JEE Main 2023 (13 April Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Matrices and Determinants: 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

Q: Why is $|\text{adj}(A)| = |A|^2$? Where does the square come from?

Great question! The general property for any $n \times n$ matrix is $|\text{adj}(A)| = |A|^{n-1}$. Since we are dealing with a $3 \times 3$ matrix, $n=3$, so $n-1 = 2$. That is why the determinant of the adjoint is the square of the determinant of the original matrix.

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the matrix

A

as a grid of numbers. The determinant is just a scalar value. Don't let the variable

α

intimidate you; it is simply a placeholder for a number we need to find. Start by expanding the determinant along the first row to create an equation in terms of

α

Hint 2

Hint 2: Recall the two fundamental properties of determinants for a

3 \times 3

matrix: first,

∣ k A ∣ = k^{3} ∣ A ∣

, and second,

∣ adj (B) ∣ = ∣ B ∣^{2}

. These are your primary tools for dismantling the nested expression

∣2 adj (2 adj (2 A)) ∣

Hint 3

Hint 3: Approach the expression

∣2 adj (2 adj (2 A)) ∣

like an onion. Start from the outermost layer and peel it back one step at a time, applying the properties of determinants carefully. Keep track of the powers of 2 at each step, and remember that

∣ A ∣ = 2

Visualized Solution

Introduction to Matrix $A$

Given matrix $A = 1 α 1 231312$
Given determinant $∣ A ∣ = 2$
Our goal is to find $α$ and then use the property of adjoints to find $n$ .

Expanding $∣ A ∣$

Expanding $∣ A ∣$ along the first row:
$∣ A ∣ = 1 (3 \times 2 - 1 \times 1) - 2 (α \times 2 - 1 \times 1) + 3 (α \times 1 - 1 \times 3)$

Simplifying the Determinant

$∣ A ∣ = 1 (6 - 1) - 2 (2 α - 1) + 3 (α - 3)$
$∣ A ∣ = 5 - 4 α + 2 + 3 α - 9$
$∣ A ∣ = - α - 2$

Solving for $α$

Given $∣ A ∣ = 2$
Substituting the expression: $- α - 2 = 2$
Adding $2$ to both sides: $- α = 4$
Therefore, $α = - 4$

Properties of Determinants

Property 1: For a $3 \times 3$ matrix $B$ , $∣ k B ∣ = k^{3} ∣ B ∣$
Property 2: For a $3 \times 3$ matrix $B$ , $∣ adj (B) ∣ = ∣ B ∣^{2}$

The Nested Expression

Expression: $∣2 adj (2 adj (2 A)) ∣$
We will peel this expression layer by layer from the outside in.

Peeling Layer 1: Outer Scalar

Applying Property 1 ( $k = 2$ ):
$= 2^{3} ∣ adj (2 adj (2 A)) ∣$

Peeling Layer 2: Outer Adjoint

Applying Property 2 ( $∣ adj (B) ∣ = ∣ B ∣^{2}$ ):
$= 2^{3} (∣2 adj (2 A) ∣)^{2}$

Peeling Layer 3: Inner Scalar

Applying Property 1 again inside the square:
$= 2^{3} (2^{3} ∣ adj (2 A) ∣)^{2}$

Peeling Layer 4: Inner Adjoint

Applying Property 2 again to $∣ adj (2 A) ∣$ :
$= 2^{3} (2^{3} (∣2 A ∣)^{2})^{2}$

Peeling Layer 5: Innermost Scalar

Applying Property 1 to $∣2 A ∣$ :
$= 2^{3} (2^{3} (2^{3} ∣ A ∣)^{2})^{2}$

Consolidating Powers of 2

Simplifying the nested powers:
$= 2^{3} (2^{3} \cdot 2^{6} ∣ A ∣^{2})^{2}$
$= 2^{3} (2^{9} ∣ A ∣^{2})^{2}$
$= 2^{3} \cdot 2^{18} ∣ A ∣^{4}$
$= 2^{21} ∣ A ∣^{4}$

Evaluating the Expression

Since $∣ A ∣ = 2$ , substitute this value:
$= 2^{21} \cdot (2)^{4}$
$= 2^{25}$

Solving for $n$

Given $∣2 adj (2 adj (2 A)) ∣ = 3 2^{n}$
We found the LHS is $2^{25}$
RHS: $3 2^{n} = (2^{5})^{n} = 2^{5 n}$
Equating powers: $2^{25} = 2^{5 n} ⟹ 5 n = 25 ⟹ n = 5$

Final Calculation

We have $n = 5$ and $α = - 4$
Calculate $3 n + α$ :
$3 (5) + (- 4) = 15 - 4 = 11$
The final answer is 11

00:00 / 00:00

The Sigma Insight: Properties of Determinants

Solution Diagram

Analyzing the Setup

We are given the matrix

A = 1 α 1 231312

with the condition that the determinant

∣ A ∣ = 2

To find the value of

α

, we expand the determinant along the first row:

∣ A ∣ = 1 (3 \times 2 - 1 \times 1) - 2 (α \times 2 - 1 \times 1) + 3 (α \times 1 - 1 \times 3)

Simplifying the expression, we obtain:

∣ A ∣ = 1 (5) - 2 (2 α - 1) + 3 (α - 3) = 5 - 4 α + 2 + 3 α - 9 = - α - 2

Given

∣ A ∣ = 2

, we set

- α - 2 = 2

, which yields $α = - 4$ .

