JEE Main 2023 (31 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: 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 ______.

Q: Why is $|A^{-1}| = 1/|A|$? I always forget this property.

Think about the relationship $AA^{-1} = I$. If you take the determinant of both sides, you get $|AA^{-1}| = |I|$. Using the property that the determinant of a product is the product of the determinants, we get $|A| \cdot |A^{-1}| = 1$. Therefore, $|A^{-1}| = 1/|A|$. It is a beautiful, logical consequence of the definition of an inverse matrix!

Enter Numerical Value:

Visualized Solution

Given Information

Matrix $A$ is of order $n \times n$ .
$∣ A ∣ = 2$
$∣ adj (2 adj (2 A^{- 1})) ∣ = 2^{84}$

The Adjoint Property

Property: For any $n \times n$ matrix $M$ , $∣ adj (M) ∣ = ∣ M ∣^{n - 1}$

Applying Adjoint Property

Let $M = 2 adj (2 A^{- 1})$
$∣2 adj (2 A^{- 1}) ∣^{n - 1} = 2^{84}$

The Scalar Multiplication Property

Property: For an $n \times n$ matrix $M$ and scalar $k$ , $∣ k M ∣ = k^{n} ∣ M ∣$

Pulling Out the Scalar

Applying $∣ k M ∣ = k^{n} ∣ M ∣$ with $k = 2$ :
$(2^{n} ∣ adj (2 A^{- 1}) ∣)^{n - 1} = 2^{84}$

Distributing the Exponent

$2^{n (n - 1)} ∣ adj (2 A^{- 1}) ∣^{n - 1} = 2^{84}$

Applying Adjoint Property Again

Apply $∣ adj (M) ∣ = ∣ M ∣^{n - 1}$ to the inner adjoint:
$2^{n (n - 1)} (∣2 A^{- 1} ∣^{n - 1})^{n - 1} = 2^{84}$

Simplifying the Exponent

Multiply the exponents: $(n - 1) \times (n - 1) = (n - 1)^{2}$
$2^{n (n - 1)} ∣2 A^{- 1} ∣^{(n - 1)^{2}} = 2^{84}$

Pulling Out the Inner Scalar

Apply $∣ k M ∣ = k^{n} ∣ M ∣$ again for $k = 2$ :
$2^{n (n - 1)} (2^{n} ∣ A^{- 1} ∣)^{(n - 1)^{2}} = 2^{84}$

Distributing the Exponent Again

$2^{n (n - 1)} \cdot 2^{n (n - 1)^{2}} \cdot ∣ A^{- 1} ∣^{(n - 1)^{2}} = 2^{84}$

Determinant of Inverse Matrix

Property: $∣ A^{- 1} ∣ = \frac{1}{∣ A ∣}$
Given $∣ A ∣ = 2$ , so $∣ A^{- 1} ∣ = \frac{1}{2} = 2^{- 1}$

Substituting the Value

Substitute $∣ A^{- 1} ∣ = 2^{- 1}$ :
$2^{n (n - 1)} \cdot 2^{n (n - 1)^{2}} \cdot (2^{- 1})^{(n - 1)^{2}} = 2^{84}$
$2^{n (n - 1)} \cdot 2^{n (n - 1)^{2}} \cdot 2^{- (n - 1)^{2}} = 2^{84}$

Combining the Exponents

Since bases are the same, add the exponents:
$2^{n (n - 1) + n (n - 1)^{2} - (n - 1)^{2}} = 2^{84}$

Equating the Exponents

Equate the powers of 2:
$n (n - 1) + n (n - 1)^{2} - (n - 1)^{2} = 84$
Factor out $(n - 1)$ :
$(n - 1) [n + n (n - 1) - (n - 1)] = 84$

Solving for n

Simplify inside the bracket:
$(n - 1) [n + n^{2} - n - n + 1] = 84$
$(n - 1) (n^{2} - n + 1) = 84$
By trial or inspection, for $n = 5$ :
$(4) (25 - 5 + 1) = 4 \times 21 = 84$
Final Answer: $n = 5$

00:00 / 00:00

The Sigma Insight: Properties of Determinants

Analyzing the Setup

We are given the matrix equation:

∣ adj (2 adj (2 A^{- 1})) ∣ = 2^{84}

where

A

is an

n \times n

matrix and

∣ A ∣ = 2

. Our goal is to determine the order

n

of the matrix.

