
you're not going to have an
inverse, it's not invertible

00:05:23.950 --> 00:05:26.750
the other thing to recognize
is seeing the patterns

00:05:26.750 --> 00:05:30.350
in the matrix itself here we
saw that the second column

00:05:30.350 --> 00:05:33.320
is just a multiple of the
first column it's twice

00:05:33.320 --> 00:05:37.230
the first column two times two
is four one times two is two

00:05:37.230 --> 00:05:38.820
you can also view it the other way round

00:05:38.820 --> 00:05:43.180
the first row is a
multiple of the second row

00:05:43.180 --> 00:05:45.070
and you can play around
with the math if you wanted

00:05:45.070 --> 00:05:46.990
to generalize it and
see if that's the case

00:05:46.990 --> 00:05:49.460
the determinant will always be zero

00:05:49.460 --> 00:05:52.800
and that's because if you
view them as transformations

00:05:52.800 --> 00:05:56.040
it's going to map to line
and you're gonna lose

00:05:56.040 --> 00:05:59.180
all your area, if you view
it as a representation

00:05:59.180 --> 00:06:01.850
for at least the left-hand
side of a system of equations

00:06:01.850 --> 00:06:04.930
you can think about it as
lines that have the same slope

00:06:04.930 --> 00:06:06.880
but we talk about that in other videos.


Invertible matrices and determinants

WEBVTT
Kind: captions
Language: en

00:00:00.130 --> 00:00:01.330
- [Instructor] So let's
dig a little bit more

00:00:01.330 --> 00:00:03.420
into matrices and their inverses.

00:00:03.420 --> 00:00:05.940
And in particular, I'm
gonna explore the situations

00:00:05.940 --> 00:00:09.180
in which there might not
be an inverse for a matrix.

00:00:09.180 --> 00:00:12.640
So just as a review, we think
about if we have some matrix,

00:00:12.640 --> 00:00:16.390
A, is there some other matrix
which we could call A inverse

00:00:16.390 --> 00:00:18.570
that when we take the composition of them,

00:00:18.570 --> 00:00:20.890
so if we viewed them
each as transformations,

00:00:20.890 --> 00:00:24.740
we would end up with the
identity transformation.

00:00:24.740 --> 00:00:26.590
Or if we take the product of the two,

00:00:26.590 --> 00:00:28.480
you get the identity matrix.

00:00:28.480 --> 00:00:30.910
And we would also think about it, well,

00:00:30.910 --> 00:00:35.650
if A inverse undoes A, then
A should undo A inverse

00:00:35.650 --> 00:00:38.270
to also get the identity matrix.

00:00:38.270 --> 00:00:42.530
And so another way to
think about it, if I take

00:00:42.530 --> 00:00:46.280
some type of region in
the coordinate plane,

00:00:46.280 --> 00:00:49.130
so this is my x axis, this is my y axis.

00:00:49.130 --> 00:00:52.340
And so let's say my original
region looks something

00:00:52.340 --> 00:00:54.350
like this right over here.

00:00:54.350 --> 00:00:58.950
And I apply the transformation
A, and I get something

00:00:58.950 --> 00:01:01.783
that looks like this, just
making up some things.

00:01:02.680 --> 00:01:05.840
So if I apply the transformation A,

00:01:05.840 --> 00:01:09.260
it takes me from that
region to that region.

00:01:09.260 --> 00:01:12.450
Then we also have a sense
that, okay, A inverse,

00:01:12.450 --> 00:01:15.100
if you transformed this purple thing,

00:01:15.100 --> 00:01:17.610
it should take you back
to where you began.

00:01:17.610 --> 00:01:20.600
Because if you start with
this little blue thing

00:01:20.600 --> 00:01:22.950
and if you have the composition,
well, then that should

00:01:22.950 --> 00:01:25.520
just be transforming it with
the identity transformation,

00:01:25.520 --> 00:01:28.870
so you should just get back to
this little blue thing here.

00:01:28.870 --> 00:01:30.870
Now this might start
triggering some thoughts

00:01:30.870 --> 00:01:34.240
about determinants,
because you might remember

00:01:34.240 --> 00:01:36.710
that the determinant of a matrix tells us

00:01:36.710 --> 00:01:40.270
how much a region's
area will be scaled by.

00:01:40.270 --> 00:01:44.610
In particular, let's say that matrix A

00:01:44.610 --> 00:01:48.490
takes a region that has
an area of, I don't know,

00:01:48.490 --> 00:01:52.350
let's call this area b and let's take,

00:01:52.350 --> 00:01:56.240
let's say it takes that area to 5 times b.

00:01:56.240 --> 00:01:59.140
So the area here is 5b.

00:01:59.140 --> 00:02:02.340
Well, we know that that scaling of 5

00:02:02.340 --> 00:02:05.310
you can determine from the
determinant of matrix A.

00:02:05.310 --> 00:02:07.430
That would tell us that the absolute value

00:02:07.430 --> 00:02:12.430
of the determinant of matrix
A is going to be equal to 5.

00:02:13.460 --> 00:02:16.480
But what does that tell us
about the absolute value

00:02:16.480 --> 00:02:19.233
of the determinant of A inverse then?

00:02:20.160 --> 00:02:25.060
Well, if A is scaling up by
5, it scales areas up by 5,

00:02:25.060 --> 00:02:27.620
then A inverse must be

00:02:27.620 --> 00:02:30.460
scaling areas down by 5.

00:02:30.460 --> 00:02:33.160
So the absolute value of the determinant

00:02:33.160 --> 00:02:36.760
of A inverse should be 1 over 5.

00:02:36.760 --> 00:02:38.490
And so now we have a general property.

00:02:38.490 --> 00:02:40.330
I just happened to use
the number five here,

00:02:40.330 --> 00:02:43.650
but generally speaking, the absolute value

00:02:43.650 --> 00:02:48.160
of the determinant of matrix
A, if it has an inverse,

00:02:48.160 --> 00:02:52.300
should be equal to 1
over the absolute value

00:02:52.300 --> 00:02:55.890
of the determinant of A inverse.

00:02:55.890 --> 00:02:57.930
And we can of course write
that the other way around.

00:02:57.930 --> 00:03:01.620
The absolute value of the
determinant of A inverse

00:03:01.620 --> 00:03:05.700
should be equal to 1 over,
or the reciprocal of,

00:03:05.700 --> 00:03:09.630
the absolute value of
the determinant of A.

00:03:09.630 --> 00:03:11.650
The sum comes straight
out of this property

00:03:11.650 --> 00:03:13.870
that the absolute value of
the determinant tells you

00:03:13.870 --> 00:03:16.290
how much you scale an area by.

00:03:16.290 --> 00:03:19.130
Well knowing that both of these
statements need to be true

00:03:19.130 --> 00:03:21.960
for any matrix A that has an inverse,

00:03:21.960 --> 00:03:24.950
it gives us a clue as to at least one way

00:03:24.950 --> 00:03:28.640
to rule out matrices that
might not have inverses.

00:03:28.640 --> 00:03:31.240
If I were to tell you that the determinant

00:03:31.240 --> 00:03:36.210
of matrix A is zero, will
that have an inverse?

00:03:36.210 --> 00:03:39.250
Well, it can't because if
this quantity right over here

00:03:39.250 --> 00:03:42.510
is zero, or this quantity
right over here is zero,

00:03:42.510 --> 00:03:44.610
that would mean that the
absolute value of the determinant

00:03:44.610 --> 00:03:47.260
of the inverse of the matrix
needs to be one over zero,

00:03:47.260 --> 00:03:49.090
which is undefined.

00:03:49.090 --> 00:03:51.630
And so we have an
interesting conclusion here.

