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**Hint:**Here we’ll use some properties of determinants and adjoint of square matrices like $\left| {{\text{AdjM}}} \right|{\text{ = }}{\left| {\text{M}} \right|^{{\text{n - 1}}}}$and $\left| {{{\text{M}}^{\text{a}}}} \right|{\text{ = }}{\left| {\text{M}} \right|^{\text{a}}}$, first we’ll find the value of |Adj${{\text{A}}^{\text{2}}}$| then again applying the same property will find the value of$\left| {{\text{Adj(Adj}}{{\text{A}}^{\text{2}}}{\text{)}}} \right|$to get the required answer.

**Complete step by step answer:**

Given data: A is a square matrix of order 3

As we all know that, if M is a square matrix of order n

Then, $\left| {{\text{AdjM}}} \right|{\text{ = }}{\left| {\text{M}} \right|^{{\text{n - 1}}}}$

Similarly, we can also say that

$\left| {{\text{Adj(AdjM)}}} \right|{\text{ = (}}{\left| {\text{M}} \right|^{{\text{n - 1}}}}{{\text{)}}^{\text{2}}}$

Now, A is a matrix of order 3, so can conclude that

\[

\left| {{\text{Adj(AdjA)}}} \right|{\text{ = (}}{\left| {\text{A}} \right|^{{\text{3 - 1}}}}{{\text{)}}^{\text{2}}} \\

{\text{ = (}}{\left| {\text{A}} \right|^{\text{2}}}{{\text{)}}^{\text{2}}} \\

{\text{ = }}{\left| {\text{A}} \right|^{\text{4}}} \\

\]

Therefore it is applicable for \[{{\text{A}}^{\text{2}}}\] as it will also be a square matrix, concluding that

\[

\left| {{\text{Adj(Adj}}{{\text{A}}^{\text{2}}}{\text{)}}} \right|{\text{ = (}}{\left| {{{\text{A}}^{\text{2}}}} \right|^{{\text{3 - 1}}}}{{\text{)}}^{\text{2}}} \\

{\text{ = (}}{\left| {{{\text{A}}^{\text{2}}}} \right|^{\text{2}}}{{\text{)}}^{\text{2}}} \\

{\text{ = }}{\left| {{{\text{A}}^{\text{2}}}} \right|^{\text{4}}} \\

\]

Since we know that for a square matrix M of order n

$\left| {{{\text{M}}^{\text{a}}}} \right|{\text{ = }}{\left| {\text{M}} \right|^{\text{a}}}$

\[

\left| {{\text{Adj(Adj}}{{\text{A}}^{\text{2}}}{\text{)}}} \right|{\text{ = }}{\left| {{{\text{A}}^{\text{2}}}} \right|^{\text{4}}} \\

{\left| {\text{A}} \right|^{\text{8}}} \\

\]

Therefore, option (C)${{\text{A}}^{\text{8}}}$ is the correct option

**Note:**An alternative solution for this question can be

Since we know that for a square matrix M of order n

$\left| {{{\text{M}}^{\text{a}}}} \right|{\text{ = }}{\left| {\text{M}} \right|^{\text{a}}}$

Now, since A is also a square matrix

$\left| {{{\text{A}}^{\text{2}}}} \right|{\text{ = }}{\left| {\text{A}} \right|^{\text{2}}}$

Now, applying the same rule as the above solution

\[

\left| {{\text{(Adj}}{{\text{A}}^{\text{2}}}{\text{)}}} \right|{\text{ = }}{\left( {{{\left| {\text{A}} \right|}^{\text{2}}}} \right)^{{\text{3 - 1}}}} \\

{\text{ = }}{\left| {\text{A}} \right|^{\text{4}}} \\

\]

Again using the same formula,

\[

\left| {{\text{Adj(Adj}}{{\text{A}}^{\text{2}}}{\text{)}}} \right|{\text{ = }}{\left( {{{\left| {\text{A}} \right|}^{\text{4}}}} \right)^{{\text{3 - 1}}}} \\

{\text{ = }}{\left| {\text{A}} \right|^{\text{8}}} \\

\]

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