Let $A$ and $B$ be two $3 \times 3$ matrices such that $AB = I$ and $|A| = \frac{1}{8}$. Then $|\operatorname{adj}(B \operatorname{adj}(2A))|$ is equal to:

  • A
    $16$
  • B
    $32$
  • C
    $64$
  • D
    $128$

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The number of $3 \times 2$ matrices $A$, which can be formed using the elements of the set $\{-2, -1, 0, 1, 2\}$ such that the sum of all the diagonal elements of $A^{T}A$ is $5$, is . . . . . . .

Let $M = \begin{bmatrix} 0 & 1 & a \\ 1 & 2 & 3 \\ 3 & b & 1 \end{bmatrix}$ and $\operatorname{adj} M = \begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \end{bmatrix}$ where $a$ and $b$ are real numbers. Which of the following options is/are correct?
$(1)$ $a+b=3$
$(2)$ $\operatorname{det}(\operatorname{adj} M^2) = 81$
$(3)$ $(\operatorname{adj} M)^{-1} + \operatorname{adj} M^{-1} = -M$
$(4)$ If $M \begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$,then $\alpha - \beta + \gamma = 3$

Let $a, b, c$ be non-real numbers satisfying the equation $x^5 = 1$ and $S$ be the set of all non-invertible matrices of the form $\begin{bmatrix} 1 & a & b \\ w & 1 & c \\ w^2 & w & 1 \end{bmatrix}$,where $w = e^{\frac{i 2\pi}{5}}$. Then the number of distinct matrices in the set $S$ is:

The total number of matrices $A = \begin{bmatrix} 0 & 2x & 2x \\ 2y & y & -y \\ 1 & -1 & 1 \end{bmatrix}$ where $x, y \in \mathbb{R}$ and $x \neq y$,for which $A^T A = 3I_3$ is:

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