Show that satisfies the equation

Solution:

Given,

$A=\left[\begin{array}{cc}5 & 3 \\ -1 & -2\end{array}\right]$

So, $A^{2}=A \cdot A=\left[\begin{array}{cc}5 & 3 \\ -1 & -2\end{array}\right] \cdot\left[\begin{array}{cc}5 & 3 \\ -1 & -2\end{array}\right]=\left[\begin{array}{cc}25-3 & 15-6 \\ -5+2 & -3+4\end{array}\right]=\left[\begin{array}{cc}22 & 9 \\ -3 & 1\end{array}\right]$

$3 A=3\left[\begin{array}{cc}5 & 3 \\ -1 & -2\end{array}\right]=\left[\begin{array}{cc}15 & 9 \\ -3 & -6\end{array}\right]$

And, $7 I=7\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}7 & 0 \\ 0 & 7\end{array}\right]$

Hence, $A^{2}-3 A-7 I=\left[\begin{array}{cc}22 & 9 \\ -3 & 1\end{array}\right]-\left[\begin{array}{cc}15 & 9 \\ -3 & -6\end{array}\right]-\left[\begin{array}{ll}7 & 0 \\ 0 & 7\end{array}\right]$

$=\left[\begin{array}{cc}22-15-7 & 9-9-0 \\ -3+3-0 & 1+6-7\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=O$

Now,

A2 – 3A – 7I = 0

Multiplying both sides with A-1, we get

A-1 [A2 – 3A – 7I] = A-1 0

A-1. A . A – 3A-1. A – 7A-1. I = 0

I . A – 3I – 7A-1 = 0 [As A-1. A = I]

A – 3I – 7A-1 = 0

7A-1 = A – 3I

$7 A^{-1}=\left[\begin{array}{cc}5 & 3 \\ -1 & -2\end{array}\right]-\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]=\left[\begin{array}{cc}2 & 3 \\ -1 & -5\end{array}\right]$

Therefore, $A^{-1}=\frac{1}{7}\left[\begin{array}{cc}2 & 3 \\ -1 & -5\end{array}\right]$