Question

If $\det \left( {{A}_{3\times 3}} \right)=6$, then $\det \left( adj2A \right)=$
A. 144
B. ${{3}^{2}}\times {{2}^{8}}$
C. ${{3}^{3}}\times {{2}^{4}}$
D. ${{3}^{2}}\times {{2}^{3}}$

Answer 1

Hint: We will start by using the fact that if $\left| {{A}_{n}} \right|=x$ then $\left| a{{A}_{n\times n}} \right|={{a}^{n}}\left| A \right|$. Then we will use this to find the value of $\det \left( 2A \right)$. After this we will use the fact that $\det \left( adjA \right)=\det {{\left( A \right)}^{n-1}}$ and using this we will find the value of $\det \left( adj2A \right)$.

Complete step-by-step answer:
Now, we have been given that $\det \left( {{A}_{3\times 3}} \right)=6$ and we have to find the value of $\det \left( adj2A \right)$.
Now, since we have given that $\det \left( {{A}_{3\times 3}} \right)=6$. We have \[\left| A \right|=6\] where A is the matrix of order 3.
Now, we know that if ${{A}_{n\times n}}$ is a matrix of order $n\times n$, then we will have,
$\det \left( a{{A}_{n\times n}} \right)={{a}^{n}}\left| A \right|$
Where a is any constant. So, we have,
$\begin{align}
  & \left| 2A \right|={{2}^{3}}\left| A \right| \\
 & ={{2}^{3}}\left( 6 \right) \\
 & =8\times 6 \\
 & =48 \\
\end{align}$
Now, we know the formula that if A is a matrix of order $n\times n$ then,
$\left| adjA \right|={{\left| A \right|}^{n-1}}$
So, we have,
$\begin{align}
  & \det \left( adj2A \right)={{\left| 2A \right|}^{3-1}} \\
 & ={{\left| 48 \right|}^{2}} \\
 & ={{\left| {{2}^{4}}\times 3 \right|}^{2}} \\
 & ={{2}^{8}}\times {{3}^{2}} \\
\end{align}$
Hence, the correct option is (B).

Note: To solve this question it is important to remember that,
\[\begin{align}
  & \left| adj{{A}_{n\times n}} \right|={{\left| A \right|}^{n-1}} \\
 & \left| a{{A}_{n}} \right|={{a}^{n}}\left| A \right| \\
\end{align}\]
Also, it is important to note that we have first find \[\left| 2A \right|\] before applying the formula for finding \[\left| adj2A \right|\] and also, it is worthwhile to see that we have been given A as a matrix of order $3\times 3$ and it’s a square matrix.