Friedburg 5.4.18

Let $A$ be an $n \times n$ matrix with charachteristic polynomial $$f(t)=(-1)^nt^n+a_{n-1}t^{n-1}+ \cdots +a_1t+a_0$$ (a) Prove that $A$ is invertible iff $a_0 \neq 0$.
(b) Prove that if $A$ is invertible, then $$A^{-1}=\frac{-1}{a_0}[(-1)^nA^{n-1}+ \cdots +a_1I]$$ (c) Use (b) to compute $A^{-1}$ for $A=\begin{pmatrix} 1&2&1\\ 0&2&3\\ 0&0&-1\\ \end{pmatrix}$
This exercise provides an interesting result. We get a formula for the inverse of a matrix $A$.

$\textit{Proof}$ : Suppose to the contrary that $a_0=0$. We will show that this implies $A^{-1} \not \exists$
If $a_0=0$ then the charachteristic polynomial of $A$ becomes $$f(t)=(-1)^nt^n+a_{n-1}t^{n-1}+ \cdots +a_1t=t[(-1)^nt^{n-1}+a_{n-1}t^{n-2}+ \cdots +a_1]$$ $\Longleftrightarrow t$ is a factor of $f(t)$ $\Longleftrightarrow t=0$ is a root of $f(t) \Longleftrightarrow 0$ is an eigenvalue of $A$ $\Longleftrightarrow$ (invertible matrix theorem) $A^{-1} \not \exists$.
Each of the above steps was "iff" so,
$\therefore$ $A$ invertible iff $a_0 \neq 0$

If $A$ is invertible, then by part (a) above, $a_0 \neq 0$ in $f(t)$. We may write the charachteristic polynomial $f(t)$ in matrix form as $f(A)$. By the Cayley-Hamilton theorem, a matrix "satisfies" its own charachteristic polynomial. So, $$f(A)=0=(-1)^nA^n+ \cdots + a_1A+a_0I$$ $$-a_0I=(-1)^nA^n+ \cdots a_1A$$ Since $A$ is invertible, we may apply $A^{-1}$ to both sides of the above equation to get $$-a_0IA^{-1}=[(-1)^nA^n+\cdots +a_1A]A^{-1}$$ $$-a_0A^{-1}=(-1)^nA^{n-1}+\cdots+a_1$$ $$A^{-1}=\frac{-1}{a_0}[(-1)^nA^{n-1}+ \cdots +a_1I]$$ $\blacksquare$
For part (c), we compute $$|A-tI|=(1-t)(2-t)(-1-t)=-t^3+2t^2+t-2$$ So, $a_0=-2$ and our formula gives $$A^{-1}=\frac{-1}{-2}(-A^2+2A+I)$$ Performing martrix operations, we get $$A^{-1}=\begin{pmatrix} 1&-1&-2\\ 0&\frac{1}{2}&\frac{3}{2}\\ 0&0&-1\\ \end{pmatrix}$$