If {a_{ij}} = frac{1}{2}(3i - 2j) and A = {[{ | vedclass

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(B) Given the formula for elements of the matrix: ${a_{ij}} = \frac{1}{2}(3i - 2j)$.For a $2 	imes 2$ matrix $A = {[{a_{ij}}]_{2 	imes 2}}$,we calcu

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$\left[ {\begin{array}{*{20}{c}}{1/2}&{ - 1/2}\2&1\end{array}} ight]$

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(B) Given the formula for elements of the matrix: ${a_{ij}} = \frac{1}{2}(3i - 2j)$.For a $2 	imes 2$ matrix $A = {[{a_{ij}}]_{2 	imes 2}}$,we calculate the elements as follows:For $i=1, j=1$: ${a_{11}} = \frac{1}{2}(3(1) - 2(1)) = \frac{1}{2}(3 - 2) = \frac{1}{2}$.For $i=1, j=2$: ${a_{12}} = \frac{1}{2}(3(1) - 2(2)) = \frac{1}{2}(3 - 4) = -\frac{1}{2}$.For $i=2, j=1$: ${a_{21}} = \frac{1}{2}(3(2) - 2(1)) = \frac{1}{2}(6 - 2) = \frac{4}{2} = 2$.For $i=2, j=2$: ${a_{22}} = \frac{1}{2}(3(2) - 2(2)) = \frac{1}{2}(6 - 4) = \frac{2}{2} = 1$.Thus,the matrix $A$ is given by:$A = \left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}\{{a_{21}}}&{{a_{22}}}\end{array}} ight] = \left[ {\begin{array}{*{20}{c}}{1/2}&{-1/2}\2&1\end{array}} ight]$.

If ${a_{ij}} = \frac{1}{2}(3i - 2j)$ and $A = {[{a_{ij}}]_{2 \times 2}}$,then $A$ is equal to

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