If A = [a_{ij}]_{3 times 3} is a matrix such | vedclass

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(D) Step $1$: Construct the matrix $A$ using $a_{ij} = |2i - 5j|$ for $i, j \in \{1, 2, 3\}$.$a_{11} = |2(1) - 5(1)| = |-3| = 3$$a_{12} = |2(1) - 5(2)

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$-1$

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(D) Step $1$: Construct the matrix $A$ using $a_{ij} = |2i - 5j|$ for $i, j \in \{1, 2, 3\}$.$a_{11} = |2(1) - 5(1)| = |-3| = 3$$a_{12} = |2(1) - 5(2)| = |-8| = 8$$a_{13} = |2(1) - 5(3)| = |-13| = 13$$a_{21} = |2(2) - 5(1)| = |-1| = 1$$a_{22} = |2(2) - 5(2)| = |-6| = 6$$a_{23} = |2(2) - 5(3)| = |-11| = 11$$a_{31} = |2(3) - 5(1)| = |1| = 1$$a_{32} = |2(3) - 5(2)| = |-4| = 4$$a_{33} = |2(3) - 5(3)| = |-9| = 9$So, $A = \begin{pmatrix} 3 & 8 & 13 \ 1 & 6 & 11 \ 1 & 4 & 9 \end{pmatrix}$.Step $2$: Calculate the determinant $|A|$.$|A| = 3(54 - 44) - 8(9 - 11) + 13(4 - 6) = 3(10) - 8(-2) + 13(-2) = 30 + 16 - 26 = 20$.Step $3$: The element in the $2^{nd}$ row and $3^{rd}$ column of $A^{-1}$ is $\frac{C_{32}}{|A|}$, where $C_{32}$ is the cofactor of $a_{32}$.$C_{32} = (-1)^{3+2} \begin{vmatrix} 3 & 13 \ 1 & 11 \end{vmatrix} = -1(33 - 13) = -20$.Step $4$: Element $= \frac{-20}{20} = -1$.

If $A = [a_{ij}]_{3 \times 3}$ is a matrix such that $a_{ij} = |2i - 5j|$, where $|.|$ denotes the modulus function, then the element in the $2^{nd}$ row and $3^{rd}$ column of $A^{-1}$ is ...

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