JEE Main 2025 (January)

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: Let A be a square matrix of order 3 such that $det (A) = - 2$ and $det (3 a d j (- 6 a d j (3 A))) = 2^{m + n} \cdot 3^{mn}$ $m > n$ . Then $4 m + 2 n$ is equal to

Q: Why does the negative sign disappear in the adjoint step?

The property is $\det(\text{adj}(X)) = (\det(X))^{n-1}$. Here, $n=3$, so the exponent is $3-1=2$. Squaring any real number, whether positive or negative, always yields a non-negative result. That is why the negative sign vanishes. Trust the algebra!

Enter Numerical Value:

Visualized Solution

Problem Setup

Given: $A$ is a square matrix of order $n = 3$ .
Given: $det (A) = - 2$ .
Target Expression: $det (3 adj (- 6 adj (3 A))) = 2^{m + n} \cdot 3^{mn}$ .

Essential Determinant Properties

Property 1: For an $n \times n$ matrix $X$ , $det (k X) = k^{n} det (X)$ .
Property 2: $det (adj (X)) = (det (X))^{n - 1}$ .
Here, $n = 3$ , so $det (k X) = k^{3} det (X)$ and $det (adj (X)) = (det (X))^{2}$ .

Calculate $det (3 A)$

Let's evaluate the innermost part: $det (3 A)$ .
Using Property 1: $det (3 A) = 3^{3} det (A)$ .
Substitute $det (A) = - 2$ : $det (3 A) = 27 \times (- 2) = - 54$ .

Calculate $det (adj (3 A))$

Next layer: $det (adj (3 A))$ .
Using Property 2: $det (adj (3 A)) = (det (3 A))^{2}$ .
Substitute $det (3 A) = - 54$ : $det (adj (3 A)) = (- 54)^{2}$ .

Calculate $det (- 6 adj (3 A))$

Next layer: $det (- 6 adj (3 A))$ .
Let $Y = adj (3 A)$ . We need $det (- 6 Y)$ .
Using Property 1: $det (- 6 Y) = (- 6)^{3} det (Y)$ .
So, $det (- 6 adj (3 A)) = (- 6)^{3} \cdot (- 54)^{2}$ .

Prime Factorization of Components

To simplify, let's use prime factorization.
$(- 6)^{3} = (- 1)^{3} \cdot (2 \cdot 3)^{3} = - 2^{3} \cdot 3^{3}$ .
$(- 54)^{2} = ((- 1) \cdot 2 \cdot 3^{3})^{2} = 2^{2} \cdot 3^{6}$ .
$det (- 6 adj (3 A)) = (- 2^{3} \cdot 3^{3}) \cdot (2^{2} \cdot 3^{6}) = - 2^{5} \cdot 3^{9}$ .

Calculate $det (adj (- 6 adj (3 A)))$

Next layer: $det (adj (- 6 adj (3 A)))$ .
Let $Z = - 6 adj (3 A)$ . We need $det (adj (Z))$ .
Using Property 2: $det (adj (Z)) = (det (Z))^{2}$ .
Substitute $det (Z) = - 2^{5} \cdot 3^{9}$ : $det (adj (Z)) = (- 2^{5} \cdot 3^{9})^{2} = 2^{10} \cdot 3^{18}$ .

Final Determinant Calculation

Outermost layer: $det (3 adj (- 6 adj (3 A)))$ .
Let $W = adj (- 6 adj (3 A))$ . We need $det (3 W)$ .
Using Property 1: $det (3 W) = 3^{3} det (W)$ .
Substitute $det (W) = 2^{10} \cdot 3^{18}$ : $det (3 W) = 3^{3} \cdot (2^{10} \cdot 3^{18}) = 2^{10} \cdot 3^{21}$ .

Comparing Exponents

We found: $det (3 adj (- 6 adj (3 A))) = 2^{10} \cdot 3^{21}$ .
Given form: $2^{m + n} \cdot 3^{mn}$ .
Comparing the powers of $2$ : $m + n = 10$ .
Comparing the powers of $3$ : $mn = 21$ .

Solving the System of Equations

We have $m + n = 10$ and $mn = 21$ .
The roots of $x^{2} - 10 x + 21 = 0$ are $7$ and $3$ .
Given condition: $m > n$ .
Therefore, $m = 7$ and $n = 3$ .

Calculate $4 m + 2 n$

Target expression: $4 m + 2 n$ .
Substitute $m = 7$ and $n = 3$ : $4 (7) + 2 (3)$ .
$28 + 6 = 34$ .
The final answer is $34$ .

