Equation with matrix

Question

$$ \begin{pmatrix} 3 & -1\\ -4 & 2 \\ \end{pmatrix}=(X^T+3I)^{-1} $$

$T$ is the transpose and $-1$ is the inverse and $I$ is the unity matrix.

I have come this far $$ \begin{pmatrix} 3 & -1\\ -4 & 2 \\ \end{pmatrix}-(3I)^{-1}=(X^T)^{-1} $$ but how do I remove the $T$ and $-1$? Made a new solution $$\left(\begin{matrix}3&-1\\-4&2\\\end{matrix}\right)=X^T+3I$$ $$\left(\begin{matrix}2&1\\4&3\\\end{matrix}\right)-3I=X^T$$ $$\left(\begin{matrix}2&1\\4&3\\\end{matrix}\right)-\left(\begin{matrix}3&0\\0&3\end{matrix}\right)=X^T$$ $$\left(\begin{matrix}-1&1\\4&0\\\end{matrix}\right)=X^T$$ $$X=\left(\begin{matrix}-1&4\\1&0\\\end{matrix}\right)$$ what is wrong with this one?

Answer 1

Lets call the left hand side $A$. You want to solve $A=(X^T+3I)^{-1}$ for $X$. Thus taking the inverse we get $A^{-1} =X^T+3I \Leftrightarrow A^{-1}-3I = X^T \Leftrightarrow X = (A^{-1}-3I)^T$

Answer 2

Watch out, there's a bad mistake, it is not true in general that $(a + b)^{-1} = a^{-1} + b^{-1}$.

Instead, invert both sides as it have been suggested by Matt, to get $$ X^{T} + 3 I = \begin{bmatrix}3&-1\\-4&2\end{bmatrix}^{-1} = \begin{bmatrix}1&{1}/{2}\\2&{3}/{2}\end{bmatrix}^{-1}, $$ and from there it should be easy.