For matrix $A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix}$,$(A^{-1})^2 = $ . . . . . .

  • A
    $\begin{bmatrix} -\frac{1}{4} & 0 & 0 \\ 0 & -\frac{1}{9} & 0 \\ 0 & 0 & -\frac{1}{16} \end{bmatrix}$
  • B
    $\begin{bmatrix} -4 & 0 & 0 \\ 0 & -9 & 0 \\ 0 & 0 & -16 \end{bmatrix}$
  • C
    $\begin{bmatrix} \frac{1}{4} & 0 & 0 \\ 0 & \frac{1}{9} & 0 \\ 0 & 0 & \frac{1}{16} \end{bmatrix}$
  • D
    $\begin{bmatrix} 4 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 16 \end{bmatrix}$

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Obtain the inverse of the following matrix using elementary row operations: $A = \left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]$

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If $A=\begin{bmatrix} 1 & 2 & 2 \\ 3 & 2 & 3 \\ 1 & 1 & 2 \end{bmatrix}$ and $A^{-1}=\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$, then $\sum_{1 \leq i, j \leq 3} a_{ij} =$

Assertion $(A)$: If $B$ is a $3 \times 3$ matrix and $|B|=6$,then $|\operatorname{Adj}(B)|=36$.
Reason $(R)$: If $B$ is a square matrix of order $n$,then $|\operatorname{Adj}(B)|=|B|^{n}$.

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