If $A=\left[a_{i j}\right]_{3 \times 3}=\left[\begin{array}{ccc}1 & 3 & 3 \\ -1 & 2 & 2 \\ 1 & 1 &…

If $A=\left[a_{i j}\right]_{3 \times 3}=\left[\begin{array}{ccc}1 & 3 & 3 \\ -1 & 2 & 2 \\ 1 & 1 & 4\end{array}\right]$ and $A_{i j}$ is a cofactor of $a_{i j}$ then the value of $a_{31} A_{31}+a_{32} A_{32}+a_{33} A_{33}$ is equal to
  1. 5
  2. 15
  3. 20
  4. 0

Solution

$\begin{aligned} & a_{31} A_{31}+a_{32} A_{32}+a_{33} A_{33}=|A| \\ & =1 \times(8-2)-3(-4-2)+3(-1-2)=6+18-9=15\end{aligned}$

Asked in: MHT CET 2022 (10 Aug Shift 2)

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