Matrix calculus: $Bx = y$ and $AB = BA = Id \implies x= A^{-1}y?$

57 Views Asked by At

As stated in the title, I have an exercise with two matrices $A$ and $B$ and a system with $Bx=y$ with $x$ and $y$ being two vectors. In a previous exercise I've already shown that $AB = BA = Id$.

Now I'm supposed to show that $x := A^{-1}y$ without doing any calculations.

Since we have $AB = BA = Id \implies B = A^{-1}$, we then get $x = B^{-1}y \iff x = Ay$

My question now is: Why is this incorrect? Since I'm supposed to show that $x = A^{-1}y$

1

There are 1 best solutions below

0
On BEST ANSWER

It's a mistake in the assignment.

From $AB=BA=I$ (where $I$ is the identity matrix), you conclude that $B=A^{-1}$, so from $$ y=Bx=A^{-1}x $$ you derive $$ Ay=x $$ and it's generally false that $x=A^{-1}y$, which would require $$ (A-A^{-1})y=0 $$ The last condition may hold for particular $A$ and $y$, but not generally.