If $n\times n$ matrix $A$ satisfies $A^2+2A-3E=O$, show that $A$ is invertible.

Question

If $n\times n$ matrix $A$ satisfies the equation $$A^2+2A-3E=O$$

where $E$ is identity matrix. Show that $A$ is invertible.

What I tried is to factorize it, and get $(A+3E)(A-E)=O$. I stuck here, because this is matrix multiplication and I can't say $A=E$ or $A=-3E$. If I take the determinant,

$$\det(A+3E)\cdot \det(A-E)=0$$

This only means $1$ or/both $-3$ are eigenvalues of $A$, which doesn't help to show $A$ is invertible. How can I proceed? Thank you in advance.

Answer 1

Write it as$$A^2+2A=3E$$

take determinant on both sides,

$$\det A\cdot \det(A+2E)=\det 3E=3^n\neq0$$

hence

$$\det A\neq0$$

Matrix $A$ is invertible.

Answer 2

This is an old nugget.

$$E=\frac{1}{3}A^2+\frac{2}{3}A=A\left(\frac{1}{3}A+\frac{2}{3}E\right)$$

so not only $A$ is invertible, its inverse is:

$$A^{-1}=\frac{1}{3}A+\frac{2}{3}E$$

Answer 3

Note that the polynomial $p(t)=t^2+2t-3=(t+3)(t-1)$ annihilates matrix $A$ so the annihilating polynomial of least degree (minimal poynomial for $A$) is $(t+3)$ or $(t-1)$ or $(t+3)(t-1)$.
For any of the above possibility of $m(t)$, the eigenvalue (ignoring multiplicity) of $A$ are $-3$ only or $1$ only or both. Since $det(A)=$ product of eigenvalues of $A$; so it is $(-3)^n$ or $1$ or $(-3)^r$ where $1\le r<n$.