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?
 A: Hint: what is $$ \pmatrix{3 & x\cr -2 & -3}^2 $$
and what do you need it to be?
A: The determinant of the matrix is 
$$
\begin{vmatrix}
3 & x \\ -2 & -3
\end{vmatrix}=2x-9.
$$
Assuming this matrix has an inverse, the determinant of that inverse will be the reciprocal of the determinant. So, we must have
$$
9+2x=\frac{1}{2x-9}\Rightarrow2x-9=\pm1.
$$
Now, $2x-9=1$ when $x=5$, and $2x-9=-1$ when $x=4$.  So, these are the only two possible values.  From here, you can check these two values to see what works.
A: If we have a $ 2 \times 2 $ matrix $A$ then the inverse can be given by
$$ A^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a  \end{bmatrix} \tag{1} $$
Then with your matrix $A$
$$A = \begin{bmatrix} 3 & x \\ -2 & -3 \end{bmatrix} \tag{2} $$
$$ A^{-1} = \frac{1}{-9+2x}\begin{bmatrix} -3 & -x \\ 2 & 3  \end{bmatrix} \tag{3} $$
You need to find where they are the same. I won't do it all I guess.
A: $A^{-1} = A$ is equivalent to $A^2 = I$ or $A^2-I = 0$.
Hence the polynomial $\lambda^2-1 = (\lambda-1)(\lambda+1)$ must annihilate $A$. Clearly $A \ne \pm I$ so the characteristic polynomial of $A$ must be equal to $\lambda^2-1$.
We have
$$\det(A - \lambda I) = \begin{vmatrix} 3-\lambda & x \\ -2 & -3-\lambda\end{vmatrix} = \lambda^2 - 9 + 2x$$
Hence $2x-9 = -1$ which gives $x = 4$.
A: 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: $A\cdot A^{-1}=I$):
$$\left[\begin{array}{cc}
  3&x\\
  -2&-3
\end{array}\right]
\left[\begin{array}{cc}
  3&x\\
  -2&-3
\end{array}\right]=
\left[\begin{array}{cc}
  1&0\\
  0&1
\end{array}\right] \Rightarrow \\
\left[\begin{array}{cc}
  9-2x&0\\
  0&-2x+9
\end{array}\right]=
\left[\begin{array}{cc}
  1&0\\
  0&1
\end{array}\right] \Rightarrow x=4.$$