The Onion Strategy

Peeling the Nested Expression

We must evaluate the expression

∣2 adj (2 adj (2 A)) ∣

. To simplify this, we utilize two fundamental properties for a

3 \times 3

matrix

B

∣ k B ∣ = k^{3} ∣ B ∣

∣ adj (B) ∣ = ∣ B ∣^{2}

We peel the expression layer by layer from the outside in:

∣2 adj (2 adj (2 A)) ∣ = 2^{3} ∣ adj (2 adj (2 A)) ∣

Applying the adjoint property

∣ adj (B) ∣ = ∣ B ∣^{2}

= 2^{3} (∣2 adj (2 A) ∣)^{2}

Extracting the scalar

2

from the inner determinant:

= 2^{3} (2^{3} ∣ adj (2 A) ∣)^{2}

Applying the adjoint property once more:

= 2^{3} (2^{3} (∣2 A ∣)^{2})^{2}

Finally, extracting the innermost scalar

2

= 2^{3} (2^{3} (2^{3} ∣ A ∣)^{2})^{2}

The Final Calculation

Consolidating the powers of

2

, we have:

2^{3} (2^{3} \cdot 2^{6} ∣ A ∣^{2})^{2} = 2^{3} (2^{9} ∣ A ∣^{2})^{2} = 2^{3} \cdot 2^{18} ∣ A ∣^{4} = 2^{21} ∣ A ∣^{4}

Substituting

∣ A ∣ = 2

2^{21} \cdot (2)^{4} = 2^{25}

The problem states this expression equals

3 2^{n}

. Since

32 = 2^{5}

, we have

3 2^{n} = (2^{5})^{n} = 2^{5 n}

Equating the powers,

2^{25} = 2^{5 n}

, which implies

5 n = 25

, or $n = 5$ .

The final required value is:

3 n + α = 3 (5) + (- 4) = 15 - 4 = 11

Common Traps & FAQs

Wait, why is

∣ k A ∣ = k^{3} ∣ A ∣

and not just

k ∣ A ∣

? Shouldn't multiplying a matrix by

k

just multiply the determinant by

k

That is a classic trap! Remember, when you multiply a matrix by a scalar

k

, every single element in the matrix is multiplied by

k

. In a

3 \times 3

matrix, this means every row is multiplied by

k

. Since there are 3 rows, you pull out a factor of

k

from each row, resulting in

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

. For an

n \times n

matrix, the property is

∣ k A ∣ = k^{n} ∣ A ∣

Why is

∣ adj (A) ∣ = ∣ A ∣^{2}

? Where does the square come from?

Great question! The general property for any

n \times n

matrix is

∣ adj (A) ∣ = ∣ A ∣^{n - 1}

. Since we are dealing with a

3 \times 3

matrix,

n = 3

, so

n - 1 = 2

. That is why the determinant of the adjoint is the square of the determinant of the original matrix.

Can I just calculate the adjoint matrix directly instead of using properties?

Technically, yes, but it would be a massive algebraic nightmare! The beauty of JEE Advanced problems is that they are designed to reward those who use properties to simplify the expression before doing any heavy lifting. Always look for the elegant path first!

Animated Solution for Mathematics - Matrices and Determinants: 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

Select Answer:

Visualized Solution

Introduction to Matrix $A$

Expanding $∣ A ∣$

Simplifying the Determinant

Solving for $α$

Properties of Determinants

The Nested Expression

Peeling Layer 1: Outer Scalar

Peeling Layer 2: Outer Adjoint

Peeling Layer 3: Inner Scalar

Peeling Layer 4: Inner Adjoint

Peeling Layer 5: Innermost Scalar

Consolidating Powers of 2

Evaluating the Expression

Solving for $n$

Final Calculation

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Onion Strategy

The Final Calculation

Similar Questions

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

Let $A = 100 0 α β 0 β α$ and $∣2 A ∣^{3} = 2^{21}$ where $α, β \in Z$ , Then a value of $α$ is

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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

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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:

For $α, β \in R$ and a natural number $n$ , let $A_{r} = r 2 r 3 r - 2 123 \frac{n ^{2}}{2} + α n^{2} - β \frac{n ( 3 n - 1 )}{2}$ . Then $2 A_{10} - A_{8}$ 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 ______.

If $A = [α 2 2 α]$ and $∣ A^{3} ∣ = 125$ then the value of $α$ is

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

Animated Solution for Mathematics - Matrices and Determinants: .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

Select Answer:

Visualized Solution

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

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

Simplifying the Determinant

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Properties of Determinants

The Nested Expression

Peeling Layer 1: Outer Scalar

Peeling Layer 2: Outer Adjoint

Peeling Layer 3: Inner Scalar

Peeling Layer 4: Inner Adjoint

Peeling Layer 5: Innermost Scalar

Consolidating Powers of 2

Evaluating the Expression

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Final Calculation

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Onion Strategy

The Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If A is a 3×3 matrix and ∣A∣=2, then ∣3 adj(∣3A∣A2)∣ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A=​100​0αβ​0βα​​ and ∣2A∣3=221 where α,β∈Z, Then a value of α is

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.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

Animated Solution for Mathematics - Matrices and Determinants: 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

Introduction to Matrix $A$

Expanding $∣ A ∣$

Solving for $α$

Solving for $n$

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

Let $A = 100 0 α β 0 β α$ and $∣2 A ∣^{3} = 2^{21}$ where $α, β \in Z$ , Then a value of $α$ is

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 $A$ 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

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

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:

For $α, β \in R$ and a natural number $n$ , let $A_{r} = r 2 r 3 r - 2 123 \frac{n ^{2}}{2} + α n^{2} - β \frac{n ( 3 n - 1 )}{2}$ . Then $2 A_{10} - A_{8}$ 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 ______.

If $A = [α 2 2 α]$ and $∣ A^{3} ∣ = 125$ then the value of $α$ is

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