# How do you solve x+3y=11 and x+4y=14 using matrices?

Jul 12, 2016

x=2; y=3

#### Explanation:

you must solve by applying the basic propositions:

x=D_x/D; y=D_y/D

where D is the determinant of the matrix

$| \quad 1 \quad 3 |$
$| \quad 1 \quad 4 |$

Its determinant is obtained by:

$D = 1 \cdot 4 - 3 \cdot 1 = 1$

${D}_{x} \mathmr{and} {D}_{y}$ are respectively the determinants of the matrices

$| \quad 11 \quad 3 |$
$| \quad 14 \quad 4 |$

$| \quad 1 \quad 11 |$
$| \quad 1 \quad 14 |$

${D}_{x} = 11 \cdot 4 - 3 \cdot 14 = 44 - 42 = 2$

${D}_{y} = 1 \cdot 14 - 11 \cdot 1 = 3$

So

$x = {D}_{x} / D = \frac{2}{1} = 2$

$y = {D}_{y} / D = \frac{3}{1} = 3$

Jul 17, 2016

$x = 2 \text{ ; } y = 3$

Two methods given

#### Explanation:

$\textcolor{b l u e}{\text{Using Gauss Jordon elimination}}$

$\text{ "color(white)(.)xcolor(white)(.)ycolor(white)(.)"answer}$
" "[[1,3,"|",11],[1,4,"|",14]]
$\text{ } R o w 2 - R o w 1$
$\text{ } \downarrow$

" "[[1,3,"|",11],[0,1,"|",3]]
$\text{ } R o w 1 - 3 \left(R o w 2\right)$
$\text{ } \downarrow$

$\text{ "[[1,0,"|",2],[0,1,"|",3]]" } \leftarrow \left[\begin{matrix}x \\ y\end{matrix}\right]$
$\textcolor{w h i t e}{.}$

$\implies \left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right] \textcolor{w h i t e}{.} \left[\begin{matrix}x \\ y\end{matrix}\right] = \left[\begin{matrix}2 \\ 3\end{matrix}\right]$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Using Linear Algebra}}$

$A = \left[\begin{matrix}1 & 3 \\ 1 & 4\end{matrix}\right]$
$\textcolor{w h i t e}{.}$

$X = \left[\begin{matrix}x \\ y\end{matrix}\right]$

$B = \left[\begin{matrix}11 \\ 14\end{matrix}\right]$

$\implies A X = B$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{Consider the Algebra method of determining } x}$
Given:$\text{ } 3 x = 4$

Divide both side by 3 $\to \frac{1}{3} \times 3 x = \frac{1}{3} \times 4$

We multiply the 3 by 3 inverse to turn the coefficient of $x$ into 1

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{Using the Algebra method shown above}}$

Multiply both sides by $A \text{ inverse. written as } \textcolor{b l u e}{{A}^{-}}$

$\textcolor{b r o w n}{\implies \textcolor{b l u e}{{A}^{-}} A X \text{ "=" } \textcolor{b l u e}{{A}^{-}} B}$

$\implies X = {A}^{-} B$ ...................................Equation(1)

Checked in Maple:

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine } {A}^{-}}$

$\textcolor{b r o w n}{\text{Method 1}}$

Write the original matrix as:

" "[[1,3,"|",1,0],[1,4,"|",0,1]]

And conduct a Gauss Jordon elimination on the LHS

$\text{ } R o w 2 - R o w 1$
$\text{ } \downarrow$

" "[[1,3,"|",1,0],[0,1,"|",-1,1]]
$\text{ } R o w 1 - 3 \left(R o w 2\right)$
$\text{ } \downarrow$

" "[[1,0,"|",4,-3],[0,1,"|",-1,1]] => A^(-)=[[4,-3],[-1,1]]
,.....................................................................................
$\textcolor{b r o w n}{\text{Method 2 - Shortcut approach}}$

Multiply a modified version of $\left[\begin{matrix}1 & 3 \\ 1 & 4\end{matrix}\right]$ by its inverted determinant.

For $\left[\begin{matrix}a & b \\ c & d\end{matrix}\right]$ the determinant is $a b - c d$
and the modified matrix is $\left[\begin{matrix}d & - b \\ - c & a\end{matrix}\right]$

$D = \left(1 \times 4\right) - \left(1 \times 3\right) = 1$

Thus ${A}^{-} = 1 \times \left[\begin{matrix}4 & - 3 \\ - 1 & 1\end{matrix}\right]$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Finish the calculation for Equation(1)}}$
$\implies X = {A}^{-} B$ ...................................Equation(1)

$\left[\begin{matrix}4 & - 3 \\ - 1 & 1\end{matrix}\right] \left[\begin{matrix}11 \\ 14\end{matrix}\right] = \left[\begin{matrix}2 \\ 3\end{matrix}\right]$