# $\text{For the matrix A = } \begin{bmatrix} 1&1&1 \\ 1& 2&- 3 \\ 2 &-1& 3 \end{bmatrix}$ $\text{Show that } A^{3} - 6A^{2} + 5A + 11I = O. \text{ Hence, find } A^{-1}$

Toolbox:
• (i)A matrix is said to be singular if $|A|$ =0.
• (ii)A matrix is said to be invertible only if $|A|\neq 0$.
• (iii) $A^{-1}=\frac{1}{|A|}adj \;A$
• (iv)The adjoint of a square matrix A=[$a_{ij}]_{n\times n}$ is defined as the transpose of the matrix$[A_{ij}]n\times n$
• where $A_{ij}$ is the cofactor of the element $[a_{ij}].$
Given $A=\begin{bmatrix}1 & 1& 1\\1 & 2& -3\\2 & -1 & 3\end{bmatrix}$
Let us find $A^2=A.A$
Matrix multiplication can be done by multiplying the rows and corresponding columns.
$A^2=A.A=\begin{bmatrix}1 & 1& 1\\1 & 2& -3\\2 & -1 & 3\end{bmatrix}\begin{bmatrix}1 & 1& 1\\1 & 2& -3\\2 & -1 & 3\end{bmatrix}$
$\;\;\;=\begin{bmatrix}1+1+2 & 1+2-1& 1-3+3\\1+2-6 & 1+4+3& 1-6-9\\2-1+6 & 2-2-3 & 2+3+9\end{bmatrix}$
$\;\;\;=\begin{bmatrix}4 & 2& 1\\-3 & 8& -14\\7 & -3 & 14\end{bmatrix}$
$A^3=A^2.A=\begin{bmatrix}4 & 2& 1\\-3 & 8& -14\\7 & -3 & 14\end{bmatrix}\begin{bmatrix}1 & 1& 1\\1 & 2& -3\\2 & -1 & 3\end{bmatrix}$
Multiplying as before
$A^3=\begin{bmatrix}4+2+2 & 4+4-1& 4-6+3\\-3+8-28 & -3+16+14& -3-24-42\\7-3+28 & 7-6-14 & 7+9+42\end{bmatrix}$
$\;\;\;=\begin{bmatrix}8 & 7& 1\\-23 & 27& -69\\32 & -13 & 58\end{bmatrix}$
Let us find $A^3-6A^2+5A+11I$
(i.e)$\begin{bmatrix}8 & 7& 1\\-23 & 27& -69\\32 & -13 & 58\end{bmatrix}-6\begin{bmatrix}4 & 2& 1\\-3 & 8& -14\\7 & -3 & 14\end{bmatrix}+5\begin{bmatrix}1 & 1& 1\\1 & 2& -3\\2 & -1 & 3\end{bmatrix}+11\begin{bmatrix}1 & 0& 0\\0 & 1& 0\\0 & 0 & 1\end{bmatrix}$
=$\begin{bmatrix}8 & 7& 1\\-23 & 27& -69\\32 & -13 & 58\end{bmatrix}-\begin{bmatrix}24 & 12& 6\\-18 & 48& -84\\42 & -18 & 84\end{bmatrix}+\begin{bmatrix}5 & 5& 5\\5 & 10& -15\\10 & -5 & 15\end{bmatrix}+\begin{bmatrix}11 & 0& 0\\0 & 11& 0\\0 & 0 & 11\end{bmatrix}$
On simplifying we get,
$\begin{bmatrix}8-24+5+11 & 7-12+5+0 & 1-6+5+0\\-23+18+5+0 & 27-48+10+11 & -69+84-15+0\\32-42+10+0 & -13+18-5+0 & 58-84+15+11\end{bmatrix}$
$\;\;\;=\begin{bmatrix} 0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}$
Hence $A^3-6A^2+5A+11I$=0
Now let us find $A^{-1}$
Given $A^3-6A^2+5A+11I$=0
Premultiply $A^{-1}$
$(AAA)A^{-1})-6(AA)A^{-1}+5(A)A^{-1}+11IA^{-1}=0$
This can be written as
$AA(AA^{-1})-6A(AA^{-1})+5(AA^{-1})+11A^{-1}=0$
But we know $AA^{-1}=I$
$AA(I)-6A(I)+5I+11IA^{-1}=0.$
$\Rightarrow A^2-6A+5I=-11A^{-1}$
$A^{-1}=\frac{-1}{11}(A^2-6A+5I)$------(1)
Let us evaluate $A^2-6A+5I.$
$\;\;\;=\begin{bmatrix} 4 & 2 & 1\\-3 & 8 & -14\\7 & -3 & 14\end{bmatrix}-6\begin{bmatrix}1 & 1& 1\\1 & 2& -3\\2 & -1 & 3\end{bmatrix}+5\begin{bmatrix}1 & 0& 0\\0 & 1& 0\\0 & 0 & 1\end{bmatrix}$
$\;\;\;=\begin{bmatrix} 4 & 2 & 1\\-3 & 8 & -14\\7 & -3 & 14\end{bmatrix}-\begin{bmatrix}6 & 6& 6\\6 & 12& -18\\12 & -6 & 18\end{bmatrix}+\begin{bmatrix}5 & 0& 0\\0 & 5& 0\\0 & 0 & 5\end{bmatrix}$
$\;\;\;=\begin{bmatrix}4-6+5 & 2-6+0& 1-6+0\\-3-6+0 & 8-12+5& -14+18+0\\7-12+0 & -3+6+0 & 14-18+5\end{bmatrix}$
$\;\;\;=\begin{bmatrix}3 & -4& -5\\-9 & 1& 4\\-5 & 3 & 1\end{bmatrix}$
Now substituting the above matrix for $A^{-1}$(eq(1)) we get,
$A^{-1}=\frac{-1}{11}\begin{bmatrix}3 & -4& -5\\-9 & 1& 4\\-5 & 3 & 1\end{bmatrix}$
$A^{-1}=\frac{1}{11}\begin{bmatrix}-3 & 4& 5\\9 & -1& -4\\5 & -3 & -1\end{bmatrix}$