Question

Given A=[x1=3, x2=x,x3=-2,x4=-3]. Find the value of x for which the matrix A is its own inverse.

Answer 1

I'm having a real hard time understanding your Q

Answer 2

Matrix 2x2 A=[3,x,-2,-2]

Answer 3

\[A=\left[\begin{matrix}3&x\\-2&-3\end{matrix}\right]\]is the bottom-right element -3 or -2?

Answer 4

-3

Answer 5

\[A^2=\left[\begin{matrix}3&x\\-2&-3\end{matrix}\right]\left[\begin{matrix}3&x\\-2&-3\end{matrix}\right]=?\]multiply the matrix my itself and see what you get

Answer 6

after all we are told that\[A=A^{-1}\]so\[AA^{-1}=AA=A^2\]right?

Answer 7

It makes sense what you are saying but I'm getting [(9-2x), 0, 0, (-2x+6)] Whats the value of x?

Answer 8

you got the last part wrong I think
it should be\[A^2=\left[\begin{matrix}3&x\\-2&-3\end{matrix}\right]\left[\begin{matrix}3&x\\-2 &-3\end{matrix}\right]=\left[\begin{matrix}9-2x&0\\0&-2x+9\end{matrix}\right]\]and since\[AA^{-1}=I\]you should know how to find a single value for x from here