Let $A = \begin{bmatrix} 2 & 2 + p & 2 + p + q \\ 4 & 6 + 2p & 8 + 3p + 2q \\ 6 & 12 + 3p & 20 + 6p + 3q \end{bmatrix}$. If $\det(\text{adj}(\text{adj}(3 A))) = 2^m \cdot 3^n, m, n \in \mathbb{N}$, then $m + n$ is equal to Options: (A) 22 (B) 24 (C) 26 (D) 20

### Analyzing the Setup We are given a $3 \times 3$ matrix $A$ and tasked with finding the value of $m+n$ given the relation: $$|\text{adj}(\text{adj}(3A))| = 2^m \cdot 3^n$$ The key to solving this problem lies in simplifying the determinant of $A$ using elementary row operations before applying the properties of the adjoint matrix. ### The Strategy of Simplification Observe the first column of matrix $A$, which contains the elements $2, 4, 6$. These are multiples of $2$, suggesting that we can create zeros to simplify the determinant calculation. We apply the following elementary row operations: $R_2 \to R_2 - 2R_1$ $R_3 \to R_3 - 3R_1$ After these operations, the $q$ terms are eliminated, significantly reducing the complexity of the matrix. ### The Determinant Reveal The determinant $|A|$ can now be expressed as: $$|A| = \begin{vmatrix} 2 & 2+p & 2+p+q \\ 0 & 2 & 4+p \\ 0 & 6 & 14+3p \end{vmatrix}$$ Expanding along the first column, we calculate the minor for the element $2$: $$|A| = 2 [ 2(14+3p) - 6(4+p) ]$$ Simplifying the expression inside the brackets: $$|A| = 2 [ 28 + 6p - 24 - 6p ] = 2 \cdot 4 = 8$$ Thus, we find that $|A| = 2^3$. ### The Adjoint Powerhouse We utilize the property for the determinant of the double adjoint of a matrix $B$ of order $n$: $$|\text{adj}(\text{adj}(B))| = |B|^{(n-1)^2}$$ For our $3 \times 3$ matrix, this becomes $|B|^{(3-1)^2} = |B|^4$. Here, $B = 3A$. First, we calculate $|3A|$ using the property $|kA| = k^n |A|$: $$|3A| = 3^3 |A| = 27 \cdot 8 = 216$$ We express this as $|3A| = 3^3 \cdot 2^3$. ### Final Calculation Now, we raise this result to the power of $4$ as required by the adjoint property: $$|\text{adj}(\text{adj}(3A))| = (3^3 \cdot 2^3)^4 = 3^{12} \cdot 2^{12}$$ Comparing this to the given form $2^m \cdot 3^n$, we identify: $m = 12$ $n = 12$ The final sum is: **$m + n = 12 + 12 = 24$**

JEE Main 2025 April

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: Let $A = 246 2 + p 6 + 2 p 12 + 3 p 2 + p + q 8 + 3 p + 2 q 20 + 6 p + 3 q$ . If $det (adj (adj (3 A))) = 2^{m} \cdot 3^{n}, m, n \in N$ , then $m + n$ is equal to

Q: Wait, why do the $p$ and $q$ terms just vanish during the row operations?

That is the beauty of the problem! When you perform $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$, you are essentially subtracting linear combinations of the first row from the others. Because the matrix is constructed such that the $p$ and $q$ terms in the second and third rows are multiples of the first row, they cancel out perfectly. It is a designed simplification, not a coincidence!

Q: Why is $|3A| = 3^3 |A|$ and not just $3|A|$?

This is a classic JEE trap! Remember that for a matrix $A$ of order $n$, the determinant of $kA$ is $k^n |A|$. Since our matrix $A$ is a $3 \times 3$ matrix, $n=3$. Therefore, $|3A| = 3^3 |A| = 27 |A|$. Always check the order of the matrix before pulling out a scalar!

Select Answer:

Visualized Solution

Analyze Matrix $A$

Given matrix $A = 246 2 + p 6 + 2 p 12 + 3 p 2 + p + q 8 + 3 p + 2 q 20 + 6 p + 3 q$
Target: Find $m + n$ from $det (adj (adj (3 A))) = 2^{m} \cdot 3^{n}$

Simplifying $∣ A ∣$ : Strategy

To find $∣ A ∣$ , we need to simplify the matrix.
Observe Column 1 ( $C_{1}$ ): Elements are $2, 4, 6$ .
We can create zeros in $C_{1}$ using row operations.

