# $\text{If A = } \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}, \text{show that } A^{2} -5A +7I = 0. \text{ Hence find } A^{-1}$

Toolbox:
• A matrix is said to be singular if $|A|$ =0.
• A matrix is said to be invertible only if $|A|\neq 0$.
•  $A^{-1}=\frac{1}{|A|}adj \;A$
• 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 $[a_{ij}].$
Let $A=\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}$

Let us find $A^2=A.A$

Matrix multiplication can be done by multiplying the rows and corresponding columns.

$A^2=A.A=\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}$

$\;\;\;=\begin{bmatrix}9-1 & 3+2\\-3-2 & -1+4\end{bmatrix}=\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}$

5A can be obtained by multiplying each element by 5

$5A=\begin{bmatrix}5\times3 & 5\times1\\5\times(-1) & 5\times 2\end{bmatrix}=\begin{bmatrix}15 & 5\\-5 & 10\end{bmatrix}$

We know$I=\begin{bmatrix}1& 0\\0 &1\end{bmatrix}$

$7I=7\begin{bmatrix}1& 0\\0 &1\end{bmatrix}=\begin{bmatrix}7& 0\\0 &7\end{bmatrix}$

Now $A^2-5A+7I=\begin{bmatrix}8& 5\\-5 &3\end{bmatrix}-\begin{bmatrix}15& 5\\-5 &10\end{bmatrix}+\begin{bmatrix}7& 0\\0 &7\end{bmatrix}$

$\;\;\;=\begin{bmatrix}8-15+7& 5-5+0\\-5-5+0 &3-10+7\end{bmatrix}=\begin{bmatrix}0& 0\\0 &0\end{bmatrix}$

Hence $A^2-5A+7I=0.$

now $A^2=5A-7I$ or $A^2-5A=-7I.$

A.A-5A=-7I.

Multiply $A^{-1}$ on both sides,

$A.A(A^{-1})-5A.A^{-1}=-7IA^{-1}$

$A(AA^{-1})-5(AA^{-1})=-7IA^{-1}$

Because $AA^{-1}=I$ and $IA^{-1}=A^{-1}$

We get,

$AI-5I=-7A^{-1}$

$A^{-1}=\frac{-1}{7}(AI-5I)$

$A^{-1}=\frac{1}{7}(5I-AI)$

Now substituting for I and A we get,

$A^{-1}=\frac{1}{7}\bigg(5\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}-\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}\bigg)$

$A^{-1}=\frac{1}{7}\bigg(\begin{bmatrix}5 & 0\\0 & 5\end{bmatrix}-\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}\bigg)$

$\;\;\;=\frac{1}{7}\begin{bmatrix}2 & -1\\1 & 3\end{bmatrix}$

Hence $A^{-1}=\frac{1}{7}\begin{bmatrix}2 & -1\\1 & 3\end{bmatrix}$

edited Feb 28, 2013