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

• A

  $A^{-1}$
• Correct answer
  B

  $-A^{-1}$
• C

  $I$
• D

  $-I$

Solution

Given, $\quad A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0\end{array}\right]$ $\therefore A^2-2 A=\left[\begin{array}{lll} 1 & 1 & 1 \\ 0 & 2 & 1 \\ 0 & 1 & 1 \end{array}\right]-\left[\begin{array}{lll} 2 & 0 & 2 \\ 0 & 2 & 2 \\ 0 & 2 & 0 \end{array}\right]$
$\begin{aligned} \text { Now, }|A| & =1(0-1)-0+1(0)=-1 \\ C_{11} & =-1, C_{12}=0, C_{13}=0 \\ C_{21} & =-(-1)=1, C_{22}=0, C_{23}=-1 \\ C_{31} & =-1, C_{32}=-(1), C_{33}=1 \\ \therefore \quad A^{-1} & =-1\left[\begin{array}{rrr} -1 & 1 & -1 \\ 0 & 0 & -1 \\ 0 & -1 & 1 \end{array}\right] \end{aligned}$ $\therefore$ From Eq. (i) $A^2-2 A=-A^{-1}$

Hence, the correct answer is option: 'B'.