# Show that

Question:

Show that $A=\left[\begin{array}{cc}-8 & 5 \\ 2 & 4\end{array}\right]$ satisfies the equation $A^{2}+4 A-42 I=O$. Hence, find $A^{-1}$.

Solution:

$A=\left[\begin{array}{cc}-8 & 5 \\ 2 & 4\end{array}\right]$

$\therefore A^{2}=\left[\begin{array}{cc}74 & -20 \\ -8 & 26\end{array}\right]$

and

$A^{2}+4 A-42 I=\left[\begin{array}{cc}74 & -20 \\ -8 & 26\end{array}\right]+\left[\begin{array}{cc}-32 & 20 \\ 8 & 16\end{array}\right]-\left[\begin{array}{cc}42 & 0 \\ 0 & 42\end{array}\right]$

$\Rightarrow A^{2}+4 A-42 I=\left[\begin{array}{cc}74-32-42 & -20+20-0 \\ -8+8-0 & 26+16-42\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=O$

Now,

$A^{2}+4 A-42 I=0$

$\Rightarrow A^{2}+4 A=42 I$

$\Rightarrow A^{-1} A^{2}+4 A^{-1} A=42 I A^{-1}$       [Pre-multiplying both sides by $A^{-1}$ ]

$\Rightarrow A+4 I=42 A^{-1}$

$\Rightarrow A^{-1}=\frac{1}{42}(A+4 I)$

$\Rightarrow A^{-1}=\frac{1}{42}\left\{\left[\begin{array}{cc}-8 & 5 \\ 2 & 4\end{array}\right]+\left[\begin{array}{ll}4 & 0 \\ 0 & 4\end{array}\right]\right\}=\frac{1}{42}\left[\begin{array}{cc}-4 & 5 \\ 2 & 8\end{array}\right]$