Question

If $A=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 1\\ 0& 1& 0\end{array}\right],\text{then}{A}^2-2A=$

• A. $A^{-1}$
• B. $-A^{-1}$
• C. $I$
• D. $2I$

Answer 1

The correct option is B $-A^{-1}$

Given $A=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 1\\ 0& 1& 0\end{array}\right]$
$A^{-1}=\frac{\text{adj}A}{|A|}$
To find $\text{adj}A,$
The cofactor matrix is $\left[\begin{array}{ccc}-1& 0& 0\\ 1& 0& -1\\ -1& -1& 1\end{array}\right]$
The transpose of the cofactor matrix is $\left[\begin{array}{ccc}-1& 1& -1\\ 0& 0& -1\\ 0& -1& 1\end{array}\right]=\text{adj}A$
$|A|=-1$
$\Rightarrow A^{-1}=\left[\begin{array}{ccc}1& -1& 1\\ 0& 0& 1\\ 0& 1& -1\end{array}\right]$

$A^2=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 1\\ 0& 1& 0\end{array}\right]\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 1\\ 0& 1& 0\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 0& 2& 1\\ 0& 1& 1\end{array}\right]$

$\Rightarrow 2A=\left[\begin{array}{ccc}2& 0& 2\\ 0& 2& 2\\ 0& 2& 0\end{array}\right]\Rightarrow A^2-2A=-A^{-1}$