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?

  • A
    $9$
  • B
    $3$
  • C
    $27$
  • D
    $2$

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Similar Questions

Using cofactors of elements of the third column,evaluate $\Delta = \left| \begin{array}{ccc} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{array} \right|$.

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 ${A_1}, {B_1}, {C_1}, \dots$ are respectively the co-factors of the elements ${a_1}, {b_1}, {c_1}, \dots$ of the determinant $\Delta = \begin{vmatrix} {a_1} & {b_1} & {c_1} \\ {a_2} & {b_2} & {c_2} \\ {a_3} & {b_3} & {c_3} \end{vmatrix}$,then $\begin{vmatrix} {B_2} & {C_2} \\ {B_3} & {C_3} \end{vmatrix} = $

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} = $

The sum of the products of the elements of any row of a determinant $A$ with the corresponding co-factors is always equal to

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