Question:
Let $A=\left[a_{i j}\right]$ be a $3 \times 3$ matrix such that $|A|=5 .$ If $C_{i j}=$ Cofactor of $a_{i j}$ in $A .$ Then $a_{11} C_{11}+a_{12} C_{12}+a_{13} C_{13}=$__________
Solution:
Given:
|A| = 5
As we know,
Sum of products of elements of row (or column) with their corresponding cofactors = Value of the determinant
and
Sum of products of elements of row (or column) with the cofactors of any other row (or column) = 0
Thus, $a_{11} C_{11}+a_{12} C_{12}+a_{13} C_{13}=|A|=5$
Hence, $a_{11} C_{11}+a_{12} C_{12}+a_{13} C_{13}=\underline{5}$.