If $A = \begin{bmatrix} x & 2 & 1 \\ -2 & y & 0 \\ 2 & 0 & -1 \end{bmatrix}$,where $x$ and $y$ are non-zero real numbers,$\text{trace}(A) = 0$,and $\det(A) = -6$,then the minor of the element $1$ (at position $a_{13}$) of $A$ is:

For determinant $A = \begin{vmatrix} 1 & 2 & 13 \\ 3 & 0 & 5 \\ 6 & 7 & 11 \end{vmatrix}$,if $p, q, r$ are co-factors of elements $13, 5$ and $11$ respectively,then $p + 3q + 6r = $ . . . . . . .

Consider the matrices $A=\begin{bmatrix} x & y & 0 \\ -3 & 1 & 2 \\ 1 & -2 & z \end{bmatrix}$ and $B=\begin{bmatrix} 1 & -2 & -2 \\ 2 & 0 & 1 \\ 2 & 1 & 0 \end{bmatrix}$. If the cofactors of the elements $z$,$1$ (in the $3^{rd}$ row,$2^{nd}$ column),and $x$ of $A$ are $9, 4, 3$ respectively,then $AB=$

If in the determinant $\Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$,$A_1, B_1, C_1$ etc. are the co-factors of $a_1, b_1, c_1$ etc.,then which of the following relations is incorrect?

Let $\alpha \beta \neq 0$ and $A = \begin{bmatrix} \beta & \alpha & 3 \\ \alpha & \alpha & \beta \\ -\beta & \alpha & 2\alpha \end{bmatrix}$. If $B = \begin{bmatrix} 3\alpha & -9 & 3\alpha \\ -\alpha & 7 & -2\alpha \\ -2\alpha & 5 & -2\beta \end{bmatrix}$ is the matrix of cofactors of the elements of $A$,then $\operatorname{det}(AB)$ is equal to.

Let ${\Delta _1} = \begin{vmatrix} {a_1} & {b_1} & {c_1} \\ {a_2} & {b_2} & {c_2} \\ {a_3} & {b_3} & {c_3} \end{vmatrix}$ and ${\Delta _2} = \begin{vmatrix} {\alpha _1} & {\beta _1} & {\gamma _1} \\ {\alpha _2} & {\beta _2} & {\gamma _2} \\ {\alpha _3} & {\beta _3} & {\gamma _3} \end{vmatrix}$. Then ${\Delta _1} \times {\Delta _2}$ can be expressed as the sum of how many determinants?