Find a homogeneous system that will give all the elements of B. If A = $\begin{bmatrix} 1 & -1 \\ 0 & -2 \end{bmatrix}$, set B = {X ∈ $M_2(\mathbb{R})$∣AX + $X^\intercal$A = 0}
Find a homogeneous system that will give all the elements of B.
My attempt:
AX + $X^\intercal$A = 0
AX = $-$ ($X^\intercal$A)
($\cfrac{1}{A}$) AX = $-$ ($X^\intercal$A)$(\cfrac{1}{A})$
X = $-$ $X^\intercal$
This would mean that all skew-symmetric matrices would satisfy the conditions.
However, I tried plugging in random skew-symmetric matrices but AX + $X^\intercal$A woouldn't equate to 0.
I may have erred in my solution: I just don't know where.

You cannot infer from the step $AX=-(X^TA)$ that $A^{-1}AX=-(X^TA)A^{-1}$.
When two square matrices are equal, say $X=Y$, you can left-multiply both sides by a matrix $M$ to get another equality $MX=MY$, or you can right-multiply both sides by $M$ to get $XM=YM$. However, you cannot infer that $MX=YM$, because matrix multiplication in general is not commutative.
