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(C) Given the matrix $A = [a_{ij}]_{3 	imes 3}$ where $a_{ij} = 2i + j$. Calculating the elements of $A$: $a_{11} = 2(1) + 1 = 3$,$a_{12} = 2(1) + 2

Vedclass · Accepted Answer

$4$

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(C) Given the matrix $A = [a_{ij}]_{3 	imes 3}$ where $a_{ij} = 2i + j$. Calculating the elements of $A$: $a_{11} = 2(1) + 1 = 3$,$a_{12} = 2(1) + 2 = 4$,$a_{13} = 2(1) + 3 = 5$ $a_{21} = 2(2) + 1 = 5$,$a_{22} = 2(2) + 2 = 6$,$a_{23} = 2(2) + 3 = 7$ $a_{31} = 2(3) + 1 = 7$,$a_{32} = 2(3) + 2 = 8$,$a_{33} = 2(3) + 3 = 9$ So,$A = \begin{bmatrix} 3 & 4 & 5 \ 5 & 6 & 7 \ 7 & 8 & 9 \end{bmatrix}$. The adjoint of a matrix $A$,denoted by $adj(A)$,is the transpose of the cofactor matrix $C = [C_{ij}]_{3 	imes 3}$. The element in the third column of the second row of $adj(A)$ is the cofactor $C_{32}$ of the element $a_{32}$ in matrix $A$. $C_{32} = (-1)^{3+2} 	imes M_{32}$,where $M_{32}$ is the minor of $a_{32}$. $M_{32} = \begin{vmatrix} 3 & 5 \ 5 & 7 \end{vmatrix} = (3 	imes 7) - (5 	imes 5) = 21 - 25 = -4$. $C_{32} = (-1)^5 	imes (-4) = -1 	imes (-4) = 4$. Thus,the required element is $4$.

The third element in the second row of the adjoint of a matrix $A = [a_{ij}]_{3 \times 3}$,where $a_{ij} = 2i + j$,is:

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