We utilize the fundamental property of the adjoint: for any

n \times n

matrix

M

∣ adj (M) ∣ = ∣ M ∣^{n - 1}

. By treating the expression

2 adj (2 A^{- 1})

M

, we simplify the outermost layer:

∣2 adj (2 A^{- 1}) ∣^{n - 1} = 2^{84}

The Scalar Trap

When extracting a scalar

k

from an

n \times n

determinant, we use the property

∣ k M ∣ = k^{n} ∣ M ∣

. Applying this to the scalar

2

inside the determinant, we obtain:

(2^{n} ∣ adj (2 A^{- 1}) ∣)^{n - 1} = 2^{84}

Distributing the exponent

(n - 1)

, we arrive at:

2^{n (n - 1)} ∣ adj (2 A^{- 1}) ∣^{n - 1} = 2^{84}

Peeling the Inner Core

We apply the adjoint property again to the inner term:

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

. Substituting this into our equation yields:

2^{n (n - 1)} (∣2 A^{- 1} ∣^{n - 1})^{n - 1} = 2^{84}

2^{n (n - 1)} ∣2 A^{- 1} ∣^{(n - 1)^{2}} = 2^{84}

Next, we extract the scalar

2

from

∣2 A^{- 1} ∣

, noting that

∣2 A^{- 1} ∣ = 2^{n} ∣ A^{- 1} ∣

. Substituting this back, the equation becomes:

2^{n (n - 1)} (2^{n} ∣ A^{- 1} ∣)^{(n - 1)^{2}} = 2^{84}

2^{n (n - 1)} \cdot 2^{n (n - 1)^{2}} \cdot ∣ A^{- 1} ∣^{(n - 1)^{2}} = 2^{84}

The Final Revelation

Given

∣ A ∣ = 2

, it follows that

∣ A^{- 1} ∣ = \frac{1}{∣ A ∣} = 2^{- 1}

. Substituting this value into the equation:

2^{n (n - 1)} \cdot 2^{n (n - 1)^{2}} \cdot (2^{- 1})^{(n - 1)^{2}} = 2^{84}

Equating the exponents of base

2

n (n - 1) + n (n - 1)^{2} - (n - 1)^{2} = 84

Factoring out

(n - 1)

, we simplify the expression:

(n - 1) [n + n (n - 1) - (n - 1)] = 84

(n - 1) (n^{2} - n + 1) = 84

Testing integer values for

n

, we find that for

n = 5

4 \times (25 - 5 + 1) = 4 \times 21 = 84

Thus, the order of the matrix is $n = 5$ .

Common Traps & FAQs

Wait, why does the scalar

k

come out as

k^{n}

and not just

k

? This feels counter-intuitive.

That is a fantastic question! Think about the definition of a determinant. A determinant is a sum of products where each term contains exactly one element from each row and each column. If you multiply every element in an

n \times n

matrix by

k

, you are essentially multiplying every row by

k

. Since there are

n

rows, you factor out a

k

from each of the

n

rows, resulting in

k^{n}

being pulled out. It is not just a rule; it is a direct consequence of the multilinearity of the determinant!

Is there a shortcut to avoid the long exponent algebra? I am terrified of making a calculation error.

I completely understand that fear! The best way to avoid errors is to be methodical. Do not try to expand everything at once. Treat each nested layer as a separate entity. Once you have the expression in terms of powers of 2, you are just solving a polynomial equation. If you are stuck, test small integer values for

n

—often, in JEE problems,

n

is a small integer, and testing values can be a valid strategy if you are pressed for time.

Why is

∣ A^{- 1} ∣ = 1/∣ A ∣

? I always forget this property.

Think about the relationship

A A^{- 1} = I

. If you take the determinant of both sides, you get

∣ A A^{- 1} ∣ = ∣ I ∣

. Using the property that the determinant of a product is the product of the determinants, we get

∣ A ∣ \cdot ∣ A^{- 1} ∣ = 1

. Therefore,

∣ A^{- 1} ∣ = 1/∣ A ∣

. It is a beautiful, logical consequence of the definition of an inverse matrix!

Animated Solution for Mathematics - Matrices and Determinants: 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 ______.

Enter Numerical Value:

Visualized Solution

Given Information

The Adjoint Property

Applying Adjoint Property

The Scalar Multiplication Property

Pulling Out the Scalar

Distributing the Exponent

Applying Adjoint Property Again

Simplifying the Exponent

Pulling Out the Inner Scalar

Distributing the Exponent Again

Determinant of Inverse Matrix

Substituting the Value

Combining the Exponents

Equating the Exponents

Solving for n

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Scalar Trap

Peeling the Inner Core

The Final Revelation

Similar Questions

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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

The trace of a square matrix is defined to be the sum of its diagonal entries. If $A$ is a $2 \times 2$ matrix such that the trace of $A$ is 3 and the trace of $A^{3}$ is $- 18$ , then the value of the determinant of $A$ is ____.

Let $P = [a_{ij}]$ be a $3 \times 3$ matrix and let $Q = [b_{ij}]$ , where $b_{ij} = 2^{i + j} a_{ij}$ for $1 \leq i, j \leq 3$ . If the determinant of $P$ is 2, then the determinant of the matrix $Q$ is

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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 $A$ 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:

Enter Numerical Value:

Visualized Solution

Given Information

The Adjoint Property

Applying Adjoint Property

The Scalar Multiplication Property

Pulling Out the Scalar

Distributing the Exponent

Applying Adjoint Property Again

Simplifying the Exponent

Pulling Out the Inner Scalar

Distributing the Exponent Again

Determinant of Inverse Matrix

Substituting the Value

Combining the Exponents

Equating the Exponents

Solving for n

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Scalar Trap

Peeling the Inner Core

The Final Revelation

Similar Questions

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