For determinant A = begin{vmatrix} 1 & 2 & 13 | vedclass

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(C) The determinant is given by $A = \begin{vmatrix} 1 & 2 & 13 \ 3 & 0 & 5 \ 6 & 7 & 11 \end{vmatrix}$.The co-factor of an element $a_{ij}$ is give

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$0$

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(C) The determinant is given by $A = \begin{vmatrix} 1 & 2 & 13 \ 3 & 0 & 5 \ 6 & 7 & 11 \end{vmatrix}$.The co-factor of an element $a_{ij}$ is given by $C_{ij} = (-1)^{i+j} M_{ij}$,where $M_{ij}$ is the minor.$1$. For element $13$ $(a_{13})$: $p = C_{13} = (-1)^{1+3} \begin{vmatrix} 3 & 0 \ 6 & 7 \end{vmatrix} = (1)(21 - 0) = 21$.$2$. For element $5$ $(a_{23})$: $q = C_{23} = (-1)^{2+3} \begin{vmatrix} 1 & 2 \ 6 & 7 \end{vmatrix} = (-1)(7 - 12) = (-1)(-5) = 5$.$3$. For element $11$ $(a_{33})$: $r = C_{33} = (-1)^{3+3} \begin{vmatrix} 1 & 2 \ 3 & 0 \end{vmatrix} = (1)(0 - 6) = -6$.Now,calculate $p + 3q + 6r$:$p + 3q + 6r = 21 + 3(5) + 6(-6)$$= 21 + 15 - 36$$= 36 - 36 = 0$.Alternatively,by the property of determinants,the sum of products of elements of a row (or column) with their corresponding co-factors is the value of the determinant,but here we are multiplying elements of the first column $(1, 3, 6)$ with co-factors of the third column $(p, q, r)$. This sum represents the expansion of the determinant along the third column,which is $0$ because the elements of the third column are replaced by the first column in this specific linear combination.

For determinant $A = \begin{vmatrix} 1 & 2 & 13 \\ 3 & 0 & 5 \\ 6 & 7 & 11 \end{vmatrix}$,if $p, q, r$ are co-factors of elements $13, 5$ and $11$ respectively,then $p + 3q + 6r = $ . . . . . . .

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