If A is a square matrix of order 4 and B = te | vedclass

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(D) Given $A$ is a square matrix of order $n = 4$. We know that $|	ext{Adj}(A)| = |A|^{n-1} = |A|^{4-1} = |A|^3$. Given $|B| = 27$,so $|A|^3 = 27$,wh

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$3^{23}$

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(D) Given $A$ is a square matrix of order $n = 4$. We know that $|	ext{Adj}(A)| = |A|^{n-1} = |A|^{4-1} = |A|^3$. Given $|B| = 27$,so $|A|^3 = 27$,which implies $|A| = 3$. We need to find $|A^{-1} 	ext{Adj}(3AB)|$. Using the property $|XY| = |X||Y|$,we have $|A^{-1} 	ext{Adj}(3AB)| = |A^{-1}| |	ext{Adj}(3AB)| = \frac{1}{|A|} |	ext{Adj}(3AB)|$. Since $3AB$ is a matrix of order $4$,$|	ext{Adj}(3AB)| = |3AB|^{4-1} = |3AB|^3$. $|3AB| = 3^4 |A| |B| = 81 	imes 3 	imes 27 = 3^4 	imes 3^1 	imes 3^3 = 3^8$. Therefore,$|	ext{Adj}(3AB)| = (3^8)^3 = 3^{24}$. Finally,$|A^{-1} 	ext{Adj}(3AB)| = \frac{1}{3} 	imes 3^{24} = 3^{23}$.

If $A$ is a square matrix of order $4$ and $B = \text{Adj}(A)$,where $|B| = 27$,then the value of $|A^{-1} \text{Adj}(3AB)|$ is,(where $A^{-1}$ denotes the inverse of matrix $A$ and $\text{Adj}(A)$ denotes the adjoint of matrix $A$):

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