If $A = [a_{ij}]_{3 \times 3} = \begin{bmatrix} 3 & 2 & 4 \\ 1 & 4 & 1 \\ 2 & 6 & 3 \end{bmatrix}$ and $A_{ij}$ is the cofactor of $a_{ij}$,then the value of $a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23}$ is equal to:

If the value of a third order determinant is $16$,then the value of the determinant formed by replacing each of its elements by its cofactor is

If the value of a third-order determinant is $11$,then the value of the square of the determinant formed by the cofactors will be:

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?

If $A = [a_{ij}]_{3 \times 3} = \begin{bmatrix} 1 & 3 & 3 \\ -1 & 2 & 2 \\ 1 & 1 & 4 \end{bmatrix}$ and $A_{ij}$ is the cofactor of $a_{ij}$,then the value of $a_{31}A_{31} + a_{32}A_{32} + a_{33}A_{33}$ is equal to

In the matrix $\begin{bmatrix} -1 & x & 3 \\ -4 & -5 & -6 \\ -7 & y & 9 \end{bmatrix}$, if the cofactors of $-6$ and $-7$ are respectively $22$ and $27$, then $5x + y = $