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?
Best Answer
Hint: what is $$ \pmatrix{3 & x\cr -2 & -3}^2 $$ and what do you need it to be?