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**alterah** i am having two problems here. The first one i am somewhat confident of:

1. Show that in general, for any square matrix satisfying:

$\displaystyle a^2 - 2a + 5i = 0$

then

$\displaystyle a^{-1} = \frac{1}{2}(2 - a)$

i proceeded as follows:

$\displaystyle a^2a^{-1} - 2aa^{-1} + 5ia^{-1} = 0a^{-1}$

$\displaystyle a - 2i + 5a^{-1} = 0$

$\displaystyle a^{-1} = \frac{1}{2}(2 - a)$

while what i did seems legitimate...how do i show that the matrices must be square? mr f says: Does a non-square matrix have an inverse ....?

2. Let a, d, and p be n x n matrices satisfying $\displaystyle p^{-1}ap = d$. Solve for a. Must it be true that a = d?

For this one i am not sure how to proceed. In order to get rid of p i need to say multiply by $\displaystyle p^{-1}$ but that doesn't seem to help me... Same sort of situation for p.