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If A is a square matrix of order 3, then $\left| {{\text{Adj(Adj}}{{\text{A}}^{\text{2}}}{\text{)}}} \right|{\text{ = }}$
A. ${{\text{A}}^{\text{2}}}$
B. ${{\text{A}}^{\text{4}}}$
C. ${{\text{A}}^{\text{8}}}$
D. ${{\text{A}}^{{\text{16}}}}$

Answer Verified Verified
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,
\[
  \left| {{\text{Adj(Adj}}{{\text{A}}^{\text{2}}}{\text{)}}} \right|{\text{ = }}{\left( {{{\left| {\text{A}} \right|}^{\text{4}}}} \right)^{{\text{3 - 1}}}} \\
  {\text{ = }}{\left| {\text{A}} \right|^{\text{8}}} \\
 \]