00:03:51.630 --> 00:03:53.980
If the determinant of a
matrix is equal to zero

00:03:53.980 --> 00:03:57.410
there is not going to be an inverse,

00:03:57.410 --> 00:03:59.720
because let's say that there
was some transformation

00:03:59.720 --> 00:04:02.130
that determinant was
zero, instead of something

00:04:02.130 --> 00:04:04.290
that's taking up two-dimensional
area to something else

00:04:04.290 --> 00:04:06.480
that takes two-dimensional
area, it would transform

00:04:06.480 --> 00:04:08.420
something that takes
up two dimensional area

00:04:08.420 --> 00:04:09.950
to something that takes no area.

00:04:09.950 --> 00:04:11.970
So maybe a curve like that,

00:04:11.970 --> 00:04:16.970
that takes up no area
or a line or a point.

00:04:17.120 --> 00:04:19.188
And if you transform to say, a line,

00:04:19.188 --> 00:04:20.730
how do you transform back?

00:04:20.730 --> 00:04:22.560
You'd have to scale up the area infinitely

00:04:22.560 --> 00:04:25.710
in order for it to take up
some two dimensional space.

00:04:25.710 --> 00:04:26.700
So big takeaway.

00:04:26.700 --> 00:04:28.960
We've just said, if the
determinant of a matrix

00:04:28.960 --> 00:04:31.050
is equal to zero, you're not
going to find an inverse.

00:04:31.050 --> 00:04:33.620
And it actually turns out the
case that any other matrix,

00:04:33.620 --> 00:04:36.490
you can find an inverse, but
I'm not going to prove that

00:04:36.490 --> 00:04:38.240
just yet, but hopefully you feel good

00:04:38.240 --> 00:04:40.983
about this principle right over here.


Volume density

WEBVTT
Kind: captions
Language: en

00:00:00.620 --> 00:00:01.453
- [Instructor] In this video,

00:00:01.453 --> 00:00:04.000
we're gonna talk a
little bit about density

00:00:04.000 --> 00:00:05.980
and we're especially
gonna talk about density

00:00:05.980 --> 00:00:08.510
in the context of volume.

00:00:08.510 --> 00:00:11.910
And one simple way to think about density

00:00:11.910 --> 00:00:15.290
is it's a quantity of something

00:00:15.290 --> 00:00:17.250
and we're going to think
about examples of it

00:00:17.250 --> 00:00:19.320
per unit volume.

00:00:19.320 --> 00:00:21.400
So per volume.

00:00:21.400 --> 00:00:26.320
So for example, let's say
that I have a cubic meter

00:00:26.320 --> 00:00:27.153
right over here,

00:00:27.153 --> 00:00:30.670
actually let me have two
different cubic meters

00:00:30.670 --> 00:00:31.940
just to give you an example.

00:00:31.940 --> 00:00:33.610
So these are both cubic meters

00:00:33.610 --> 00:00:38.610
and let's say in the one on
the left, I have a quantity of

00:00:38.760 --> 00:00:42.850
let's say six of these
dots per cubic meter

00:00:42.850 --> 00:00:44.820
and over here, I only have three

00:00:44.820 --> 00:00:47.160
of these dots per cubic meter.

00:00:47.160 --> 00:00:50.560
Well here, I have a higher
density and in general,

00:00:50.560 --> 00:00:51.980
we're gonna take the quantity

00:00:51.980 --> 00:00:53.900
and divide it by the volume

00:00:53.900 --> 00:00:58.870
and the units are going to be
some quantity per unit volume.

00:00:58.870 --> 00:01:01.410
Now you're typically going
to see mass per unit volume

00:01:01.410 --> 00:01:03.760
but density, especially
in the volume context,

00:01:03.760 --> 00:01:07.100
can refer to any quantity per unit volume.

00:01:07.100 --> 00:01:07.933
Now with that out of the way,

00:01:07.933 --> 00:01:11.110
let's give ourselves a
little bit of an example.

00:01:11.110 --> 00:01:13.390
So here we're told that stone spheres

00:01:13.390 --> 00:01:15.830
thought to be carved by the Diquis people,

00:01:15.830 --> 00:01:17.670
I'm not sure if I'm
pronouncing that correctly,

00:01:17.670 --> 00:01:19.570
more than a thousand years ago

00:01:19.570 --> 00:01:22.230
are a national symbol of Costa Rica.

00:01:22.230 --> 00:01:24.740
One such sphere has a diameter

00:01:24.740 --> 00:01:29.740
of about 1.8 meters and
masses about 8,300 kilograms.

00:01:31.960 --> 00:01:34.690
Based on these measurements,
what is the density

00:01:34.690 --> 00:01:39.690
of this sphere in
kilograms per cubic meter,

00:01:39.700 --> 00:01:43.790
round to the nearest hundred
kilograms per cubic meter.

00:01:43.790 --> 00:01:46.540
So pause this video and see
if you can figure that out.

00:01:48.250 --> 00:01:51.800
So we're gonna want to get
kilograms per cubic meter.

00:01:51.800 --> 00:01:54.820
So we know the total number of kilograms

00:01:54.820 --> 00:01:59.410
in one point in a sphere that
has a diameter of 1.8 meters.

00:01:59.410 --> 00:02:01.470
So that's the total number of kilograms,

00:02:01.470 --> 00:02:03.640
but we don't know the volume just yet.

00:02:03.640 --> 00:02:05.790
So we have a sphere like this.

00:02:05.790 --> 00:02:07.480
This would be a cross section of it.

00:02:07.480 --> 00:02:11.760
Its diameter is 1.8 meters.

00:02:11.760 --> 00:02:14.690
Now you may or may not
already know that the volume

00:02:14.690 --> 00:02:19.690
of a sphere is given by 4/3 PI R cubed.

00:02:22.550 --> 00:02:26.220
And so the radius here is 0.9 meters.

00:02:26.220 --> 00:02:28.780
And so that would be
the R right over here.

00:02:28.780 --> 00:02:33.780
So the volume of one of these
spheres is going to be...

00:02:33.970 --> 00:02:34.803
Let me write it over here.

00:02:34.803 --> 00:02:38.430
The volume is going to be 4/3 PI

00:02:38.430 --> 00:02:43.300
times 0.9 to the third power.

00:02:43.300 --> 00:02:45.130
And we know what the mass is,

00:02:45.130 --> 00:02:48.780
the mass in that volume
is 8,300 kilograms.

00:02:48.780 --> 00:02:51.610
So we would know that the density,

00:02:51.610 --> 00:02:53.960
the density in this situation

00:02:53.960 --> 00:02:56.480
is going to be 8,300 kilograms,

00:02:56.480 --> 00:03:01.480
8,300 kilograms per
this many cubic meters,

00:03:02.180 --> 00:03:07.180
4/3 PI times 0.9 to the
third power cubic meters.

00:03:09.760 --> 00:03:11.640
And we're going to need
a calculator for this

00:03:11.640 --> 00:03:15.640
and we're gonna round to the
nearest hundred kilograms.

00:03:15.640 --> 00:03:20.640
So we have 8,300 divided by,

00:03:21.160 --> 00:03:23.800
let me just open parenthesis here.

00:03:23.800 --> 00:03:28.800
Four divided by three times PI

00:03:31.370 --> 00:03:36.370
times 0.9 to the third power,

00:03:36.700 --> 00:03:40.390
and then I'm going to
close my parentheses,

00:03:40.390 --> 00:03:45.140
is equal to this right over here.

00:03:45.140 --> 00:03:48.760
We want to round to the
nearest hundred kilogram.

00:03:48.760 --> 00:03:53.760
So approximately 2,700
kilograms per cubic meter,

00:03:54.340 --> 00:03:59.340
2,700 or 2,700 kilograms per cubic meter.

00:04:00.930 --> 00:04:02.770
And we are done.


Area density

WEBVTT
Kind: captions
Language: en

00:00:00.320 --> 00:00:01.160
- [Narrator] In this video,

00:00:01.160 --> 00:00:05.800
we're going to talk about
density in the context of area.

