(C) We know that the sum of the product of elements of any row (or column) with their corresponding cofactors is equal to the determinant of the matri

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(C) We know that the sum of the product of elements of any row (or column) with their corresponding cofactors is equal to the determinant of the matrix,i.e.,$a_{i1} A_{i1} + a_{i2} A_{i2} + a_{i3} A_{i3} = |A|$.Here,$a_{31} A_{31} + a_{32} A_{32} + a_{33} A_{33} = |A|$.Now,we calculate the determinant of $A$ by expanding along the first row:$|A| = 1(1 \times 7 - 5 \times 4) - 2(1 \times 7 - 5 \times 2) + 3(1 \times 4 - 1 \times 2)$$|A| = 1(7 - 20) - 2(7 - 10) + 3(4 - 2)$$|A| = 1(-13) - 2(-3) + 3(2)$$|A| = -13 + 6 + 6$$|A| = -1$Therefore,the value is $-1$.

Class 12 Mathematics 3 and 4 .Determinants and Matrices Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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If matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7 \end{bmatrix}$,then the value of $a_{31} A_{31} + a_{32} A_{32} + a_{33} A_{33}$ is

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A
$1$
B
$13$
C
$-1$
D
$-13$

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If matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7 \end{bmatrix}$,then the value of $a_{31} A_{31} + a_{32} A_{32} + a_{33} A_{33}$ is

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