00:00 / 00:00

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The fortress we must dismantle is the expression

det (3 adj (- 6 adj (3 A)))

, where

A

is a

3 \times 3

matrix with

det (A) = - 2

To solve this, we rely on two fundamental properties of determinants for an

n \times n

matrix:

1. Scalar Multiplication:

det (k X) = k^{n} det (X)

. For our

3 \times 3

case, this is

det (k X) = k^{3} det (X)

2. Adjoint Property:

det (adj (X)) = (det (X))^{n - 1}

. For our

3 \times 3

case, this simplifies to

det (adj (X)) = (det (X))^{2}

Peeling the Layers

First, we evaluate the innermost term

det (3 A)

det (3 A) = 3^{3} det (A) = 27 \times (- 2) = - 54

Next, we move to

det (adj (3 A))

det (adj (3 A)) = (det (3 A))^{2} = (- 54)^{2}

Now, consider the term

det (- 6 adj (3 A))

. Let

Y = adj (3 A)

. Using the scalar property:

det (- 6 Y) = (- 6)^{3} det (Y) = (- 6)^{3} \cdot (- 54)^{2}

The Power of Primes

To avoid massive integers, we express the components using prime factors of

2

and

3

(- 6)^{3} = (- 1)^{3} \cdot 2^{3} \cdot 3^{3} = - 2^{3} \cdot 3^{3}

(- 54)^{2} = ((- 1) \cdot 2 \cdot 3^{3})^{2} = 2^{2} \cdot 3^{6}

Multiplying these gives:

det (- 6 adj (3 A)) = (- 2^{3} \cdot 3^{3}) \cdot (2^{2} \cdot 3^{6}) = - 2^{5} \cdot 3^{9}

Now, we apply the adjoint property to the outer layer

det (adj (- 6 adj (3 A)))

det (adj (- 6 adj (3 A))) = (- 2^{5} \cdot 3^{9})^{2} = 2^{10} \cdot 3^{18}

Finally, we account for the outermost scalar

3

det (3 adj (- 6 adj (3 A))) = 3^{3} \cdot (2^{10} \cdot 3^{18}) = 2^{10} \cdot 3^{21}

Final Calculation

We are given that the result equals

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

. By comparing exponents, we establish the system:

m + n = 10

mn = 21

These are the roots of the quadratic equation

x^{2} - 10 x + 21 = 0

, which factors to

(x - 7) (x - 3) = 0

. Given the constraint

m > n

, we identify $m = 7$ and $n = 3$ .

The final value is:

4 m + 2 n = 4 (7) + 2 (3) = 28 + 6 = 34

Common Traps & FAQs

Why does the scalar

k

come out as

k^{3}

and not just

k

Remember, a determinant represents a volume scaling factor. In a

3 \times 3

matrix, scaling the matrix by

k

effectively scales the volume in all three dimensions (length, width, and height). Thus, the determinant scales by

k \times k \times k = k^{3}

. Always check the order of the matrix before pulling out constants!

Why does the negative sign disappear in the adjoint step?

The property is

det (adj (X)) = (det (X))^{n - 1}

. Here,

n = 3

, so the exponent is

3 - 1 = 2

. Squaring any real number, whether positive or negative, always yields a non-negative result. That is why the negative sign vanishes. Trust the algebra!

Is there a shortcut to avoid calculating the huge numbers?

Absolutely! Never calculate the full integer value. Always keep your numbers in prime factor form (powers of

2

and

3

). This is the 'JEE Advanced' way—it prevents arithmetic errors and makes the final comparison with

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

immediate. If you see exponents, think prime factorization.

Animated Solution for Mathematics - Matrices and Determinants: Let A be a square matrix of order 3 such that $det (A) = - 2$ and $det (3 a d j (- 6 a d j (3 A))) = 2^{m + n} \cdot 3^{mn}$ $m > n$ . Then $4 m + 2 n$ is equal to

Enter Numerical Value:

Visualized Solution

Problem Setup

Essential Determinant Properties

Calculate $det (3 A)$

Calculate $det (adj (3 A))$

Calculate $det (- 6 adj (3 A))$

Prime Factorization of Components

Calculate $det (adj (- 6 adj (3 A)))$

Final Determinant Calculation

Comparing Exponents

Solving the System of Equations

Calculate $4 m + 2 n$

The Sigma Insight: Properties of Determinants

Analyzing the Setup

Peeling the Layers

The Power of Primes

Final Calculation

Similar Questions

Let $A$ be a non-singular matrix of order 3. If $det (3 adj (2 adj ((det A) A))) = 3^{- 13} \cdot 2^{- 10}$ and $det (3 adj (2 A)) = 2^{m} \cdot 3^{n}$ , then $∣3 m + 2 n ∣$ is equal to

If $A$ is a square matrix of order 3 such that $det (A) = 3$ and $det (adj (- 4 adj (- 3 adj (3 adj ((2 A)^{- 1}))))) = 2^{m} 3^{n}$ , then $m + 2 n$ is equal to:

Let $A = 1 α 1 231312$ for $α \in R$ . If $∣ A ∣ = 2$ . If $∣2 adj (2 adj (2 A)) ∣ = 3 2^{n}$ , then $3 n + α$ is equal to

Let $A$ be a matrix of order $3 \times 3$ and $det (A) = 2$ . Then $det (det (A) adj (5 adj (A^{3})))$ is equal to

If $A$ is a $3 \times 3$ matrix and $∣ A ∣ = 2$ , then $∣3 adj (∣3 A ∣ A^{2}) ∣$ is equal to

Let $A$ and $B$ be two square matrices of order 3 such that $∣ A ∣ = 3$ and $∣ B ∣ = 2$ . Then $∣ A^{T} A (adj (2 A))^{- 1} (adj (4 B)) (adj (A B))^{- 1} A A^{T} ∣$ is equal to:

Let $A$ be a $n \times n$ matrix such that $∣ A ∣ = 2$ . If the determinant of the matrix $adj (2 adj (2 A^{- 1}))$ is $2^{84}$ , then $n$ is equal to ______.

