# Solve this problem

Question:

If $A=\frac{1}{9}\left[\begin{array}{ccc}-8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4\end{array}\right]$, prove that $A^{-1}=A^{3}$

Solution:

$A=\frac{1}{9}\left[\begin{array}{lll}-8 & 1 & 4\end{array}\right.$

$\begin{array}{lll}4 & 4 & 7\end{array}$

$\left.\begin{array}{lll}1 & -8 & 4\end{array}\right]=\left[\begin{array}{lll}-8 / 9 & 1 / 9 & 4 / 9\end{array}\right.$

$\begin{array}{ccc}4 / 9 & 4 / 9 & 7 / 9 \\ 1 / 9 & -8 / 9 & 4 / 9]\end{array}$

$\Rightarrow A^{T}=\left[\begin{array}{lll}-8 / 9 & 4 / 9 & 1 / 9\end{array}\right.$

$\begin{array}{lll}1 / 9 & 4 / 9 & -8 / 9\end{array}$

$4 / 9 \quad 7 / 9 \quad 4 / 9]=\frac{1}{9}[-8 \quad 4 \quad 1$

$\begin{array}{lll}1 & 4 & -8\end{array}$

$\left.\begin{array}{lll}4 & 7 & 4\end{array}\right] \quad \ldots(1)$

$|A|=\mid \begin{array}{lll}-8 / 9 & 1 / 9 & 4 / 9\end{array}$

$4 / 9 \quad 4 / 9 \quad 7 / 9$

$\begin{array}{lll}1 / 9 & -8 / 9 & 4 / 9\end{array}=\frac{1}{9 \times 9 \times 9} \mid-8 \quad 1 \quad 4$

$\begin{array}{lll}4 & 4 & 7\end{array}$

$\begin{array}{lll}1 & -8 & 4\end{array}$

$=\frac{1}{9 \times 9 \times 9}[(-8 \times 72)-(1 \times 9)+\{4 \times(-36)\}]$

$=\frac{1}{9 \times 9 \times 9} \times 9 \times\{-64-1-16\}=-\frac{9 \times 81}{9 \times 9 \times 9}=-1$

If $C_{i j}$ is a cofactor of $a_{i j}$ such that $A=\left[a_{i j}\right]$, then we have

$\begin{array}{lll}C_{11}=\frac{8}{9} & C_{12}=\frac{-1}{9} & C_{13}=\frac{-4}{9}\end{array}$

$\begin{array}{clc}C_{21}=\frac{-4}{9} & C_{22}=\frac{-4}{9} & C_{23}=\frac{-7}{9} \\ C_{31}=\frac{-1}{9} & C_{32}=\frac{8}{9} & C_{33}=\frac{-4}{9}\end{array}$

Now,

$\operatorname{adj} A=\left[\begin{array}{lll}8 / 9 & -1 / 9 & -4 / 9\end{array}\right.$

$\begin{array}{lll}-4 / 9 & -4 / 9 & -7 / 9\end{array}$

$-1 / 9 \quad 8 / 9 \quad-4 / 9]^{T}=\left[\begin{array}{lll}8 / 9 & -4 / 9 & -1 / 9\end{array}\right.$

$\begin{array}{lll}-1 / 9 & -4 / 9 & 8 / 9\end{array}$

$\left.\begin{array}{lll}-4 / 9 & -7 / 9 & -4 / 9\end{array}\right]$

$\therefore A^{-1}=\frac{1}{|A|} a d j A=-1 \times \frac{1}{9}\left[\begin{array}{lll}8 & -4 & -1\end{array}\right.$

$\begin{array}{lll}-1 & -4 & 8\end{array}$

$-4 \quad-7 \quad-4]=\frac{1}{9}[-8 \quad 4 \quad 1$

$\begin{array}{lll}1 & 4 & -8\end{array}$

$4 \quad 7 \quad 4]=A^{T} \quad[$ From $(1)]$

$\Rightarrow A^{-1}=A^{T}$