If A = begin{bmatrix} 1 & 2 3 & 4 end{bmatri | vedclass

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(A) Given,$A = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}$.The determinant of $A$ is calculated as $|A| = (1)(4) - (2)(3) = 4 - 6 = -2$.The adjoint

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$-\frac{1}{2} \begin{bmatrix} 4 & -2 \ -3 & 1 \end{bmatrix}$

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(A) Given,$A = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}$.The determinant of $A$ is calculated as $|A| = (1)(4) - (2)(3) = 4 - 6 = -2$.The adjoint of $A$,denoted as $\operatorname{adj}(A)$,is obtained by swapping the diagonal elements and changing the signs of the off-diagonal elements:$\operatorname{adj}(A) = \begin{bmatrix} 4 & -2 \ -3 & 1 \end{bmatrix}$.The inverse of a matrix is given by the formula $A^{-1} = \frac{1}{|A|} \operatorname{adj}(A)$.Substituting the values,we get $A^{-1} = \frac{1}{-2} \begin{bmatrix} 4 & -2 \ -3 & 1 \end{bmatrix} = -\frac{1}{2} \begin{bmatrix} 4 & -2 \ -3 & 1 \end{bmatrix}$.

If $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$,then $A^{-1}$ is equal to

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