Browse Questions

# If $A = \begin{bmatrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix},$ show that $A^2-7A+10I=0.$ Hence , find $A^{-1}$

Toolbox:
• Matrix multiplication is possible if the two matrices have same number of columns.
• The rows of first matrix can be multiplied by the column of the second matrix.
• $A^{-1}=\frac{1}{|A|}$adjoint of A
Step 1:
$A^2=\begin{bmatrix}3 & 2 & 0\\1 & 4 & 0\\0 & 0 & 5\end{bmatrix}\begin{bmatrix}3 & 2 & 0\\1 & 4 & 0\\0 & 0 & 5\end{bmatrix}$
$\quad=\begin{bmatrix}9+2+0 & 6+8+0 & 0+0+0\\3+4+0 & 2+16+0 & 0+0+0\\0+0+0 & 0+0+0 & 0+0+25\end{bmatrix}$
$\quad=\begin{bmatrix}-11 & 14 & 0\\7 & 18 & 0\\0 & 0 & 25\end{bmatrix}$
Step 2:
$7A=7\begin{bmatrix}3 & 2 & 0\\1 & 4 & 0\\0 & 0 & 5\end{bmatrix}=\begin{bmatrix}21 & 14 & 0\\7 & 28 & 0\\0 & 0 & 35\end{bmatrix}$
Step 3:
$10I=10\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}=\begin{bmatrix}10 & 0 & 0\\0 & 10 & 0\\0 & 0 & 10\end{bmatrix}$
Step 4:
$A^2-7A+10I=\begin{bmatrix}-11 & 14 & 0\\7 & 18 & 0\\0 & 0 & 25\end{bmatrix}-\begin{bmatrix}21 & 14 & 0\\7 & 28 & 0\\0 & 0 & 35\end{bmatrix}+\begin{bmatrix}10 & 0 & 0\\0 & 10 & 0\\0 & 0 & 10\end{bmatrix}$
$\qquad\qquad\qquad=\begin{bmatrix}11-21+10 & 14-14+0 & 0-0+0\\7-7+0 & 18-28+10 & 0-0+0\\0-0+0 & 0-0+0 & 25-35+10\end{bmatrix}=\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}$
$A^2-7A+10I=0$
$\Rightarrow A^2-7A=-10I$
Premultiplying by $A^{-1}$ on both sides we get,
$-A^2.A^{-1}+7A.A^{-1}=10IA^{-1}$
This can be written as
$-A(AA^{-1})+7I=10A^{-1}$ $\qquad[ AA^{-1}=I$]
Step 5:
$A^{-1}=\frac{1}{10}(-AI+7I)$
$\;\;=\frac{1}{10}\bigg(-\begin{bmatrix}3 & 2 & 0\\1 & 4 & 0\\0 & 0 & 5\end{bmatrix}\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}+7\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}\bigg)$
$A^{-1}=\frac{1}{10}\begin{bmatrix}-3+0+0 & -0-2+0 & 0+0+0\\-1+0+0 & 0-4+0 & 0+0+0\\0+0+0 & 0+0-5 & 0+0-5\end{bmatrix}$
$A^{-1}=\frac{1}{10}\begin{bmatrix}-3 & -2 & 0\\-1 & -4 & 0\\0 & -5 & -5\end{bmatrix}$