00:00:05.800 --> 00:00:08.370
And the simplest way of
thinking about it is density

00:00:08.370 --> 00:00:13.370
is going to be some
quantity per unit area.

00:00:13.700 --> 00:00:16.880
So for example, let's say

00:00:16.880 --> 00:00:20.670
that I have a football
field right over here

00:00:20.670 --> 00:00:24.640
and I have another identical
football field right over here.

00:00:24.640 --> 00:00:26.620
Now they have the same area,

00:00:26.620 --> 00:00:31.620
but if I have let's say five
people on this football field

00:00:32.300 --> 00:00:34.150
actually six people on
this football field.

00:00:34.150 --> 00:00:37.300
And I only have three people
on this football field

00:00:37.300 --> 00:00:41.020
the density of people
per average unit area

00:00:41.020 --> 00:00:43.870
or the density of people I
should say per football field

00:00:43.870 --> 00:00:46.230
is going to be higher
in this left example.

00:00:46.230 --> 00:00:49.460
So it's always going to
be quantity per area.

00:00:49.460 --> 00:00:51.760
Now, with that out of the
way, let's do a worked example

00:00:51.760 --> 00:00:55.060
that helps us understand this
idea a little bit better.

00:00:55.060 --> 00:00:57.470
So here we're told the town of Tigersville

00:00:57.470 --> 00:01:02.470
has a population density of
13 cats per square kilometer.

00:01:02.570 --> 00:01:07.090
So they're giving us the density
let me write that 13 cats.

00:01:07.090 --> 00:01:12.090
So the quantity is quantity
and cats per square kilometer.

00:01:12.120 --> 00:01:13.900
That's the density right over there.

00:01:13.900 --> 00:01:18.750
The town is shaped like a
perfect isosceles trapezoid.

00:01:18.750 --> 00:01:20.330
So it looks something

00:01:20.330 --> 00:01:25.150
like this say perfect isosceles trapezoid.

00:01:25.150 --> 00:01:26.820
It's gonna look something like that,

00:01:26.820 --> 00:01:30.700
with two parallel boundaries,
12 kilometers apart.

00:01:30.700 --> 00:01:34.580
So this distance right
over here is 12 kilometers

00:01:34.580 --> 00:01:36.210
one measuring eight kilometers.

00:01:36.210 --> 00:01:38.680
So this side over here is eight kilometers

00:01:38.680 --> 00:01:41.143
the other is 16 that's
a longer one over there.

00:01:42.330 --> 00:01:44.720
How many cats are in Tigersville?

00:01:44.720 --> 00:01:47.750
So they give us the density
here and they give us,

00:01:47.750 --> 00:01:50.290
I think enough information
to figure out the area.

00:01:50.290 --> 00:01:52.750
And they want us to figure
out how many cats we have.

00:01:52.750 --> 00:01:54.310
So what is the quantity?

00:01:54.310 --> 00:01:57.173
So pause this video and see
if you can figure that out.

00:01:58.530 --> 00:01:59.910
Well, just as we said the density

00:01:59.910 --> 00:02:02.830
is equal to quantity divided by area.

00:02:02.830 --> 00:02:06.370
If we multiply both sides
of this equation by area

00:02:06.370 --> 00:02:11.370
you get area times density is
going to be equal to quantity.

00:02:14.400 --> 00:02:16.320
And we know the density.

00:02:16.320 --> 00:02:19.460
It's 13 cats per square
kilometer, and we can figure out

00:02:19.460 --> 00:02:22.090
the area and then just multiply the two.

00:02:22.090 --> 00:02:24.540
So what's the area of
this right over here?

00:02:24.540 --> 00:02:27.710
Well, the area of a
trapezoid is going to be,

00:02:27.710 --> 00:02:30.580
write here area is going
to be 12 kilometers.

00:02:30.580 --> 00:02:32.490
The height of the trapezoid

00:02:32.490 --> 00:02:37.050
times the average of
the two parallel sides

00:02:37.050 --> 00:02:37.883
I guess you could say.

00:02:37.883 --> 00:02:40.896
So the average of those it's
going to be eight kilometers

00:02:40.896 --> 00:02:45.810
plus 16 kilometers over two.

00:02:45.810 --> 00:02:48.810
So this is going to be
equal to 12 kilometers

00:02:48.810 --> 00:02:53.300
times eight plus 16 is
24 divided by two is 12.

00:02:53.300 --> 00:02:55.850
So times 12 kilometers.

00:02:55.850 --> 00:03:00.850
So this gives us 144 square kilometers.

00:03:00.960 --> 00:03:04.620
Now we know we have 13
cats per square kilometer.

00:03:04.620 --> 00:03:06.580
So let me do this year in another color.

00:03:06.580 --> 00:03:11.580
So if I multiply 13 cats
per kilometer squared

00:03:12.930 --> 00:03:17.870
and I multiply that times
this business right over here,

00:03:17.870 --> 00:03:22.150
times 144 square kilometers.

00:03:22.150 --> 00:03:24.690
And you might also notice
that the units cancel out

00:03:24.690 --> 00:03:26.550
the same way that variables might.

00:03:26.550 --> 00:03:28.310
So that cancels out with that.

00:03:28.310 --> 00:03:31.340
You're going to get 13 times 144.

00:03:31.340 --> 00:03:34.570
And the units that you're
left with is just cats.

00:03:34.570 --> 00:03:39.390
So 144 times 13, three times four is 12

00:03:41.200 --> 00:03:44.060
that gives us to 13,
three times a hundred,

00:03:44.060 --> 00:03:46.670
300 plus another hundred is 400.

00:03:46.670 --> 00:03:49.580
Now I'm just gonna multiply
144 essentially by 10

00:03:49.580 --> 00:03:51.920
which is going to be 1440.

00:03:51.920 --> 00:03:54.150
And so if I add up all of that together

00:03:54.150 --> 00:03:55.710
I'm gonna jump down to here.

00:03:55.710 --> 00:03:57.513
I get 1872.

00:03:59.190 --> 00:04:04.190
So this is 1872 cats in
total and we are done.


Dividing rational expressions

WEBVTT
Kind: captions
Language: en

00:00:00.230 --> 00:00:01.610
- [Instructor] The goal of this video

00:00:01.610 --> 00:00:03.820
is to take this big, hairy expression,

00:00:03.820 --> 00:00:07.310
where we are essentially
dividing rational expressions

00:00:07.310 --> 00:00:10.470
and see if we can
essentially do the division

00:00:10.470 --> 00:00:12.870
and then write it in reduced terms.

00:00:12.870 --> 00:00:14.550
So if you are so inspired,

00:00:14.550 --> 00:00:16.190
I encourage you to pause the video

00:00:16.190 --> 00:00:19.123
and work on this on your own
before we do this together.

00:00:20.140 --> 00:00:22.050
All right, now let's do this together.

00:00:22.050 --> 00:00:26.630
So this is completely analogous
to dividing fractions.

00:00:26.630 --> 00:00:31.220
So if we were to divide
the fraction six over 25

00:00:32.260 --> 00:00:37.260
by the fraction 15 over nine,

00:00:38.090 --> 00:00:39.900
we know that we can rewrite this

00:00:39.900 --> 00:00:44.900
as six over 25 divided by 15 over nine,

00:00:46.000 --> 00:00:49.150
which we know is the
same thing as six over 25

00:00:49.150 --> 00:00:52.480
times nine over 15.

00:00:52.480 --> 00:00:54.857
And then we can factor
the various numerators

00:00:54.857 --> 00:00:55.930
and the denominators.

00:00:55.930 --> 00:00:57.770
This is two times three.

00:00:57.770 --> 00:00:59.810
This is three times three.

00:00:59.810 --> 00:01:01.910
This is five times five.

