Let for A = pmatrix 1 & 2 & 3 1 & 2 & 3 1 & 1 & 2 pmatrix, |A| = 2. If |2 adj(2A)| = 32, then 3n + is equal to:

Let for A = pmatrix 1 & 2 & 3 1 & 2 & 3 1 & 1 & 2 pmatrix, |A| =… — Solution | JEE Mathematics

MCQMediumJEE 2023Inverse & Adjoint of a Matrix

Let for $A = \begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 1 & 1 & 2 \end{pmatrix}, \quad |A| = 2. \quad \text{If} \quad |2 \, adj(2A)| = 32, \quad \text{then} \quad 3n + \alpha \text{ is equal to:}$

A
$10$
B
$9$
C
$12$
D
$11$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: $|A| = 2$ and $|2 \, adj(2A)| = 32$ .

Find: $3n + \alpha$ .

From the solution, use the determinant property for a matrix of order $3$ :

|kA| = k^3 |A|

and

|adj(A)| = |A|^{3-1} = |A|^2.

So,

|adj(A)| = 2^2 = 4.

The solution further simplifies the given expression and writes

|2 \, adj(2 \, adj(2A))| = 2^7 \cdot 4 = 2^9 = 32^n.

Hence,

32^n = 2^9 = (2^5)^n

which gives

n = 5.

Now using the determinant condition written in the solution,

(6-1) - 2(2\alpha - 1) + 3(\alpha - 3) = 2.

Simplifying,

5 - 4\alpha + 2 + 3\alpha - 9 = 2

-2 - \alpha = 2

\alpha = -4.

Therefore,

3n + \alpha = 3(5) + (-4) = 11.

So the correct option is D.

Note: The solution displays option C, but its own final calculation gives $11$ , which matches option D. Therefore the defensible answer is D.

Common mistakes

Using $|kA| = k^2|A|$ for a $3 \times 3$ matrix. This is wrong because the determinant scales as $k^m$ where $m$ is the order of the matrix. Here the correct factor is $k^3$ .
Forgetting the identity $|adj(A)| = |A|^{n-1}$ . For a matrix of order $3$ , this becomes $|adj(A)| = |A|^2$ , not $|A|$ .
Trusting the option letter shown on the solution without checking the working. The displayed option letter is inconsistent with the final computed value. Always verify the numerical result and then match it with the options.

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Answer

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