If $A=\left[\begin{array}{ccc}1 & 1 & 1 \ 2 & 1 & -3 \ -1 & 2 & 3\end{array} ight]$, then $a_{11} A_{31}+a_{12} A_{32}+a_{13} A_{33}=$ where $A_{i j}$, where $A=\left[a_{i j} ight]_{3 imes 3}$

Question

If $A=\left[\begin{array}{ccc}1 & 1 & 1 \ 2 & 1 & -3 \ -1 & 2 & 3\end{array}ight]$, then $a_{11} A_{31}+a_{12} A_{32}+a_{13} A_{33}=$ where $A_{i j}$, where $A=\left[a_{i j}ight]_{3 	imes 3}$

Aicharya · Accepted Answer

We know that sum of the product of elements of one row of a determinant and cofactor of corresponding element of any other row is zero. $\begin{aligned} & \Rightarrow a_{11} A_{31}+a_{12} A_{32}+a_{13} A_{33}=0 \ & \Rightarrow 1 \cdot A_{31}+1 \cdot A_{32}+1 \cdot A_{33}=0 \Rightarrow A_{31}+A_{32}+A_{33}=0\end{aligned}$

If $A=\left[\begin{array}{ccc}1 & 1 & 1 \\ 2 & 1 & -3 \\ -1 & 2 & 3\end{array}\right]$, then $a_{11}…

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