# Show that

Question:

Show that $A=\left[\begin{array}{cc}5 & 3 \\ -1 & -2\end{array}\right]$ satisfies the equation $x^{2}-3 x-7=0$. Thus, find $A^{-1}$.

Solution:

$A=\left[\begin{array}{ll}5 & 3\end{array}\right.$

$-1-2]$

$A^{2}=\left[\begin{array}{ll}22 & 9\end{array}\right.$

$\left.\begin{array}{ll}-3 & 1\end{array}\right]$

If $I_{2}$ is the identity matrix of order 2, then

$A^{2}-3 A-7 I_{2}=\left[\begin{array}{ll}22 & 9\end{array}\right.$

$-3 \quad 1]-3\left[\begin{array}{ll}5 & 3\end{array}\right.$

$-1-2]-7\left[\begin{array}{ll}1 & 0\end{array}\right.$

$\left.\begin{array}{ll}0 & 1\end{array}\right]$

$\Rightarrow A^{2}-3 A-7 I_{2}=\left[\begin{array}{ll}22-15-7 & 9-9-0\end{array}\right.$

$-3+3+0 \quad 1+6-7]=\left[\begin{array}{ll}0 & 0\end{array}\right.$

$0 \quad 0]=0$

$\Rightarrow A^{2}-3 A-7 I_{2}=0$

Thus, $A$ satisfies $x^{2}-3 x-7=0$.

Now,

$A^{2}-3 A-7 I_{2}=0$

$\Rightarrow A^{2}-3 A=7 I_{2}$

$\Rightarrow A^{-1}\left(A^{2}-3 A\right)=A^{-1} \times 7 I_{2} \quad\left[\right.$ Pre $-$ multiplying both sides by $\left.A^{-1}\right]$

$\Rightarrow A-3 I_{2}=7 A^{-1}$

$\Rightarrow\left[\begin{array}{ll}5 & 3\end{array}\right.$

$-1-2]-3\left[\begin{array}{ll}1 & 0\end{array}\right.$

$0 \quad 1]=7 A^{-1}$

$\Rightarrow A^{-1}=\frac{1}{7}[5-3 \quad 3-0$

$-1-0 \quad-2-3]=\frac{1}{7}\left[\begin{array}{ll}2 & 3\end{array}\right.$

$-1 \quad-5]$