If A is a 3 × 3 matrix and |A| = 2 , then |3 adj (|3A| A ^ 2 )| is...

We need to find the value of $\left|3adj\left(\left|3A\right|{A}^{2}\right)\right|$

We know that \left|kA\right| = {k}^{n}\left|A\right|$where$n$is the order of the matrix and$k$is a constant.</p><p>$⇒ \left|3A\right| = {3}^{3}\left(2\right)$</p><p>$⇒ \left|3adj\left(\left|3A\right|{A}^{2}\right)\right| = {3}^{3}\left|adj\left(\left({3}^{3} \cdot 2\right){A}^{2}\right)\right|$</p><p>We know that$adj\left(kA\right) = {k}^{n - 1}adj\left(A\right)$</p><p>$= {3}^{3}\left|{\left({3}^{3} \cdot 2\right)}^{2}adj\left({A}^{2}\right)\right|$</p><p>Now we know that \left|adjA\right| = {\left|A\right|}^{n - 1}$

$= {3}^{3}{\left({\left({3}^{3} \cdot 2\right)}^{2}\right)}^{3}\left|adj\left({A}^{2}\right)\right|$

$= {3}^{3}{\left({3}^{3} \cdot 2\right)}^{6}{\left|{A}^{2}\right|}^{3 - 1}$

$= {3}^{3}{\left({3}^{3} \cdot 2\right)}^{6}{\left(2\right)}^{4}$

$= {2}^{10} \cdot {3}^{21}$

Hence, the required answer is ${2}^{10} \cdot {3}^{21}$$= {6}^{10} \cdot {3}^{11}$