Matrix equation confusion

Question

I have to solve the following matrix equation

$$X(A - A^T + E) = A^{-1}$$

where $E$ is the $3 \times 3$ identity matrix and $A$ is the matrix

$$A = \begin{bmatrix}2&3&2\\0&1&2\\0&0&-1\end{bmatrix}$$

Any idea on how to solve it? I would like a step-by-step explanation.

Answer 1

Hint: Find the matrix $B := A - A^T + E$ so you will have the equation

$$XB = A^{-1}$$

Find $A^{-1}$. See if $B$ is invertible, i.e., see if $\det(B) \neq 0$ and then find $B^{-1}$ Such that you can use

$$XBB^{-1} = XE = X = A^{-1}B^{-1}$$

Answer 2

Left-multiplying both sides of the matrix equation by $\mathrm A$, we obtain

$$\mathrm A \mathrm X (\mathrm A - \mathrm A^{\top} + \mathrm I_3) = \mathrm I_3$$

Vectorizing both sides of the matrix equation above, we obtain a linear system of $9$ equations in $9$ unknowns

$$((\mathrm A^{\top} - \mathrm A + \mathrm I_3) \otimes \mathrm A) \, \mbox{vec} (\mathrm X) = \mbox{vec} (\mathrm I_3)$$

whose augmented matrix is

$$\left[\begin{array}{ccc|c} \mathrm A & -3 \mathrm A & -2 \mathrm A & \mathrm e_1\\ 3 \mathrm A & \mathrm A & -2 \mathrm A & \mathrm e_2\\ 2 \mathrm A & 2 \mathrm A & \mathrm A & \mathrm e_3\end{array}\right]$$