00:01:01.910 --> 00:01:04.550
This is five times three.

00:01:04.550 --> 00:01:06.010
Let's see, three in the numerator,

00:01:06.010 --> 00:01:07.830
three in the denominator.

00:01:07.830 --> 00:01:09.760
And actually, that's about
as far as we can get.

00:01:09.760 --> 00:01:12.150
So then we'll have two
times three times three,

00:01:12.150 --> 00:01:14.920
which is going to be 18 in the numerator.

00:01:14.920 --> 00:01:16.100
And then in the denominator,

00:01:16.100 --> 00:01:21.100
we have five times five
times five, which is 125.

00:01:21.180 --> 00:01:23.790
So we're going to do the
exact same thing up here,

00:01:23.790 --> 00:01:25.950
but there's one extra complication.

00:01:25.950 --> 00:01:28.730
We have to keep track of the x values

00:01:28.730 --> 00:01:32.250
that would make this expression
undefined in any way,

00:01:32.250 --> 00:01:34.170
because as we reduce to lowest terms,

00:01:34.170 --> 00:01:35.720
we might lose that information,

00:01:35.720 --> 00:01:37.350
but if we lose that information,

00:01:37.350 --> 00:01:38.970
then we have changed the expression.

00:01:38.970 --> 00:01:40.330
So we have to keep track

00:01:40.330 --> 00:01:43.340
of how we are constraining this domain.

00:01:43.340 --> 00:01:45.300
So first, I can just rewrite this

00:01:45.300 --> 00:01:50.300
as x squared minus three x minus four,

00:01:50.500 --> 00:01:55.483
all of that over negative
three x minus 15.

00:01:57.390 --> 00:02:00.210
And that's getting divided by,

00:02:00.210 --> 00:02:02.750
divided by this business.

00:02:02.750 --> 00:02:07.020
X squared minus 16 over,

00:02:07.020 --> 00:02:12.020
we have x squared minus x minus 30.

00:02:12.760 --> 00:02:14.100
Now, the next thing we can do

00:02:14.100 --> 00:02:17.660
is we can factor the various
numerators and denominators

00:02:17.660 --> 00:02:21.110
and think about which x values
could get us into trouble.

00:02:21.110 --> 00:02:23.600
So x squared minus three x minus four.

00:02:23.600 --> 00:02:25.990
Let's see, negative one.

00:02:25.990 --> 00:02:29.290
Let's see, negative four times plus one

00:02:29.290 --> 00:02:30.270
would be negative four,

00:02:30.270 --> 00:02:31.930
and then these would add
up to negative three.

00:02:31.930 --> 00:02:35.730
So I can rewrite this as x minus four

00:02:35.730 --> 00:02:38.390
times x plus one.

00:02:38.390 --> 00:02:40.240
Rewrite it that way.

00:02:40.240 --> 00:02:43.190
And then I can rewrite
what I have down here,

00:02:43.190 --> 00:02:45.110
I could factor out a negative three.

00:02:45.110 --> 00:02:50.110
So I could write that as
negative three times x plus five.

00:02:50.210 --> 00:02:52.820
And then I could write this one here.

00:02:52.820 --> 00:02:54.440
This is a difference of squares.

00:02:54.440 --> 00:02:59.440
It's going to be x plus
four times x minus four.

00:02:59.490 --> 00:03:02.840
And then last but not
least, this one over here.

00:03:02.840 --> 00:03:06.490
Let's see, if I have a five and a six,

00:03:06.490 --> 00:03:09.090
negative six plus five is negative one,

00:03:09.090 --> 00:03:11.590
negative six times five is negative 30.

00:03:11.590 --> 00:03:13.980
So it's going to be x minus six

00:03:13.980 --> 00:03:16.710
times x plus five.

00:03:16.710 --> 00:03:18.190
Now, before I go any further,

00:03:18.190 --> 00:03:20.100
and the reason why I
factored at that point,

00:03:20.100 --> 00:03:22.130
is now we can think about what x values

00:03:22.130 --> 00:03:23.540
will get us in trouble.

00:03:23.540 --> 00:03:25.100
We know that any x values

00:03:25.100 --> 00:03:28.400
that make any of the
denominators equal to zero,

00:03:28.400 --> 00:03:29.600
that would be undefined.

00:03:29.600 --> 00:03:31.670
So we wanna constrain
our domain in that way.

00:03:31.670 --> 00:03:33.340
So we know, for example,

00:03:33.340 --> 00:03:35.900
that x cannot be equal to negative five.

00:03:35.900 --> 00:03:38.760
That would make this
denominator equal to zero.

00:03:38.760 --> 00:03:40.080
Let me write that here.

00:03:40.080 --> 00:03:43.520
So x cannot be equal to negative five.

00:03:43.520 --> 00:03:47.150
We also know that x
cannot be equal to six.

00:03:47.150 --> 00:03:49.550
X cannot be equal to six.

00:03:49.550 --> 00:03:51.000
And this would also tell us

00:03:51.000 --> 00:03:52.770
that x cannot be equal to negative five.

00:03:52.770 --> 00:03:54.950
So I don't have to rewrite that again.

00:03:54.950 --> 00:03:56.270
But we're not done.

00:03:56.270 --> 00:03:57.850
So we've figured out the x values

00:03:57.850 --> 00:04:00.200
that make these
denominators equal to zero,

00:04:00.200 --> 00:04:02.120
but remember, we're also dividing by

00:04:02.120 --> 00:04:04.530
this entire expression here.

00:04:04.530 --> 00:04:05.490
So anything that would make

00:04:05.490 --> 00:04:08.180
the entire expression equal to
zero would also be a problem,

00:04:08.180 --> 00:04:09.840
'cause you can't divide by zero.

00:04:09.840 --> 00:04:13.460
So anything that would make
this numerator equal to zero,

00:04:13.460 --> 00:04:15.670
which was this numerator right over here,

00:04:15.670 --> 00:04:17.810
would make us also divide by zero.

00:04:17.810 --> 00:04:18.940
So we have to constrain there.

00:04:18.940 --> 00:04:20.780
Not this numerator here, that one's fine.

00:04:20.780 --> 00:04:22.710
That one could be equal to zero.

00:04:22.710 --> 00:04:24.580
You can divide zero by other things.

00:04:24.580 --> 00:04:27.460
So let's see, we could
see that x cannot be equal

00:04:27.460 --> 00:04:29.750
to negative four, and actually,

00:04:29.750 --> 00:04:33.290
x cannot be equal to positive four.

00:04:33.290 --> 00:04:35.560
So now we've fully constrained our domain

00:04:35.560 --> 00:04:37.730
and now we can proceed.

00:04:37.730 --> 00:04:41.280
So let me box this off right over here,

00:04:41.280 --> 00:04:42.130
and then we continue.

00:04:42.130 --> 00:04:43.860
I can rewrite all of this.

00:04:43.860 --> 00:04:48.860
So we're going to have x
minus four times x plus one,

00:04:50.300 --> 00:04:55.300
all of that over negative
three times x plus five,

00:04:56.780 --> 00:04:59.610
and now I'm just going to,
instead of divide by this,

00:04:59.610 --> 00:05:01.760
I'm gonna multiply by the reciprocal.

00:05:01.760 --> 00:05:03.840
So this is going to be times,

00:05:03.840 --> 00:05:05.910
and I'm just gonna take the reciprocal,

00:05:05.910 --> 00:05:10.230
x minus six times x plus five,

00:05:10.230 --> 00:05:11.930
all of that over,

00:05:11.930 --> 00:05:16.930
we have x plus four times x minus four.

00:05:17.520 --> 00:05:20.120
And once again, our domain
is constrained in this way,

00:05:20.120 --> 00:05:22.900
but we see we have an x minus
four in the numerator now,

00:05:22.900 --> 00:05:24.880
x minus four in the denominator,

00:05:24.880 --> 00:05:26.750
x plus five in the denominator,

00:05:26.750 --> 00:05:29.230
x plus five in the numerator.

