Let $A = [a_{ij}] = \begin{bmatrix} \log_5 128 & \log_4 5 \\ \log_5 8 & \log_4 25 \end{bmatrix}$. If $A_{ij}$ is the cofactor of $a_{ij}$,$C_{ij} = \sum_{k=1}^2 a_{ik} A_{jk}$,$1 \leq i, j \leq 2$,and $C = [C_{ij}]$,then $8|C|$ is equal to:

The minors of $-4$ and $9$ and the co-factors of $-4$ and $9$ in the determinant $\left| \begin{array}{ccc} -1 & -2 & 3 \\ -4 & -5 & -6 \\ -7 & 8 & 9 \end{array} \right|$ are respectively:

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 $A = [a_{ij}]_{3 \times 3} = \begin{bmatrix} 3 & 2 & 4 \\ 1 & 4 & 1 \\ 2 & 6 & 3 \end{bmatrix}$ 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:

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 $A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 0 & 3 & -5 \end{bmatrix}$,where $A_{ij}$ is the cofactor of the element $a_{ij}$ of matrix $A$,then $a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23} = $