Weirdly written matrix equation

Question

The matrix $A$ is a $3 \times 3$ matrix.

$$A = \left [\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 2 & 1 \\ 0 & 1 & 1 \\ \end{array} \right ]$$

How can I solve the following equation:

$$X\,A = A+2\,X$$

• Since $X$ is positioned left of $A$, $X$ must be a row vector, am I right? Wolfram refuses to add the vector $2X = 2\cdot(x_1,x_2,x_3)$ to $A$, claims they are of unequal length.

Answer 1

$$XA=A+2X\implies X(A-2I)=A\implies X=A(A-2I)^{-1} $$

Observe the rightmost matrix is invertible since $\;2\;$ is not an eigenvalue of $\;A\;$

Answer 2

The $X$ must be a matrix of the same dimension as $A$. You can rewrite your equation for exmple as $$X(A - 2 I) =A$$ where $I$ is the identity matrix. You are now faced with the task of solving three times a system of (three) linear equations (yet note the coefficient matrix is always the same).