00:05:29.230 --> 00:05:30.680
And now we can say that this is going

00:05:30.680 --> 00:05:35.680
to be equal to x plus
one times x minus six,

00:05:37.710 --> 00:05:42.710
all of that over negative
three, negative three,

00:05:43.010 --> 00:05:46.730
times x plus four, x plus four.

00:05:46.730 --> 00:05:48.960
So the way it's written
now, we would still,

00:05:48.960 --> 00:05:51.780
it's clear that x cannot
be equal to negative four.

00:05:51.780 --> 00:05:53.250
So this information,

00:05:53.250 --> 00:05:56.000
you can say that it's already
there in this expression

00:05:56.000 --> 00:05:58.620
now that we have reduced
it to lowest terms,

00:05:58.620 --> 00:06:00.830
but this other information
right over here,

00:06:00.830 --> 00:06:02.250
this has been lost.

00:06:02.250 --> 00:06:03.830
So if you want this expression

00:06:03.830 --> 00:06:06.520
to truly be equivalent to
this expression up here,

00:06:06.520 --> 00:06:10.970
you would also have to say
comma x cannot be equal

00:06:10.970 --> 00:06:15.970
to negative five, six, or positive four.

00:06:16.040 --> 00:06:18.910
You could throw the negative
four in there if you like,

00:06:18.910 --> 00:06:22.823
but that one's already in
the expression, so to speak.


Multiplying rational expressions

WEBVTT
Kind: captions
Language: en

00:00:00.200 --> 00:00:01.920
- [Instructor] So what I
have here is an expression

00:00:01.920 --> 00:00:05.240
where I'm multiplying
rational expressions.

00:00:05.240 --> 00:00:08.550
And we want to do this multiplication

00:00:08.550 --> 00:00:10.310
and then reduce to lowest terms.

00:00:10.310 --> 00:00:13.450
So if you feel so inspired,
I encourage you to pause

00:00:13.450 --> 00:00:15.850
this video and see if you
can have a go at that.

00:00:16.690 --> 00:00:19.470
All right, now let's work
through this together.

00:00:19.470 --> 00:00:22.280
So multiplying rational
expressions like this,

00:00:22.280 --> 00:00:25.830
it's very analogous to
multiplying fractions.

00:00:25.830 --> 00:00:30.830
For example, if I were to multiply 6/25

00:00:32.010 --> 00:00:37.010
times 15/9, there's a
few ways you could do it.

00:00:38.300 --> 00:00:40.750
You could just multiply six
times 15 in the numerator,

00:00:40.750 --> 00:00:42.850
and 25 times nine in the denominator.

00:00:42.850 --> 00:00:44.480
But the way that many of us approach it

00:00:44.480 --> 00:00:47.230
so that it's easier to
reduce to lowest terms

00:00:47.230 --> 00:00:49.490
is to factor things, to realize that look,

00:00:49.490 --> 00:00:53.000
six is two times three.

00:00:53.000 --> 00:00:58.000
Nine is three times three,
15 is three times five,

00:00:59.050 --> 00:01:02.440
and 25 is five times five.

00:01:02.440 --> 00:01:04.980
And then you can realize
in your eventual product

00:01:04.980 --> 00:01:05.900
you're going to have a five

00:01:05.900 --> 00:01:07.940
in the numerator, five in the denominator.

00:01:07.940 --> 00:01:10.590
You can, five divided by five is one,

00:01:10.590 --> 00:01:13.040
three divided by three is one,

00:01:13.040 --> 00:01:15.820
and then three divided by three is one.

00:01:15.820 --> 00:01:18.690
And so all you'd be left with is that two

00:01:18.690 --> 00:01:19.670
and then that five.

00:01:19.670 --> 00:01:22.140
So this is going to be equal to 2/5.

00:01:22.140 --> 00:01:23.680
So we'll do the analogous thing

00:01:23.680 --> 00:01:25.420
here with these rational expressions.

00:01:25.420 --> 00:01:26.570
We're going to factor them,

00:01:26.570 --> 00:01:29.150
all of them in the
numerators and denominators.

00:01:29.150 --> 00:01:32.240
And then we'll see if we
can divide the numerator

00:01:32.240 --> 00:01:34.220
and the denominator by the same thing.

00:01:34.220 --> 00:01:35.850
Now, the one thing we have to make sure

00:01:35.850 --> 00:01:39.150
of as we do that is we
keep track of the domain.

00:01:39.150 --> 00:01:41.160
'Cause these rational expressions here,

00:01:41.160 --> 00:01:42.570
they might have x values

00:01:42.570 --> 00:01:45.260
that make their
denominators equal to zero.

00:01:45.260 --> 00:01:49.040
And so even if we reduce to lowest terms,

00:01:49.040 --> 00:01:50.770
and we get rid of those expressions,

00:01:50.770 --> 00:01:53.600
in order for the expressions
to be the same expression,

00:01:53.600 --> 00:01:57.120
we have to constrain the
domain in the same way.

00:01:57.120 --> 00:01:59.110
So let's get started.

00:01:59.110 --> 00:02:01.300
So this is going to be equal to,

00:02:01.300 --> 00:02:02.480
I'll just rewrite everything,

00:02:02.480 --> 00:02:04.140
x squared minus nine.

00:02:04.140 --> 00:02:05.240
How do we factor that?

00:02:05.240 --> 00:02:06.910
Well, that's going to be
a difference of squares.

00:02:06.910 --> 00:02:11.910
We could write that as x plus
three times x minus three.

00:02:12.430 --> 00:02:16.750
And then that is going
to be over this business.

00:02:16.750 --> 00:02:19.670
And let's see, five squared is 25,

00:02:19.670 --> 00:02:22.490
negative five plus negative
five is negative 10.

00:02:22.490 --> 00:02:26.380
So this is going to be x minus five

00:02:27.480 --> 00:02:30.010
times x minus five.

00:02:30.010 --> 00:02:31.850
If what I'm just doing
here with the factoring

00:02:31.850 --> 00:02:33.110
is not making sense,

00:02:33.110 --> 00:02:36.030
I encourage you to review
factoring on Khan Academy.

00:02:36.030 --> 00:02:39.120
And then we multiply
that times, let's see,

00:02:39.120 --> 00:02:41.690
in this numerator here,
I can factor out of four.

00:02:41.690 --> 00:02:45.840
So that's going to be
four times x minus five.

00:02:45.840 --> 00:02:46.770
Which is going to be useful.

00:02:46.770 --> 00:02:49.450
I have an x minus five
there, x minus five there.

00:02:49.450 --> 00:02:51.840
And then that's going to be over.

00:02:51.840 --> 00:02:55.930
Let's see, this expression over here.

00:02:55.930 --> 00:02:58.690
Two plus three is five,
two times three to six.

00:02:58.690 --> 00:03:03.690
So it's going to be x plus
two times x plus three.

00:03:04.250 --> 00:03:07.500
Now before I start
reducing to lowest terms,

00:03:07.500 --> 00:03:09.520
let's think about the domain here.

00:03:09.520 --> 00:03:11.310
And the domain is going to be constrained

00:03:11.310 --> 00:03:14.740
by things that make these
denominators equal to zero.

00:03:14.740 --> 00:03:17.760
So the domain would be all real numbers

00:03:17.760 --> 00:03:20.300
except x cannot equal five.

00:03:20.300 --> 00:03:21.410
Let me write it over here.

00:03:21.410 --> 00:03:25.740
X cannot equal five
because if that happened,

00:03:25.740 --> 00:03:28.260
then this denominator would become zero.

00:03:28.260 --> 00:03:31.583
X cannot be equal to negative two.

