Inverse Matrices Help

I multiplied out to get A² and got $\begin{pmatrix} 2-x & 0 & 1\\ 0 & -1+x & -1\\ 1 & x & 1 \end{pmatrix}$ (and I know this is the same as the inverse).

How do I solve for $x$?

I believe you can do this:

$\mathbf{A^2} = \mathbf{A^{-1}}$

$\Rightarrow \ \mathbf{A} \mathbf{A^2} = \mathbf{A} \mathbf{A^{-1}}$

$\Rightarrow \ \mathbf{A} \mathbf{A^2} = \mathbf{I}$

$\Rightarrow \ \mathbf{A} \begin{pmatrix} 2-x & 0 & 1 \\0 & 1-x & -1 \\1 & x & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \end{pmatrix}$

I know from here you can multiply each side by A^-1 to get IA² = A^-1I and the cancel the Is, but I don't know how to solve for x from here:

$\begin{pmatrix} 2-x & 0 & 1\\ 0 & -1+x & -1\\ 1 & x & 1 \end{pmatrix} = \begin{pmatrix} 1 & x & 1\\ -1 & -1 & 0\\ 1 & 0 & 0 \end{pmatrix}$

The most straight forward way would be to note A^3=I, so it has determinant 1. Simply solve the determinant equation you get in terms of x.

Ok so if I multiply by A to get A³ = I I get this:

$\begin{pmatrix} 3-x & x^2 & 2-x\\ -2+x & 1-x & 0\\ 2-x & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}$

We've only done half a lesson or so on inverse matrices so they're not my strong point- how do I go about solving the determinant when it's equal to 1?

Work through component by component. For the top row:

3-x = 1 so x=2
x^2 = 0 so x=0
2-x=0 so x=2

You have conflicting values of x coming out, which suggest you haven't found the right value of A^3.

edit: I get 2x-x^2 for the top middle component.

I just tried the multiplication again and definitely got the same answer... How did you get $2x-x^2$?

Most of the solutions are $x=2$ but what about the quadratic where $x=0$? Also what does $0=0$ and $1=1$ mean in this context?

EDIT: not sure why the image has rotated...

Well you're trying to end up with the identity matrix - with 1s on the diagonal and 0s elsewhere. So if you end up with 0s and 1s where they should be, that's great.

All you have to do now is make sure that you can find an x value that puts 1s and 0s in the other places where you need them. If one of your x values doesn't do this you can reject it!