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Question: If \(\det \left( {{A}_{3\times 3}} \right)=6\), then \(\det \left( adj2A \right)=\) A. 144 B. \(...

If det⁡(A3×3)=6\det \left( {{A}_{3\times 3}} \right)=6, then det⁡(adj2A)=\det \left( adj2A \right)=
A. 144
B. 32×28{{3}^{2}}\times {{2}^{8}}
C. 33×24{{3}^{3}}\times {{2}^{4}}
D. 32×23{{3}^{2}}\times {{2}^{3}}

Explanation

Solution

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

Complete step-by-step answer:
Now, we have been given that det⁡(A3×3)=6\det \left( {{A}_{3\times 3}} \right)=6 and we have to find the value of det⁡(adj2A)\det \left( adj2A \right).
Now, since we have given that det⁡(A3×3)=6\det \left( {{A}_{3\times 3}} \right)=6. We have ∣A∣=6\left| A \right|=6 where A is the matrix of order 3.
Now, we know that if An×n{{A}_{n\times n}} is a matrix of order n×nn\times n, then we will have,
det⁡(aAn×n)=an∣A∣\det \left( a{{A}_{n\times n}} \right)={{a}^{n}}\left| A \right|
Where a is any constant. So, we have,
∣2A∣=23∣A∣ =23(6) =8×6 =48 \begin{aligned} & \left| 2A \right|={{2}^{3}}\left| A \right| \\\ & ={{2}^{3}}\left( 6 \right) \\\ & =8\times 6 \\\ & =48 \\\ \end{aligned}
Now, we know the formula that if A is a matrix of order n×nn\times n then,
∣adjA∣=∣A∣n−1\left| adjA \right|={{\left| A \right|}^{n-1}}
So, we have,
det⁡(adj2A)=∣2A∣3−1 =∣48∣2 =∣24×3∣2 =28×32 \begin{aligned} & \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{aligned}
Hence, the correct option is (B).

Note: To solve this question it is important to remember that,

& \left| adj{{A}_{n\times n}} \right|={{\left| A \right|}^{n-1}} \\\ & \left| a{{A}_{n}} \right|={{a}^{n}}\left| A \right| \\\ \end{aligned}$$ Also, it is important to note that we have first find $$\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.