00:03:32.480 --> 00:03:34.410
X could not be equal to negative two

00:03:34.410 --> 00:03:36.330
because that would make
the denominator here zero,

00:03:36.330 --> 00:03:38.050
which would make the
denominator here zero.

00:03:38.050 --> 00:03:41.940
And x cannot be equal to negative three.

00:03:41.940 --> 00:03:44.320
So the domain is constrained in this way.

00:03:44.320 --> 00:03:46.050
We have to carry this throughout.

00:03:46.050 --> 00:03:48.000
No matter what we do to the expression,

00:03:48.000 --> 00:03:50.900
this is the constraints on our domain.

00:03:50.900 --> 00:03:55.290
With that out of the way, now
we can reduce to lowest terms.

00:03:55.290 --> 00:03:58.980
So x, we have an x plus
three in the numerator,

00:03:58.980 --> 00:04:00.410
x plus three in the denominator.

00:04:00.410 --> 00:04:02.320
X minus five in the numerator.

00:04:02.320 --> 00:04:04.310
X minus five in the denominator.

00:04:04.310 --> 00:04:07.070
And I think we've gone
about as far as we can.

00:04:07.070 --> 00:04:09.330
And so when we multiply the numerators,

00:04:09.330 --> 00:04:10.570
we are going to get,

00:04:10.570 --> 00:04:14.960
this business is going to
be four times x minus three.

00:04:14.960 --> 00:04:19.430
Four times x minus three,

00:04:19.430 --> 00:04:22.513
over, we have an x minus five here.

00:04:23.510 --> 00:04:25.610
X minus five.

00:04:25.610 --> 00:04:28.523
And then we have an x plus two.

00:04:29.710 --> 00:04:32.410
We have an x plus two.

00:04:32.410 --> 00:04:34.500
And we could leave it
like this if you want.

00:04:34.500 --> 00:04:35.720
In some cases,

00:04:35.720 --> 00:04:39.020
people like to multiply the
things out, but we're done.

00:04:39.020 --> 00:04:41.580
We've just finished multiplying
these rational expressions.

00:04:41.580 --> 00:04:42.650
And we have to remind ourselves

00:04:42.650 --> 00:04:46.100
that x cannot be equal
to any of these things.

00:04:46.100 --> 00:04:48.060
Now, the way that we've simplified it,

00:04:48.060 --> 00:04:50.960
we still have an x minus
five right over here.

00:04:50.960 --> 00:04:53.490
So it might be redundant
to say that x cannot

00:04:53.490 --> 00:04:56.350
be equal to five because
that's still the case

00:04:56.350 --> 00:05:00.200
in our reduced terms expression here.

00:05:00.200 --> 00:05:03.380
And that's true also of
the x cannot be equal

00:05:03.380 --> 00:05:04.540
to negative two.

00:05:04.540 --> 00:05:06.510
We still have an x plus two here.

00:05:06.510 --> 00:05:08.960
So still, even in this
expression it's pretty clear,

00:05:08.960 --> 00:05:11.100
that x cannot be equal to negative two,

00:05:11.100 --> 00:05:12.930
but the x cannot equal negative three

00:05:12.930 --> 00:05:15.890
isn't so obvious when you
just look at this expression.

00:05:15.890 --> 00:05:17.250
But in order for this expression

00:05:17.250 --> 00:05:19.560
to be completely
equivalent to the original,

00:05:19.560 --> 00:05:21.800
it has to have the same domain.

00:05:21.800 --> 00:05:24.400
And so you might want to explicitly say

00:05:24.400 --> 00:05:26.910
that x cannot be equal
to negative three here.

00:05:26.910 --> 00:05:28.630
You could also say the other two,

00:05:28.630 --> 00:05:31.100
but those are, that's still very clear

00:05:31.100 --> 00:05:32.750
when you look at this expression.


Inverse matrices and matrix equations

WEBVTT
Kind: captions
Language: en

00:00:00.000 --> 00:00:02.270
- [Instructor] In a previous
video, we talked about

00:00:02.270 --> 00:00:05.510
how you can represent
a system of equations

00:00:05.510 --> 00:00:08.140
as essentially a matrix equation.

00:00:08.140 --> 00:00:10.460
So for example, here I have two equations

00:00:10.460 --> 00:00:13.490
with two unknowns x, y,
well let's just assume

00:00:13.490 --> 00:00:16.460
that we know what a,
b, p, c, d, and q are,

00:00:16.460 --> 00:00:20.610
and you can represent
this type of system as a

00:00:20.610 --> 00:00:24.040
matrix vector equation like
this, where the coefficients

00:00:24.040 --> 00:00:26.400
on the x's are this first column,

00:00:26.400 --> 00:00:28.850
coefficients on the y's
are the second column.

00:00:28.850 --> 00:00:31.090
And then we see our unknown variables,

00:00:31.090 --> 00:00:34.760
what we would want to solve
for maybe, as this vector here,

00:00:34.760 --> 00:00:37.290
so you could do that as the
unknown two-dimensional vector.

00:00:37.290 --> 00:00:39.900
And then we know when
we either think about it

00:00:39.900 --> 00:00:42.150
as a transformation on this unknown vector

00:00:42.150 --> 00:00:45.450
we get this known vector,
we get the vector pq,

00:00:45.450 --> 00:00:48.110
or you can think about it
as matrix multiplication.

00:00:48.110 --> 00:00:51.220
When you multiply this
vector by this matrix

00:00:51.220 --> 00:00:54.560
you get this pq vector.

00:00:54.560 --> 00:00:56.340
And in other videos we also talked

00:00:56.340 --> 00:00:58.530
about this idea of inverses.

00:00:58.530 --> 00:01:03.100
So for example, if we call
this right over here, matrix A,

00:01:03.100 --> 00:01:06.260
you can imagine that, or what
we're seeing here is matrix A,

00:01:06.260 --> 00:01:10.470
times the vector xy,

00:01:10.470 --> 00:01:12.110
I'll just write it like that,

00:01:12.110 --> 00:01:16.570
is equal to the vector pq.

00:01:16.570 --> 00:01:19.650
I'll just stick to one color
for convenience right now.

00:01:19.650 --> 00:01:23.790
And we talked about this idea
that if you have an inverse

00:01:23.790 --> 00:01:25.480
of a matrix times the matrix

00:01:25.480 --> 00:01:27.670
it's going to give you
the identity matrix.

00:01:27.670 --> 00:01:31.030
So one idea for trying to quote solve,

00:01:31.030 --> 00:01:34.620
this matrix vector equation
is what if we multiplied

00:01:34.620 --> 00:01:38.140
both sides on the left
by the inverse of A?

00:01:38.140 --> 00:01:40.130
So if I had A inverse here

00:01:40.130 --> 00:01:43.770
and if I multiplied by A
inverse here, what would happen?

00:01:43.770 --> 00:01:46.240
Well, assuming A inverse exists,

00:01:46.240 --> 00:01:48.530
and that's actually going to
be the focus of this video,

00:01:48.530 --> 00:01:51.500
if A inverse exists,
then this right over here

00:01:51.500 --> 00:01:55.100
is just going to become
the identity matrix.

00:01:55.100 --> 00:01:57.800
So that's just the matrix
that if I try to transform

00:01:57.800 --> 00:02:00.530
anything or if I multiply
anything by it it's just gonna

00:02:00.530 --> 00:02:02.570
give us that thing that we had before,

00:02:02.570 --> 00:02:04.780
in the two by two scenario,
the identity matrix,

00:02:04.780 --> 00:02:07.940
it looks like that.

00:02:07.940 --> 00:02:10.190
And then on the right-hand
side, we'll be multiplying

00:02:10.190 --> 00:02:12.493
a two by two matrix times the vector pq.

