HI I'm really close to the answer, but do not know where I went wrong..

Consider the two equations 4x+8y=1 and 2x-ay=11 ( One is on top of the other) so in matrix form it is

$\displaystyle \begin{bmatrix} 4 & 8 \\2 & -a \end{bmatrix}$$\displaystyle \begin{bmatrix} x \\y \end{bmatrix}$ = $\displaystyle \begin{bmatrix} 1 \\11 \end{bmatrix}$

For what values of a does the system have a unique solution?

Arranging in augmented matrix and using row operations, I reduced the matrix to

$\displaystyle \begin{bmatrix} 4 & 8 \\0 & 2a+8 \end{bmatrix}$$\displaystyle \begin{bmatrix} x \\y \end{bmatrix}$ = $\displaystyle \begin{bmatrix} 1 \\-22 \end{bmatrix}$

I had to invert the matrix premultiplied to the matrix containing x and y, so I had to get the det, I got 8a+32

$\displaystyle \begin{bmatrix} x \\y \end{bmatrix}$ = $\displaystyle \frac{1}{8a+32}\begin{bmatrix} 2a+8 & -8 \\0 & 4 \end{bmatrix}\begin{bmatrix} 1\\-22\end{bmatrix}$

Then I multiplied the determinant into the inversed matrix, simplified, was left with

$\displaystyle \begin{bmatrix}\frac{a+4}{4a+16} &\frac{-1}{a+4} \\0 &\frac{1}{2a+8} \end{bmatrix}\begin{bmatrix} 1 \\-22 \end{bmatrix}$

multiplying them, I get

$\displaystyle \begin{bmatrix}\frac{a+4}{4a+16}+\frac{22}{a+4}\\0 +\frac{-22}{2a+8}\end{bmatrix}$

So my answer is x = $\displaystyle \frac{a+92}{4a+16}$.. and y = $\displaystyle \frac{-22}{2a+8}$

but it's wrong.. can someone show me where I went wrong?

Will really appreciate it, thank you so much!