Question

Find matrix A if (4A-I)^-1 = (2 -1
3 5)
I simplified the left side expression but I end up with a long stupid thing that I can't seem to solve for.

Answer 1

Let A=[a b \\ c d]. Perform 4A-I and then find its inverse.

Answer 2

thats what I did but I am stuck with like a=(-bc+1)/(4-2d)

Answer 3

Approach it this way (this way, you avoid computing \(\large (4A-I)^{-1}\) for a general matrix \(\large A\)). Note that \[\large \begin{aligned} (4A-I)^{-1} = \begin{bmatrix}2 & -1\\ 3 & 5\end{bmatrix} &\implies ((4A-I)^{-1})^{-1} = \begin{bmatrix}2 & -1\\ 3 & 5\end{bmatrix}^{-1} \\ &\implies 4A-I = \begin{bmatrix}2 & -1\\ 3 & 5\end{bmatrix}^{-1} \\ &\implies \ldots\end{aligned}\] Doing things this way makes your life easier because it's easier to compute \(\large \begin{bmatrix}2 & -1\\ 3 & 5\end{bmatrix}^{-1}\) instead of \(\large (4A-I)^{-1}\) for general \(\large A = \begin{bmatrix} a & b\\ c & d\end{bmatrix}\).

Can you see how to proceed from here? :-)

Answer 4

ooohh that is much more easier thanks a lot!