00:02:13.518 --> 00:02:14.760
And so on the left-hand side,

00:02:14.760 --> 00:02:16.820
essentially the identity matrix times xy

00:02:16.820 --> 00:02:21.030
is just going to give us xy,
and on the right hand side

00:02:21.030 --> 00:02:22.780
we would know what that equals to.

00:02:22.780 --> 00:02:25.410
So that would essentially
solve this system

00:02:25.410 --> 00:02:27.940
when it is represented that way.

00:02:27.940 --> 00:02:32.660
But this gives us a clue about
thinking when is it solvable?

00:02:32.660 --> 00:02:35.190
Because when it's solvable,
you're going to have

00:02:35.190 --> 00:02:37.610
a situation where you
do have an inverse here.

00:02:37.610 --> 00:02:39.810
And when it's not
solvable, you're gonna have

00:02:39.810 --> 00:02:41.920
a situation where there
is no inverse here,

00:02:41.920 --> 00:02:46.120
where this matrix A does
not have an inverse.

00:02:46.120 --> 00:02:49.490
So when we go back to what we've
learned in previous algebra

00:02:49.490 --> 00:02:52.080
classes about solving
systems of equations,

00:02:52.080 --> 00:02:54.700
we know there's two scenarios where we get

00:02:54.700 --> 00:02:57.510
either no solutions or an
infinite number of solutions.

00:02:57.510 --> 00:02:59.560
Let me draw a little coordinate axis here.

00:02:59.560 --> 00:03:01.360
We know that the lines
have different slopes.

00:03:01.360 --> 00:03:02.676
So one line looks like this

00:03:02.676 --> 00:03:04.880
and then the other line, it
could look like anything.

00:03:04.880 --> 00:03:06.570
As long as it has a different slope,

00:03:06.570 --> 00:03:09.100
they are going to intersect
at exactly one point.

00:03:09.100 --> 00:03:11.500
Two lines with different
slopes are gonna intersect

00:03:11.500 --> 00:03:13.680
in exactly one point.

00:03:13.680 --> 00:03:16.810
The situation which you
have no solutions is if

00:03:16.810 --> 00:03:19.520
they are the same slope.

00:03:19.520 --> 00:03:23.420
So parallel lines like
this would not intersect.

00:03:23.420 --> 00:03:26.470
Another weird situation that
you get when you solve systems

00:03:26.470 --> 00:03:29.810
is that they have the same
slope, but they're the same line.

00:03:29.810 --> 00:03:31.720
So that would be something like this.

00:03:31.720 --> 00:03:34.420
There would not be a unique
xy, there would actually

00:03:34.420 --> 00:03:37.940
be an infinite xy's that
would satisfy the equation.

00:03:37.940 --> 00:03:40.440
So those are both situations
where we're not finding

00:03:40.440 --> 00:03:43.250
a nice, clean, unique
solution to our system.

00:03:43.250 --> 00:03:45.540
And if we think about the
matrix world, where we're not

00:03:45.540 --> 00:03:48.320
going to find an inverse of
A that when we multiply it

00:03:48.320 --> 00:03:50.830
times pq it gives us
a nice clean solution,

00:03:50.830 --> 00:03:52.180
in either of those scenarios

00:03:52.180 --> 00:03:55.190
where the two equations
have the same slope.

00:03:55.190 --> 00:03:56.820
Now let's think about what do we know

00:03:56.820 --> 00:03:59.820
about abcd when we have the same slope?

00:03:59.820 --> 00:04:01.800
So if we were to try to
put this top equation

00:04:01.800 --> 00:04:04.840
into slope intercept form,
what would that look like?

00:04:04.840 --> 00:04:08.190
Well, let's see we can
subtract ax from both sides.

00:04:08.190 --> 00:04:09.370
So you could have something like this,

00:04:09.370 --> 00:04:13.420
by is equal to p minus ax.

00:04:13.420 --> 00:04:15.100
I just subtract ax from both sides.

00:04:15.100 --> 00:04:17.050
And then if you divide both sides by b,

00:04:17.050 --> 00:04:22.050
you get y is equal to p
over b minus a over bx.

00:04:22.430 --> 00:04:24.980
And so you can see in this first equation

00:04:24.980 --> 00:04:27.700
our slope is negative a over b.

00:04:27.700 --> 00:04:29.410
Now what about the second equation?

00:04:29.410 --> 00:04:31.500
Well, by the same logic,
if you do the same thing,

00:04:31.500 --> 00:04:35.070
you subtract cx from both
sides and then divide by d,

00:04:35.070 --> 00:04:39.240
you're going to get y is equal to q over d

00:04:39.240 --> 00:04:43.780
minus c over dx.

00:04:43.780 --> 00:04:47.430
And so we see the slope
here is negative c over d.

00:04:47.430 --> 00:04:50.260
So these strange scenarios, not strange,

00:04:50.260 --> 00:04:52.270
but these scenarios where
you don't get a nice clean

00:04:52.270 --> 00:04:54.890
unique solution are the
ones where these slopes

00:04:54.890 --> 00:04:56.370
are equivalent to each other.

00:04:56.370 --> 00:04:58.640
So we're talking about
the scenario in which

00:04:58.640 --> 00:05:03.640
negative a over b is equal
to negative c over d.

00:05:04.170 --> 00:05:06.150
Now, to make a little bit
of sense of that, let's say

00:05:06.150 --> 00:05:09.630
we multiply both sides of
this equation by negative bd

00:05:09.630 --> 00:05:11.730
to get rid of these things
out of the denominator.

00:05:11.730 --> 00:05:12.840
So let me do that.

00:05:12.840 --> 00:05:14.140
And I'm multiplying by a negative

00:05:14.140 --> 00:05:18.000
to get rid of the negatives, negative bd.

00:05:18.000 --> 00:05:21.430
So on the left-hand side,
the b cancels with the b,

00:05:21.430 --> 00:05:23.010
negative times a negative is a positive,

00:05:23.010 --> 00:05:26.470
we're gonna get ad, and
on the right-hand side,

00:05:26.470 --> 00:05:30.103
negatives cancel out, d goes
away, and then you have cb.

00:05:32.020 --> 00:05:34.390
Or another way to think about it is,

00:05:34.390 --> 00:05:38.080
ad minus cb is going to be equal to zero.

00:05:38.080 --> 00:05:42.190
When ad minus cb is equal to zero,

00:05:42.190 --> 00:05:47.190
this system of equations does
not have a unique solution.

00:05:47.270 --> 00:05:50.150
Now, things, bells might be
ringing in your head right now,

00:05:50.150 --> 00:05:54.280
because ad, so that's a times d

00:05:54.280 --> 00:05:57.610
minus c times b,

00:05:57.610 --> 00:05:59.780
minus c times b,

00:05:59.780 --> 00:06:02.700
that's the determinant
of this matrix A here.

00:06:02.700 --> 00:06:06.000
So this is going to be
true only in a situation,

00:06:06.000 --> 00:06:09.340
if and only if, the
determinant of our matrix A

00:06:09.340 --> 00:06:10.660
is equal to zero.

00:06:10.660 --> 00:06:12.610
So just like that, we
have a pretty neat clue

00:06:12.610 --> 00:06:15.690
about when you're not going
to see a nice neat solution

00:06:15.690 --> 00:06:17.610
to a systems of equations represented

00:06:17.610 --> 00:06:19.770
as a matrix vector equation like this.

00:06:19.770 --> 00:06:21.240
You're not gonna have a nice, clean

00:06:21.240 --> 00:06:23.400
unique solution when the determinant

00:06:23.400 --> 00:06:27.060
of your matrix A is equal to zero.

00:06:27.060 --> 00:06:29.470
And since you're not going to
have a nice clean solution,

00:06:29.470 --> 00:06:31.530
you must not be able
to find an inverse here

00:06:31.530 --> 00:06:34.030
because if you had an inverse,
you could just multiply it.

