Let $A$ be a square matrix of order $n$ over $\mathbb{R}$ such that $A^2-3A+I_n=0_n$. Then $A^{-1}=3I_n-A$ 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.
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.
i guess this method's easier. Just expand the 'I'