Question

If $A=\left[\begin{array}{ccc}9& 0& 0\\ 0& 10& 0\\ 0& 0& 8\end{array}\right]$ ,then $A^{-1}$:

• A. $\left[\begin{array}{ccc}\frac{1}{9}& 0& 0\\ 0& 10& 0\\ 0& 0& \frac{1}{8}\end{array}\right]$
• B. $\left[\begin{array}{ccc}9& 0& 0\\ 0& \frac{1}{10}& 0\\ 0& 0& 8\end{array}\right]$
• C. $\left[\begin{array}{ccc}9& 0& 0\\ 0& 10& 0\\ 0& 0& \frac{1}{8}\end{array}\right]$
• D. $\left[\begin{array}{ccc}\frac{1}{9}& 0& 0\\ 0& \frac{1}{10}& 0\\ 0& 0& \frac{1}{8}\end{array}\right]$

Answer 1

The correct option is D $\left[\begin{array}{ccc}\frac{1}{9}& 0& 0\\ 0& \frac{1}{10}& 0\\ 0& 0& \frac{1}{8}\end{array}\right]$

$A=\left[\begin{array}{ccc}9& 0& 0\\ 0& 10& 0\\ 0& 0& 8\end{array}\right]$
$|A|=720$
$adjA=C^T={\left[\begin{array}{ccc}80& 0& 0\\ 0& 72& 0\\ 0& 0& 90\end{array}\right]}^T$
$\Rightarrow adjA=\left[\begin{array}{ccc}80& 0& 0\\ 0& 72& 0\\ 0& 0& 90\end{array}\right]$
Hence, $A^{-1}=\frac{1}{720}\left[\begin{array}{ccc}80& 0& 0\\ 0& 72& 0\\ 0& 0& 90\end{array}\right]$
$\Rightarrow A^{-1}=\left[\begin{array}{ccc}1/9& 0& 0\\ 0& 1/10& 0\\ 0& 0& 1/8\end{array}\right]$