00:06:34.030 --> 00:06:37.800
So this is a situation where
this is only going to be true,

00:06:37.800 --> 00:06:39.440
and I haven't proven it rigorously,

00:06:39.440 --> 00:06:42.230
but hopefully it gives you a
little bit of a justification.

00:06:42.230 --> 00:06:46.840
This is a situation where
A inverse doesn't exist.

00:06:46.840 --> 00:06:48.670
So there's a lot that's interesting here.

00:06:48.670 --> 00:06:51.130
In a previous video, we
thought about a matrix A

00:06:51.130 --> 00:06:52.710
and we thought about
it as a transformation

00:06:52.710 --> 00:06:56.320
and how its determinant tells
us how we are scaling areas.

00:06:56.320 --> 00:06:58.630
But if its determinant is
zero, that means you're

00:06:58.630 --> 00:07:00.470
taking things that have
two dimensional area

00:07:00.470 --> 00:07:02.620
and you're scaling them
down to having zero area.

00:07:02.620 --> 00:07:04.520
It'd be very hard to go
the other way around,

00:07:04.520 --> 00:07:07.890
which is what A inverse
transformation matrix would do.

00:07:07.890 --> 00:07:11.070
Here, we're getting the same
result, not viewing matrix A

00:07:11.070 --> 00:07:14.640
as a transformation, but
viewing it as a representation

00:07:14.640 --> 00:07:17.920
of a system of linear equations like this.

00:07:17.920 --> 00:07:20.090
But once again, we got the same idea

00:07:20.090 --> 00:07:22.610
that the determinant
of A is equal to zero.

00:07:22.610 --> 00:07:24.910
You're not going to get a
nice clean solution here.

00:07:24.910 --> 00:07:27.743
And so the inverse does not exist.


Composing 3x3 matrices

WEBVTT
Kind: captions
Language: en

00:00:00.160 --> 00:00:02.700
- [Instructor] So we have two,
three by three matrices here,

00:00:02.700 --> 00:00:04.700
matrix A and matrix B.

00:00:04.700 --> 00:00:05.870
And we can of course,

00:00:05.870 --> 00:00:08.940
view each of them as a transformation

00:00:08.940 --> 00:00:11.200
in three dimensional space.

00:00:11.200 --> 00:00:12.710
Now, what we're going to
think about in this video

00:00:12.710 --> 00:00:15.630
is the composition of A of Bs.

00:00:15.630 --> 00:00:17.730
So you can think of this
as the transformation

00:00:17.730 --> 00:00:21.720
where you apply B first, and
then you apply A after that.

00:00:21.720 --> 00:00:23.210
And then we can represent that

00:00:23.210 --> 00:00:25.280
by another three by three matrix,

00:00:25.280 --> 00:00:26.810
which is partially completed here.

00:00:26.810 --> 00:00:29.000
We have the first and
the third column here.

00:00:29.000 --> 00:00:30.450
And so my question to you is

00:00:30.450 --> 00:00:33.790
what is this middle column
where I have these three blanks?

00:00:33.790 --> 00:00:36.740
Pause the video and try
to work through that.

00:00:36.740 --> 00:00:39.770
All right, now let's work
through this together.

00:00:39.770 --> 00:00:43.160
So one way to think about
how to construct A of B

00:00:43.160 --> 00:00:44.420
is that what you're doing is

00:00:44.420 --> 00:00:46.950
you're taking each of the columns of B

00:00:46.950 --> 00:00:47.783
and you're thinking

00:00:47.783 --> 00:00:50.450
about what would they be
under the transformation A.

00:00:50.450 --> 00:00:53.430
So if you were to apply
the transformation A,

00:00:53.430 --> 00:00:56.130
to this column, right over here,

00:00:56.130 --> 00:00:58.360
you would get this column.

00:00:58.360 --> 00:01:02.130
If you apply the transformation
A to this column,

00:01:02.130 --> 00:01:03.330
right over here,

00:01:03.330 --> 00:01:05.170
you would get this column.

00:01:05.170 --> 00:01:06.410
So what we really need to do

00:01:06.410 --> 00:01:10.110
is apply the transformation
A to this column,

00:01:10.110 --> 00:01:12.050
to the middle column, right over here.

00:01:12.050 --> 00:01:13.240
And just as a reminder,

00:01:13.240 --> 00:01:15.100
how this transformation works.

00:01:15.100 --> 00:01:17.230
A vector zero to three,

00:01:17.230 --> 00:01:22.230
you can think of it as zero
of the one zero, zero vector,

00:01:26.010 --> 00:01:28.420
the unit vector in the X direction.

00:01:28.420 --> 00:01:33.420
Plus two of the zero, one, zero vector

00:01:33.830 --> 00:01:38.830
plus three of the zero, zero one vector.

00:01:40.350 --> 00:01:41.730
Now, when you're applying transformation,

00:01:41.730 --> 00:01:44.460
instead of using these unit vectors,

00:01:44.460 --> 00:01:47.760
you use the image of them
under this transformation.

00:01:47.760 --> 00:01:49.340
And now in this situation,

00:01:49.340 --> 00:01:51.600
instead of one zero, zero vector,

00:01:51.600 --> 00:01:54.200
we are going to be using this thing,

00:01:54.200 --> 00:01:56.030
instead of a zero one zero,

00:01:56.030 --> 00:01:58.950
we're going to be using this thing.

00:01:58.950 --> 00:02:00.860
Instead of a zero, zero one,

00:02:00.860 --> 00:02:03.990
we are going to be using this thing.

00:02:03.990 --> 00:02:05.900
So this middle column,

00:02:05.900 --> 00:02:07.520
when it's transformed by this vector

00:02:07.520 --> 00:02:11.660
is going to be zero, instead
of the one zero, zero one,

00:02:11.660 --> 00:02:15.610
it's going to be zero
of the negative three,

00:02:15.610 --> 00:02:18.370
negative three, three vector.

00:02:18.370 --> 00:02:23.370
And then we have plus
two, plus two of the zero.

00:02:25.380 --> 00:02:27.130
Let me do that in that purple color

00:02:27.130 --> 00:02:32.130
of the zero, negative two, three vector.

00:02:33.520 --> 00:02:34.760
And then last but not least,

00:02:34.760 --> 00:02:38.650
you're going to have
three of the plus three

00:02:38.650 --> 00:02:40.420
of then I'll do that in yellow,

00:02:40.420 --> 00:02:45.160
the zero negative for one vector.

00:02:45.160 --> 00:02:47.540
Now we just do the math.

00:02:47.540 --> 00:02:49.420
So when you play zero times, all of this,

00:02:49.420 --> 00:02:51.410
you're just gonna have a
zero, zero, zero vector.

00:02:51.410 --> 00:02:53.000
So we can, those all go away.

00:02:53.000 --> 00:02:55.780
And then you are left with, let's see,

00:02:55.780 --> 00:02:59.040
this one is going to be
two times zero is zero.

00:02:59.040 --> 00:03:00.960
Two times negative two is negative four,

00:03:00.960 --> 00:03:02.890
two times three is six.

00:03:02.890 --> 00:03:07.700
You're gonna have that,
plus three times zero

00:03:07.700 --> 00:03:11.010
is zero, three times
negative four is negative 12,

00:03:11.010 --> 00:03:12.870
three times one is three.

00:03:12.870 --> 00:03:14.650
I could have written
this a little bit neater,

00:03:14.650 --> 00:03:16.220
but hopefully you get the idea.

00:03:16.220 --> 00:03:18.100
And then when we add those two things,

00:03:18.100 --> 00:03:21.760
zero plus zero, is zero.

00:03:21.760 --> 00:03:26.260
Negative four plus
negative 12 is negative 16,

00:03:26.260 --> 00:03:31.210
six plus three is nine, and we're done.

00:03:31.210 --> 00:03:36.053
We have just completed
the composition of A of B.