The cofactor of the element $4$ in the determinant $\left| \begin{array}{cccc} 1 & 3 & 5 & 1 \\ 2 & 3 & 4 & 2 \\ 8 & 0 & 1 & 1 \\ 0 & 2 & 1 & 1 \end{array} \right|$ is

Write the minors and cofactors of the elements of the following determinant: $\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|$

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:

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 $A = \begin{bmatrix} 5 & 6 & 3 \\ -4 & 3 & 2 \\ -4 & -7 & 3 \end{bmatrix}$,then the cofactors of all elements of the second row are respectively:

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?