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# If $A_{ij}$ is the cofactor of the element $a_{ij}$ of the determinant $\begin{vmatrix}2 & -3 & 5\\6 & 0 & 4\\1 & 5 & -7\end{vmatrix}$,then write the value of $a_{32}.A_{32}$.

This question appeared in 65-1,65-2 and 65-3 versions of the paper in 2013.

Toolbox:
• If $A$ is a square matrix, then the minor of its entry $a_{ij}$, is denoted by $M_{ij}$ and is defined to be the determinant of the submatrix obtained by removing from $A$ its $i$-th row and $j$-th column.
• Given a square matrix, the cofactor of $a_{ij}$ is denoted by $A_{ij} =(-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor of the entry $a_{ij}$.
Given $\begin{vmatrix}2 & -3 & 5\\6 & 0 & 4\\1 & 5 & -7\end{vmatrix}$
If $A$ is a square matrix, then the minor of its entry $a_{ij}$, is denoted by $M_{ij}$ and is defined to be the determinant of the submatrix obtained by removing from $A$ its $i$-th row and $j$-th column.
Therefore, $M_{32}=\begin{vmatrix}2 & 5\\6 & 4\end{vmatrix}$
Given a square matrix, the cofactor of $a_{ij}$ is denoted by $A_{ij} =(-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor of the entry $a_{ij}$.
Therefore, $A_{32}=(-1)^{3+2}\begin{vmatrix}2 & 5\\6 & 4\end{vmatrix}=-(8-30)=22.$
Since, $a_{32}=5$, $a_{32}A_{32}=5\times 22=110.$
edited Mar 21, 2013