Proof linear algebra - Inverse Verify that A=$\begin{bmatrix}1&-1\\0&2\end{bmatrix}$
satisfies $A^2-3A+2I=0$.
Use this fact to show that $A^{-1}= \frac{1}{2}(3I-A)$
I know it does satisfies the first equation/already proved it.
 A: By the given equation, we have $A^2-3A=-2I$, thus
$$\frac12(3A-A^2)=I$$
A: An $n\times n$ square matrix $A$ is said to be invertible iff there exists an $n\times n$ matrix $B$ such that $AB = I_{n\times n}$.
If such a matrix $B$ exists, it is known that it will be unique and we can give it the name "The inverse of $A$" and a special symbol, $A^{-1}$, to further emphasize the relationship between it and the original matrix $A$.  It is also known that for square matrices that $AB=I$ will imply that $BA=I$.  (Warning: the same is not true for non-square matrices)
For your problem, once having verified that our matrix $A$ satisfies the matrix equation mentioned, all that remains to prove that $\frac{1}{2}(3I-A)$ is in fact the inverse of $A$ is to show that $A\cdot \frac{1}{2}(3I-A) = I$.
By basic algebraic manipulation we have that $$A\cdot \frac{1}{2}(3I-A) = \frac{1}{2}(3A-A^2)$$
From here, we can use the fact that "adding zero" doesn't change anything and that from the earlier part of the problem we learned that $0 = A^2-3A+2I$, so this continues:
$$\frac{1}{2}(3A-A^2) = \frac{1}{2}(3A-A^2+0) = \frac{1}{2}(3A-A^2+A^2-3A+2I) = \frac{1}{2}(2I)=I$$
We have thus shown that $A\cdot \frac{1}{2}(3I-A)=I$, thereby proving that $\frac{1}{2}(3I-A)=A^{-1}$
