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(D) The characteristic equation of a $3 	imes 3$ matrix $A$ is given by $|A - \lambda I| = 0$,which results in $-\lambda^3 + 	ext{tr}(A)\lambda^2 -

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$4$

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(D) The characteristic equation of a $3 	imes 3$ matrix $A$ is given by $|A - \lambda I| = 0$,which results in $-\lambda^3 + 	ext{tr}(A)\lambda^2 - (\dots)\lambda + |A| = 0$. Given the characteristic equation $\lambda^{3}-5 \lambda^{2}-3 \lambda+2=0$,we compare this with the standard form $\lambda^3 - 	ext{tr}(A)\lambda^2 + (\dots)\lambda - |A| = 0$. Comparing the constant terms,we get $-|A| = 2$,which implies $|A| = -2$. We know that for a square matrix $A$ of order $n$,$|	ext{adj}(A)| = |A|^{n-1}$. Here,$n = 3$,so $|	ext{adj}(A)| = |A|^{3-1} = |A|^2$. Substituting the value of $|A|$,we get $|	ext{adj}(A)| = (-2)^2 = 4$.

The characteristic equation of a matrix $A$ is $\lambda^{3}-5 \lambda^{2}-3 \lambda+2=0$. Then $|\text{adj}(A)|$ is equal to:

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