Let A = [2 2+p 2+p+q; 4 6+2p 8+3p+2q; 6 12+3p 20+6p+3q]. If det(adj(adj(3A))) = 2^m * 3^n, m, n ∈ N, then m+n is equal to

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q4 – JEE-MAIN 2025

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Home Maths MatricesQuestion #303

Let

A = [\begin{matrix} 2 & 2 + p & 2 + p + q \\ 4 & 6 + 2 p & 8 + 3 p + 2 q \\ 6 & 12 + 3 p & 20 + 6 p + 3 q \end{matrix}]

. If

det (adj (adj (3 A))) = 2^{m} \cdot 3^{n}

m, n \in N

, then

m + n

is equal to

Clue 1 of 3: Matrix Determinant Properties

The problem involves properties of determinants and adjoints of matrices, specifically how they behave under scalar multiplication and repeated adjoint operations.

🧠 Full Solution Path

Step 1: Calculate the Determinant of A✦ Active

Perform elementary row operations on matrix $A$ to simplify its form and find its determinant. Apply $R_{2} \to R_{2} - 2 R_{1}$ and $R_{3} \to R_{3} - 3 R_{1}$ :

A \sim [\begin{matrix} 2 & 2 + p & 2 + p + q \\ 0 & 2 & 4 + p \\ 0 & 6 & 14 + 3 p \end{matrix}]

Next, apply $R_{3} \to R_{3} - 3 R_{2}$ :

A \sim [\begin{matrix} 2 & 2 + p & 2 + p + q \\ 0 & 2 & 4 + p \\ 0 & 0 & 2 \end{matrix}]

The determinant of an upper triangular matrix is the product of its diagonal elements. Thus, $det (A) = 2 \cdot 2 \cdot 2 = 8$ .

Step 2: Apply Determinant and Adjoint Properties○ Expand

For an $n \times n$ matrix (here $n = 3$ ), the properties are $det (k A) = k^{n} det (A)$ and $det (adj (A)) = (det (A))^{n - 1}$ . We need to find $det (adj (adj (3 A)))$ . This can be generalized as $det (adj (adj (X))) = (det (X))^{(n - 1)^{2}}$ .

Let $X = 3 A$ . Then $det (X) = det (3 A) = 3^{3} det (A) = 3^{3} \cdot 8 = 3^{3} \cdot 2^{3}$ .

Using the generalized formula for $n = 3$ :

det (adj (adj (3 A))) = (det (3 A))^{(3 - 1)^{2}} = (det (3 A))^{2^{2}} = (det (3 A))^{4}

Substitute the value of $det (3 A)$ :

det (adj (adj (3 A))) = (3^{3} \cdot 2^{3})^{4} = (3^{3})^{4} \cdot (2^{3})^{4} = 3^{12} \cdot 2^{12}

💡 Teacher's Secret Hint

Ensure to correctly apply the power $(n - 1)^{2}$ for the double adjoint, not just $(n - 1)$ twice.

Step 3: Determine m+n○ Expand

We are given that $det (adj (adj (3 A))) = 2^{m} \cdot 3^{n}$ . Comparing this with our calculated value:

2^{12} \cdot 3^{12} = 2^{m} \cdot 3^{n}

By comparing the exponents, we find $m = 12$ and $n = 12$ . The required sum is $m + n = 12 + 12 = 24$ .

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Maths Question 4 – JEE-MAIN 2025

Clue 1 of 3: Matrix Determinant Properties

🧠 Full Solution Path