I have a question about Inverse Matrix. I would appreciate if anyone could provide some help.
Question:
Suppose that a matrix A satisfies the equation $A^2-4A+3I=0$. Find an expression for $A^-1$.
When I working on this question
$A^2-4A=-3I$
$(-1/3)(A^2-4A)=I$
$(-1/3)A(A-4I)=I$
I don't understand why $A^2-4A$ become $A(A-4I)$
Then I saw my teacher wrote this:
4*I is a matrix and 4 is a number so they aren't the same, but 4A = 4AI = A(4I)
so A2−4A=A.A−A.4I=A(A−4I)
As I know $AA^-1=I$, if $A(A-4I)$ put back the A inside the (),
it will become $A^2-4AI$ -> $A^2-4AAA^-1$
It’s blowing my mind. Can anyone explain this?
That $A^2-4A = A(A-4I)$ is just laws of matrix multiplication. You can really just prove the law $A(B+C)=AB+AC$ for all matrices s.t. the dimensions fit if you wish to.