Let $A$ be a $3 \times 3$ matrix with $det (A) = 4$ . Let $R_{i}$ denote the $i^{t h}$ row of $A$ . If a matrix $B$ is obtained by performing the operation $R_{2} \to 2 R_{2} + 5 R_{3}$ on $2 A$ , then $det (B)$ is equal to:

Let $A = x y z y z x z x y$ , where $x, y$ and $z$ are real numbers such that $x + y + z > 0$ and $x y z = 2$ . If $A^{2} = I_{3}$ , then the value of $x^{3} + y^{3} + z^{3}$ is

If $A = [2335]$ , then the determinant of the matrix $(A^{2025} - 3 A^{2024} + A^{2023})$ is

Enter Numerical Value:

Visualized Solution

Problem Setup

Essential Determinant Properties

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } det(3A)

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } det(adj(3A))

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } det(−6 adj(3A))

Prime Factorization of Components

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } det(adj(−6 adj(3A)))

Final Determinant Calculation

Comparing Exponents

Solving the System of Equations

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4m+2n

The Sigma Insight: Properties of Determinants

Analyzing the Setup

Peeling the Layers

The Power of Primes

Final Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If A is a square matrix of order 3 such that det(A)=3 and det(adj(−4adj(−3adj(3adj((2A)−1)))))=2m3n, then m+2n is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A=​1α1​231​312​​ for α∈R. If ∣A∣=2. If ∣2 adj(2 adj(2A))∣=32n, then 3n+α is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A be a matrix of order 3×3 and det(A)=2. Then det(det(A)adj(5adj(A3))) is equal to

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A and B be two square matrices of order 3 such that ∣A∣=3 and ∣B∣=2. Then ∣ATA(adj(2A))−1(adj(4B))(adj(AB))−1AAT∣ is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A be a n×n matrix such that ∣A∣=2. If the determinant of the matrix adj(2adj(2A−1)) is 284, then n is equal to ______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A=​xyz​yzx​zxy​​, where x,y and z are real numbers such that x+y+z>0 and xyz=2. If A2=I3​, then the value of x3+y3+z3 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If A=[23​35​], then the determinant of the matrix (A2025−3A2024+A2023) is

Calculate $det (3 A)$

Calculate $det (adj (3 A))$

Calculate $det (- 6 adj (3 A))$

Calculate $det (adj (- 6 adj (3 A)))$

Calculate $4 m + 2 n$

Let $A$ be a non-singular matrix of order 3. If $det (3 adj (2 adj ((det A) A))) = 3^{- 13} \cdot 2^{- 10}$ and $det (3 adj (2 A)) = 2^{m} \cdot 3^{n}$ , then $∣3 m + 2 n ∣$ is equal to

If $A$ is a square matrix of order 3 such that $det (A) = 3$ and $det (adj (- 4 adj (- 3 adj (3 adj ((2 A)^{- 1}))))) = 2^{m} 3^{n}$ , then $m + 2 n$ is equal to:

Let $A = 1 α 1 231312$ for $α \in R$ . If $∣ A ∣ = 2$ . If $∣2 adj (2 adj (2 A)) ∣ = 3 2^{n}$ , then $3 n + α$ is equal to

Let $A$ be a matrix of order $3 \times 3$ and $det (A) = 2$ . Then $det (det (A) adj (5 adj (A^{3})))$ is equal to

If $A$ is a $3 \times 3$ matrix and $∣ A ∣ = 2$ , then $∣3 adj (∣3 A ∣ A^{2}) ∣$ is equal to

Let $A$ and $B$ be two square matrices of order 3 such that $∣ A ∣ = 3$ and $∣ B ∣ = 2$ . Then $∣ A^{T} A (adj (2 A))^{- 1} (adj (4 B)) (adj (A B))^{- 1} A A^{T} ∣$ is equal to:

Let $A$ be a $n \times n$ matrix such that $∣ A ∣ = 2$ . If the determinant of the matrix $adj (2 adj (2 A^{- 1}))$ is $2^{84}$ , then $n$ is equal to ______.

Let $A = x y z y z x z x y$ , where $x, y$ and $z$ are real numbers such that $x + y + z > 0$ and $x y z = 2$ . If $A^{2} = I_{3}$ , then the value of $x^{3} + y^{3} + z^{3}$ is

If $A = [2335]$ , then the determinant of the matrix $(A^{2025} - 3 A^{2024} + A^{2023})$ is