Matrices and number of solutions

[t_n_p]

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]

Actually, infinitely solutions exist for your system when the system of equations is dependent, and for three equations, this means that anyone 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?

 

[t_n_p]

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

 

[HallsofIvy]

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.