Row Operation: $R_{2} \to R_{2} - 2 R_{1}$

Apply $R_{2} \to R_{2} - 2 R_{1}$
$R_{2} = [4 - 2 (2), (6 + 2 p) - 2 (2 + p), (8 + 3 p + 2 q) - 2 (2 + p + q)]$
Simplified $R_{2} = [0, 2, 4 + p]$

Row Operation: $R_{3} \to R_{3} - 3 R_{1}$

Apply $R_{3} \to R_{3} - 3 R_{1}$
$R_{3} = [6 - 3 (2), (12 + 3 p) - 3 (2 + p), (20 + 6 p + 3 q) - 3 (2 + p + q)]$
Simplified $R_{3} = [0, 6, 14 + 3 p]$

The Simplified Determinant $∣ A ∣$

The determinant becomes:
$∣ A ∣ = 200 2 + p 26 2 + p + q 4 + p 14 + 3 p$
Expand along $C_{1}$ (Column 1).

Evaluating $∣ A ∣$

$∣ A ∣ = 2 [2 (14 + 3 p) - 6 (4 + p)]$
$∣ A ∣ = 2 [28 + 6 p - 24 - 6 p]$
$∣ A ∣ = 2 [4] = 8 = 2^{3}$

Property of Double Adjoint

Recall the property: $∣ adj (adj (B)) ∣ = ∣ B ∣^{(n - 1)^{2}}$
Here, the matrix order is $n = 3$ .
Let $B = 3 A$ .
$∣ adj (adj (3 A)) ∣ = ∣3 A ∣^{(3 - 1)^{2}} = ∣3 A ∣^{4}$

Property of Scalar Multiplication

Recall the property: $∣ k A ∣ = k^{n} ∣ A ∣$
For $k = 3$ and order $n = 3$ :
$∣3 A ∣ = 3^{3} ∣ A ∣$

Calculating $∣3 A ∣$

Substitute $∣ A ∣ = 2^{3}$ into $∣3 A ∣ = 3^{3} ∣ A ∣$
$∣3 A ∣ = 3^{3} \cdot 2^{3}$

Final Substitution

Substitute $∣3 A ∣$ back into the adjoint expression:
$∣ adj (adj (3 A)) ∣ = (∣3 A ∣)^{4}$
$= (3^{3} \cdot 2^{3})^{4}$
Using $(a \cdot b)^{x} = a^{x} \cdot b^{x}$ :
$= 3^{12} \cdot 2^{12}$

Comparing Exponents

Given: $∣ adj (adj (3 A)) ∣ = 2^{m} \cdot 3^{n}$
We calculated: $2^{12} \cdot 3^{12}$
Equating the two: $2^{m} \cdot 3^{n} = 2^{12} \cdot 3^{12}$
Comparing the powers of 2 and 3:
$m = 12$ and $n = 12$

The Final Answer

We need to find $m + n$ .
$m + n = 12 + 12$
$m + n = 24$
Correct Option: 24

00:00 / 00:00

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

We are given a

3 \times 3

matrix

A

and tasked with finding the value of

m + n

given the relation:

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

The key to solving this problem lies in simplifying the determinant of

A

using elementary row operations before applying the properties of the adjoint matrix.

The Strategy of Simplification

Observe the first column of matrix

A

, which contains the elements

2, 4, 6

. These are multiples of

2

, suggesting that we can create zeros to simplify the determinant calculation.

We apply the following elementary row operations:

R_{2} \to R_{2} - 2 R_{1}

R_{3} \to R_{3} - 3 R_{1}

After these operations, the

q

terms are eliminated, significantly reducing the complexity of the matrix.

