Let $A$ be a square matrix of order $n$ over $\mathbb{R}$ such that $A^2-3A+I_n=0_n$. Then $A^{-1}=3I_n-A$

Question

Let $A$ be a square matrix of order $n$ over $\mathbb{R}$ such that $A^2-3A+I_n=0_n$. Then $A^{-1}=3I_n-A$.

The answer given is true.

Suppose $A^{-1}$ exists. Then $A^{-1}(A^2-3A+I_n=A^{-1}(0_n)$ which implies $A-3I_n+A^{-1}=0_n$. So clearly, $A^{-1}=3I_n-A$.

But I can't show that the inverse of $A$ exists.

Answer 1

$$ 3A-A^2 = I_n \implies (3I_n-A)A =I_n \implies 3I_n-A = A^{-1} $$

Also, $$ 3A-A^2 = I_n \implies A(3I_n-A) =I_n \implies 3I_n-A = A^{-1} $$

So the inverse of $A$ exists, it is both a right and left inverse.

Answer 2

For square matrices, a one-sided inverse is automatically the (unique) inverse. Let's exhibit a one-sided inverse... $3\,{\rm Id}_n-A$, say? We check: $$A(3\,{\rm Id}_n - A) = 3A - A^2 \stackrel{(\ast)}{=} {\rm Id}_n,$$where in $(\ast)$ we use the hypothesis.

Answer 3

i guess this method's easier. Just expand the 'I'