# Matrices and number of solutions

Q1. Find the value of a for which there are infinitely many solutions to the equations
2x + ay − z = 0
3x + 4y − (a + 1)z = 13
10x + 8y + (a − 4)z = 26

Now I know that for there to be infinitely many solutions the determinant of the coefficient matrix must = 0.

I did this on a calculator and found 2 possibilities, 0 and 2.

Additionally, I know that there are infinitely many solutions when 2 of the equations are indentical (by a factor). By trying both 0 and 2 I cannot see how any of the 2 equations will be identical. (Apparently the answer is a=0)

Q2. Find a value of p for which the system of equations
3x + 2y − z = 1 and x + y + z = 2 and px + 2y − z = 1
has more than one solution.

Not sure where to start here, more than one solutions hints at what?

Char. Limit
Gold Member
Actually, infinitely solutions exist for your system when the system of equations is dependent, and for three equations, this means that any one equation can be written as a linear combination of the other two. For example, if you set a=2, then 8 times the first equation plus -2 times the second equation results in the third equation. I figured this out by doing something like this:

b(2x+2y-z) + c(3x+4y-3z) = 10x+8y-2z = (2b+3c)x + (2b+4c)y + (-b-3c)z

We know that the x components are equal, the y components are equal, and the z components are equal. This, we can prove that b=8 and c=-2. It's easy to prove something similar for a=0. Thus, the system of equations is dependent, and there is more than one solution.

For your second question, have you learned yet about augmented matrices and elementary row operations?

For the first question, I just wish to find the value of a for which there are infinitely many solutions. The answer actually specifies a=0 alone, rather than a = 0 or a = 2. So I'm wondering how they eliminate a = 2.

For second question:

more than one solution means det = 0

det of the coefficient matrix is 3(p-3)

thus 3(p-3) = 0, p = 3

hope its as simple as that

Last edited:
HallsofIvy
Homework Helper
If the determinant of the coefficient matrix is not 0 there is a unique solution. If is is 0, there are either an infinite number of solutions or no solution.

It's easy to see that if a= 0, then the equations become
2x − z = 0
3x + 4y − z = 13
10x + 8y − 4z = 26

Of you multiply the second equation by 2 and subtract that from the third equation you get 4x- 2z= 0 which is just a multiple of the first equation. That's why a= 0 gives an infinite number of solutions. You can choose x to be any number at all, take z= 2x, y= (13- x)/4 and you have a solution.

If a= 2 the equations become
2x + 2y − z = 0
3x + 4y − 3z = 13
10x + 8y − 2z = 26
Now, if you multiply the second equation by 2 and subtract that from the third equation you get 4x + 4z= 0. If you multiply the first equation by 2 and subtract that from the second equation you get -x- z= 13. No values of x and z make both of those equations true. That's why there is no solution.

If you wanted to do this with matrices, you could write the "augmented matrix",
$$\begin{bmatrix}2 & a & -1 & 0\\ 3 & 4 & -(a+1) & 13\\ 10 & 8 & a- 4 & 26\end{bmatrix}$$
and row reduce. If a= 0, the resulting reduced matrix will have all "0"s in the last row. If a= 0, the last row will have "0"s in the first three columns but not in the fourth column.