The Determinant Reveal

The determinant

∣ A ∣

can now be expressed as:

∣ A ∣ = 200 2 + p 26 2 + p + q 4 + p 14 + 3 p

Expanding along the first column, we calculate the minor for the element

2

∣ A ∣ = 2 [2 (14 + 3 p) - 6 (4 + p)]

Simplifying the expression inside the brackets:

∣ A ∣ = 2 [28 + 6 p - 24 - 6 p] = 2 \cdot 4 = 8

Thus, we find that

∣ A ∣ = 2^{3}

The Adjoint Powerhouse

We utilize the property for the determinant of the double adjoint of a matrix

B

of order

n

∣ adj (adj (B)) ∣ = ∣ B ∣^{(n - 1)^{2}}

For our

3 \times 3

matrix, this becomes

∣ B ∣^{(3 - 1)^{2}} = ∣ B ∣^{4}

. Here,

B = 3 A

First, we calculate

∣3 A ∣

using the property

∣ k A ∣ = k^{n} ∣ A ∣

∣3 A ∣ = 3^{3} ∣ A ∣ = 27 \cdot 8 = 216

We express this as

∣3 A ∣ = 3^{3} \cdot 2^{3}

Final Calculation

Now, we raise this result to the power of

4

as required by the adjoint property:

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

Comparing this to the given form

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

, we identify:

m = 12

n = 12

The final sum is: $m + n = 12 + 12 = 24$

Common Traps & FAQs

Wait, why do the

p

and

q

terms just vanish during the row operations?

That is the beauty of the problem! When you perform

R_{2} \to R_{2} - 2 R_{1}

and

R_{3} \to R_{3} - 3 R_{1}

, you are essentially subtracting linear combinations of the first row from the others. Because the matrix is constructed such that the

p

and

q

terms in the second and third rows are multiples of the first row, they cancel out perfectly. It is a designed simplification, not a coincidence!

Why is

∣3 A ∣ = 3^{3} ∣ A ∣

and not just

3∣ A ∣

This is a classic JEE trap! Remember that for a matrix

A

of order

n

, the determinant of

k A

k^{n} ∣ A ∣

. Since our matrix

A

is a

3 \times 3

matrix,

n = 3

. Therefore,

∣3 A ∣ = 3^{3} ∣ A ∣ = 27∣ A ∣

. Always check the order of the matrix before pulling out a scalar!

Can we just expand the determinant directly without row operations?

Technically, yes, but it would be an algebraic nightmare! You would be dealing with complex expressions involving

p

and

q

that would likely lead to calculation errors. The row operations are a strategic choice to reduce the matrix to a simpler form, making the determinant calculation trivial. Always look for ways to simplify before you start calculating.

Animated Solution for Mathematics - Matrices and Determinants: Let $A = 246 2 + p 6 + 2 p 12 + 3 p 2 + p + q 8 + 3 p + 2 q 20 + 6 p + 3 q$ . If $det (adj (adj (3 A))) = 2^{m} \cdot 3^{n}, m, n \in N$ , then $m + n$ is equal to

Select Answer:

Visualized Solution

Analyze Matrix $A$

Simplifying $∣ A ∣$ : Strategy

Row Operation: $R_{2} \to R_{2} - 2 R_{1}$

Row Operation: $R_{3} \to R_{3} - 3 R_{1}$

The Simplified Determinant $∣ A ∣$

Evaluating $∣ A ∣$

Property of Double Adjoint

Property of Scalar Multiplication

Calculating $∣3 A ∣$

Final Substitution

Comparing Exponents

The Final Answer

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

The Strategy of Simplification

The Determinant Reveal

The Adjoint Powerhouse

Final Calculation

Similar Questions

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Let $P = 323 - 1 0 - 5 - 2 α 0$ , where $α \in R$ . Suppose $Q = [q_{ij}]$ is a matrix such that $P Q = k I$ , where $k \in R, k \neq = 0$ and $I$ is the identity matrix of order $3$ . If $q_{23} = - k /8$ and $det (Q) = k^{2} /2$ , then

Select Answer:

Visualized Solution

Analyze Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣A∣: Strategy

Row Operation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R2​→R2​−2R1​

Row Operation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R3​→R3​−3R1​

The Simplified Determinant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣A∣

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣A∣

Property of Double Adjoint

Property of Scalar Multiplication

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣3A∣

Final Substitution

Comparing Exponents

The Final Answer

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

The Strategy of Simplification

The Determinant Reveal

The Adjoint Powerhouse

Final Calculation

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Analyze Matrix $A$

Simplifying $∣ A ∣$ : Strategy

Row Operation: $R_{2} \to R_{2} - 2 R_{1}$

Row Operation: $R_{3} \to R_{3} - 3 R_{1}$

The Simplified Determinant $∣ A ∣$

Evaluating $∣ A ∣$

Calculating $∣3 A ∣$

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