Finding a value for A so that a system of linear equations is inconsistent

Consider the system of linear equations

x − 3z = −3

−2x − Ay + z = 2

x + 2y + Az = 1

Find the respective values of A such that the system is inconsistent..

I've tried rewriting this as an augemented matrix, then doing row operations to get the bottom row as [0 0 {(A+3)/2 - 5/A} | 2-(4/A)]

so for it to be inconsistent i know that it has to look like [0 0 0 | K] for any real k

so (A+3)/2 - 5/A should equal zero, but 2-4A shouldn't because that would make the system have infinitely many solutions, not NO solutions.

but i can't find any values of A that solve both these equations!

Please help, I'm so confused (Crying)

Re: Finding a value for A so that a system of linear equations is inconsistent

Note that the system of linear equations does not have a solution only if the determinant of $\displaystyle \begin{bmatrix}1 & 0 & -3 \\ -2 & -A & 1 \\ 1 & 2 & A \end{bmatrix} $ is 0, the determinant of this matrix is $\displaystyle -A^2 - 3A + 10 $ which is 0 at (using quadratic formula) A = -5, A = 2

so to see plug in A = -5, and to verify if the system is inconsistent, first note that By simple examination i see that $\displaystyle -3 * \begin{bmatrix}1 \\ -2 \\ 1 \end{bmatrix} + -1 * \begin{bmatrix}0 \\ 5 \\ 2 \end{bmatrix} = \begin{bmatrix}-3 \\ 1 \\ -5 \end{bmatrix} $ so the last column vector is linearly dependent, so we can throw it out.

Now $\displaystyle \begin{bmatrix}1 & 0 & -3 \\ -2 & 5 & 1 \\ 1 & 2 & -5 \end{bmatrix} x = \begin{bmatrix} -3 \\ 2 \\ 1 \end{bmatrix} $ is aking to solving

$\displaystyle \begin{bmatrix}1 & 0 \\ -2 & 5 \\ 1 & 2 \end{bmatrix} x = \begin{bmatrix} -3 \\ 2 \\ 1 \end{bmatrix} $

so by observation we note that a solution $\displaystyle \begin{bmatrix}x \\ y \end{bmatrix} $ of this system must have x = -3, which constraints y = $\displaystyle \frac{-4}{5}$ so our soln, if we have one is $\displaystyle \begin{bmatrix}-3 \\ -\frac{4}{5} \end{bmatrix} $ but

$\displaystyle \begin{bmatrix}1 & 0 \\ -2 & 5 \\ 1 & 2 \end{bmatrix} \begin{bmatrix}-3 \\ -\frac{4}{5} \end{bmatrix} = \begin{bmatrix}-3 \\ 2 \\ \frac{-23}{5} \end{bmatrix} $ which does not equal $\displaystyle \begin{bmatrix}-3 \\ 2 \\ 1 \end{bmatrix}$ Thus for A = -5, this system is inconsistent (has no solutions).

Now for A = 2, using the above analysis, it has an infinitely many soln.

So only for A = -5, the system of lin eqn is inconsistent.