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


Hint: Take the determinant of both sides.

  • $\begingroup$ I'm getting 1 = -5. (1/detA) (1/det B) ( det B) (det A) = 7(1) - 6(2). Am I doing this wrong? $\endgroup$ – Compsci Jul 10 '17 at 17:32
  • 2
    $\begingroup$ You are doing it correctly. What does that tell you? $\endgroup$ – Ross Millikan Jul 10 '17 at 17:35
  • $\begingroup$ there's no solution $\endgroup$ – Compsci Jul 10 '17 at 18:04
  • $\begingroup$ @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 $A,B$.) $\endgroup$ – Semiclassical Jul 10 '17 at 20:59

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