If $M$ is any square matrix of order $3$ over $\mathbb{R}$ and if $M^{\prime}$ is the transpose of $M$, then $\text{adj}(M^{\prime}) - (\text{adj } M)^{\prime}$ is equal to

Let $I$ be a unit matrix of order $6$. Let $A = (a_{ij})$ be a square matrix of order $6$ such that $a_{ij} = \begin{cases} 1, & \text{if } i+j=7 \\ 0, & \text{if } i+j \neq 7 \end{cases}$. Then $(A(\text{adj } A) A^{-1}) A^2 = $

If the adjoint of a $3 \times 3$ matrix $P$ is $\begin{bmatrix} 1 & 4 & 4 \\ 2 & 1 & 7 \\ 1 & 1 & 3 \end{bmatrix}$,then the possible value$(s)$ of the determinant of $P$ is (are):

Matrix $A = \begin{bmatrix} x & 3 & 2 \\ 1 & y & 4 \\ 2 & 2 & z \end{bmatrix}$. If $xyz = 60$ and $8x + 4y + 3z = 20$,then $A (adj A)$ is equal to:

Let $A = \begin{bmatrix} 1 & 2 \\ -5 & 1 \end{bmatrix}$ and $A^{-1} = xA + yI_2$,(where $I_2$ is the unit matrix of order $2$),then

If the inverse of $\begin{bmatrix} 1 & 2 & x \\ 4 & -1 & 7 \\ 2 & 4 & -6 \end{bmatrix}$ does not exist,then $x=$