How to find a value for a variable that makes a matrix (with said variable) equal to its own inverse I'm given $$\begin{bmatrix}3&x\\-2&-3\\\end{bmatrix}$$
and am asked to find x such that it's inverse would equal itself. To attempt this I first tried to put the question into an augmented matrix and got this:
$$\begin{bmatrix}1&x/3&1/3&0\\0&(x/3)-(3/2)&1/3&1/2\\ \end{bmatrix}$$
I found that my answer was wrong so I tried:
$$\begin{bmatrix}3&x\\-2&-3\\\end{bmatrix}$$
times
$$\begin{bmatrix}x_1&x_2\\x_3&x_4\\\end{bmatrix}$$ to try and solve for x but found similar dissatisfactory results.
The answer is listed as x = 4; how might I go about solving this? Did I just make a mistake with my methods or is this the entirely wrong way about?

Both methods you attempted must give correct answer. 
Method 1. Finding the inverse from the augmented matrix:
$$ \left[
\begin{array}{cc|cc}
  1&0&3&x\\
  0&1&-2&-3
\end{array}
\right] \stackrel{R_1/3}=
\left[\begin{array}{cc|cc}
  \frac13&0&1&\frac x3\\
  0&1&-2&-3
\end{array}
\right] \stackrel{2R_1+R_2\to R_2}=\\
\left[\begin{array}{cc|cc}
  \frac13&0&1&\frac x3\\
  \frac23&1&0&\frac {2x}3-3
\end{array}
\right] \stackrel{\frac{3}{2x-9}\cdot R_2}=
\left[\begin{array}{cc|cc}
  \frac13&0&1&\frac x3\\
  \frac{2}{2x-9}&\frac{3}{2x-9}&0&1
\end{array}
\right] \stackrel{-\frac{x}{3}\cdot R_2+R_1\to R_1}=\\
\left[\begin{array}{cc|cc}
  \frac13-\frac{2x}{3(2x-9)}&-\frac{x}{2x-9}&1&0\\
  \frac{2}{2x-9}&\frac{3}{2x-9}&0&1
\end{array}
\right].$$
So, it must be:
$$\left[\begin{array}{cc}
  \frac13-\frac{2x}{3(2x-9)}&-\frac{x}{2x-9}\\
  \frac{2}{2x-9}&\frac{3}{2x-9}
\end{array}\right]=
\left[\begin{array}{cc}
  3&x\\
  -2&-3
\end{array}\right] \Rightarrow x=4.$$
Method 2. Multiply by its inverse (by condition it must be equal to the original matrix and remember the result is an identity matrix: $A\cdot A^{-1}=I$):
$$\left[\begin{array}{cc}
  3&x\\
  -2&-3
\end{array}\right]
\left[\begin{array}{cc}
  3&x\\
  -2&-3
\end{array}\right]=
\left[\begin{array}{cc}
  1&0\\
  0&1
\end{array}\right] \Rightarrow \\
\left[\begin{array}{cc}
  9-2x&0\\
  0&-2x+9
\end{array}\right]=
\left[\begin{array}{cc}
  1&0\\
  0&1
\end{array}\right] \Rightarrow x=4.$$
