If A = begin{bmatrix} 3 & 4 5 & 7 end{bmatri | vedclass

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(C) We know that for any square matrix $A$ of order $n$,the property $A(adj A) = (adj A)A = |A|I$ holds true,where $I$ is the identity matrix of the s

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$|A|I$

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(C) We know that for any square matrix $A$ of order $n$,the property $A(adj A) = (adj A)A = |A|I$ holds true,where $I$ is the identity matrix of the same order.Given $A = \begin{bmatrix} 3 & 4 \ 5 & 7 \end{bmatrix}$.First,calculate the determinant $|A| = (3 	imes 7) - (4 	imes 5) = 21 - 20 = 1$.The adjoint of $A$ is obtained by swapping the diagonal elements and changing the signs of the off-diagonal elements:$adj A = \begin{bmatrix} 7 & -4 \ -5 & 3 \end{bmatrix}$.Now,calculate $A(adj A)$:$A(adj A) = \begin{bmatrix} 3 & 4 \ 5 & 7 \end{bmatrix} \begin{bmatrix} 7 & -4 \ -5 & 3 \end{bmatrix} = \begin{bmatrix} (21 - 20) & (-12 + 12) \ (35 - 35) & (-20 + 21) \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = I$.Since $|A| = 1$,we have $I = |A|I$.Therefore,$A(adj A) = |A|I$.

If $A = \begin{bmatrix} 3 & 4 \\ 5 & 7 \end{bmatrix}$,then $A(adj A) = $

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