Find minors and cofactors of the elements of the determinant $\begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix}$ and find $a_{11} A_{31} + a_{12} A_{32} + a_{13} A_{33}$.

0

The correct answer is Option (2) → 0 ##

We have $M_{11} = \begin{vmatrix} 0 & 4 \\ 5 & -7 \end{vmatrix} = 0 - 20 = -20$; $\quad A_{11} = (-1)^{1+1} (-20) = -20$

$M_{12} = \begin{vmatrix} 6 & 4 \\ 1 & -7 \end{vmatrix} = -42 - 4 = -46; \quad A_{12} = (-1)^{1+2} (-46) = 46$

$M_{13} = \begin{vmatrix} 6 & 0 \\ 1 & 5 \end{vmatrix} = 30 - 0 = 30; \quad A_{13} = (-1)^{1+3} (30) = 30$

$M_{21} = \begin{vmatrix} -3 & 5 \\ 5 & -7 \end{vmatrix} = 21 - 25 = -4; \quad A_{21} = (-1)^{2+1} (-4) = 4$

$M_{22} = \begin{vmatrix} 2 & 5 \\ 1 & -7 \end{vmatrix} = -14 - 5 = -19; \quad A_{22} = (-1)^{2+2} (-19) = -19$

$M_{23} = \begin{vmatrix} 2 & -3 \\ 1 & 5 \end{vmatrix} = 10 + 3 = 13; \quad A_{23} = (-1)^{2+3} (13) = -13$

$M_{31} = \begin{vmatrix} -3 & 5 \\ 0 & 4 \end{vmatrix} = -12 - 0 = -12; \quad A_{31} = (-1)^{3+1} (-12) = -12$

$M_{32} = \begin{vmatrix} 2 & 5 \\ 6 & 4 \end{vmatrix} = 8 - 30 = -22; \quad A_{32} = (-1)^{3+2} (-22) = 22$

and $M_{33} = \begin{vmatrix} 2 & -3 \\ 6 & 0 \end{vmatrix} = 0 + 18 = 18; \quad A_{33} = (-1)^{3+3} (18) = 18$

Now $a_{11} = 2, a_{12} = -3, a_{13} = 5; \quad A_{31} = -12, A_{32} = 22, A_{33} = 18$

So $a_{11} A_{31} + a_{12} A_{32} + a_{13} A_{33}$

$= 2 (-12) + (-3) (22) + 5 (18) = -24 - 66 + 90 = 0$