# Solving for matrices $A$ and $B$

$$A^{-1}B^{-1}A\,B=\begin{bmatrix}7&6\\2&1\end{bmatrix}$$

Is there a shortcut to solve for $A$ and $B$? Or is there a solution for $A$ and $B$ at all?

## 1 Answer

Hint: Take the determinant of both sides.

• I'm getting 1 = -5. (1/detA) (1/det B) ( det B) (det A) = 7(1) - 6(2). Am I doing this wrong? – Compsci Jul 10 '17 at 17:32
• You are doing it correctly. What does that tell you? – Ross Millikan Jul 10 '17 at 17:35
• there's no solution – Compsci Jul 10 '17 at 18:04
• @Compsci Correct. More generally, this argument shows that $\det{M}=1$ is a necessary condition for there to be matrices $A,B$ such that $M=A^{-1}B^{-1}A B$. (Interestingly, under fairly weak conditions this is also sufficient: "The commutator subgroup of the general linear group $GL_n(k)$ over a field or a division ring $k$ equals the special linear group $SL_n(k)$ provided that $n\neq 2$ or $k$ is not the field with two elements." source. So if the determinant were 1, then indeeed there should be such $A,B$.) – Semiclassical Jul 10 '17 at 20:59