Find all matrices $X$ such that $ABXB^tA^t=I$

Question

Find all matrices $X$ such that:

$$ABXB^tA^t=I$$ if $A=\begin{pmatrix} 1 &-2 &2\\ 3 &-5 &6\\ -1 &2 &-1 \end{pmatrix}$ and $B=\begin{pmatrix} -3 &-2 &-2\\ 2 & 1 &1\\ 6 &3 &4 \end{pmatrix}$.

So I managed to get that $AB=\begin{pmatrix} 5 &2 &4\\ 17 &7 &13\\ 1&1&0 \end{pmatrix}$ and $B^tA^t=(AB)^t=\begin{pmatrix} 5 &17 &1\\ 2 &7 &1\\ 4 &13 &0 \end{pmatrix}$

So we have $\begin{pmatrix} 5 &2 &4\\ 17 &7 &13\\ 1&1 &0 \end{pmatrix}\cdot X\cdot\begin{pmatrix} 5 &17 &1\\ 2&7&1\\ 4&13&0 \end{pmatrix}=I$

Now how do I get $X$ from here?

Answer 1

Matrices multiplication is not commutative. You are allowed to multiply an equation by a non-zero matrix, but you have to mind the side from which you are applying the multiplication.

Let us say that $P=AB$ and $Q=B^TA^T$.

Your equation takes the form of $P \cdot X \cdot Q = I$.

You can multiply each side by $P^{-1}$ from the left and $Q^{-1}$ from the right to get (note that $\det(P)\ne0\ne\det(Q)$, which makes $P$ and $Q$ invertible):

$A^{-1} \cdot P \cdot X \cdot Q \cdot Q^{-1} = P^{-1} \cdot I \cdot Q^{-1}$

This can be simplified based on $M^{-1} \cdot M = M \cdot M^{-1} = I$:

$I \cdot X \cdot I = P^{-1} \cdot I \cdot Q^{-1}$

This can further be simplified based on $M \cdot I = M$:

$X=P^{-1} \cdot Q^{-1}$

Mind that:

$M^{-1}=\frac{1}{\det(M)}\left(M^*\right)^T$

Where $M^*$ is the cofactor matrix.

Answer 2

If either $AB$ or $B^TA^T$ is singular, there is no chance to get a solution. Therefore the $3 \times 3$ matrices, have to be nonsingular (and they are, I just checked). So, to get $X$ you need to invert both of them and get to $X=B^{-1}A^{-1}A^{-T}B^{-T} = B^{-1}(A^TA)^{-1}B^{-T}$

Usually you would decompose $AB$ into $LR$ or $QR$ decompositions and pull them to the right hand side.

Answer 3

From $ABXB^tA^t=I$, we see that $det(A) \ne 0 \ne det(B)$. Hence $A$ and $B$ are invertible.

Again , from $ABXB^tA^t=I$, we see

$X=(AB)^{-1}((AB)^{-1})^t$