JEE Main 2021 (20 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: Let $A = {a_{ij}}$ be a $3 \times 3$ matrix, where $a_{ij} = ⎩ ⎨ ⎧ (- 1)^{j - i} 2 (- 1)^{i + j} if i < j if i = j if i > j$ then $det (3 Adj (2 A^{- 1}))$ is equal to

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: The Spark. To solve this, you must first construct the matrix

A

by carefully applying the piecewise definition for each element

a_{ij}

. Pay close attention to the conditions

i < j

i = j

, and

i > j

. Do not rush the sign calculations, as a single error here will propagate through the entire determinant calculation.

Hint 2

Hint 2: The Arsenal. You do not need to calculate the inverse or the adjoint matrix directly. Instead, leverage the powerful properties of determinants:

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

det (Adj (M)) = (det M)^{n - 1}

, and

det (A^{- 1}) = \frac{1}{d e t ( A )}

Hint 3

Hint 3: The Execution. Start by expanding the target expression

det (3 Adj (2 A^{- 1}))

using the scalar property. It becomes

3^{3} det (Adj (2 A^{- 1}))

. Then, apply the adjoint property to simplify the expression further into a form involving

det (A)

Visualized Solution

Defining the Matrix $A$

Given a $3 \times 3$ matrix $A = {a_{ij}}$
Element definition: $a_{ij} = ⎩ ⎨ ⎧ (- 1)^{j - i} 2 (- 1)^{i + j} i < j i = j i > j$

Diagonal Elements ( $i = j$ )

For $i = j$ : $a_{11} = a_{22} = a_{33} = 2$

Upper Triangle Elements ( $i < j$ )

For $i < j$ : $a_{ij} = (- 1)^{j - i}$
Calculations: $a_{12} = (- 1)^{1} = - 1$ , $a_{13} = (- 1)^{2} = 1$ , $a_{23} = (- 1)^{1} = - 1$

Lower Triangle Elements ( $i > j$ )

For $i > j$ : $a_{ij} = (- 1)^{i + j}$
Calculations: $a_{21} = (- 1)^{3} = - 1$ , $a_{31} = (- 1)^{4} = 1$ , $a_{32} = (- 1)^{5} = - 1$

Calculating $det (A)$

$A = 2 - 1 1 - 1 2 - 1 1 - 1 2$
$det (A) = 2 (4 - 1) - (- 1) (- 2 + 1) + 1 (1 - 2)$

Simplifying $det (A)$

$det (A) = 6 - 1 - 1 = 4$

The Target Expression

Target: $det (3 Adj (2 A^{- 1}))$

Property: Scalar Multiplication

Property: $det (k M) = k^{n} det (M)$
Here $n = 3, k = 3$ , so $det (3 Adj (2 A^{- 1})) = 3^{3} det (Adj (2 A^{- 1})) = 27 det (Adj (2 A^{- 1}))$

Property: Adjoint Determinant

Property: $det (Adj (M)) = (det M)^{n - 1}$
Here $n = 3$ , so $det (Adj (2 A^{- 1})) = (det (2 A^{- 1}))^{2}$

Scaling Inside the Determinant

Inside the square: $det (2 A^{- 1}) = 2^{3} det (A^{- 1}) = 8 det (A^{- 1})$

Property: Inverse Determinant

Property: $det (A^{- 1}) = \frac{1}{d e t ( A )}$
Since $det (A) = 4$ , $det (A^{- 1}) = \frac{1}{4}$

Final Substitution

Expression: $27 \times (8 \times \frac{1}{4})^{2}$
$= 27 \times (2)^{2}$

The Final Answer

$27 \times 4 = 108$
Final Answer: 108

00:00 / 00:00

The Sigma Insight: Adjoint and Inverse of a Matrix

Solution Diagram

Analyzing the Setup

To begin, we define the

3 \times 3

matrix

A

based on the provided piecewise rules. For the diagonal elements where

i = j

, the rule is

a_{ij} = 2

, yielding:

a_{11} = a_{22} = a_{33} = 2

For the upper triangle where

i < j

, we use

a_{ij} = (- 1)^{j - i}

. This results in:

a_{12} = (- 1)^{2 - 1} = - 1, a_{13} = (- 1)^{3 - 1} = 1, a_{23} = (- 1)^{3 - 2} = - 1

For the lower triangle where

i > j

, we use

a_{ij} = (- 1)^{i + j}

. This gives:

a_{21} = (- 1)^{2 + 1} = - 1, a_{31} = (- 1)^{3 + 1} = 1, a_{32} = (- 1)^{3 + 2} = - 1

The resulting matrix

A

is:

A = 2 - 1 1 - 1 2 - 1 1 - 1 2

The Heartbeat

Calculating the Determinant

We calculate the determinant of

A

by expanding along the first row:

det (A) = 2 (4 - 1) - (- 1) (- 2 + 1) + 1 (1 - 2)

Simplifying the expression, we obtain:

det (A) = 2 (3) + 1 (- 1) + 1 (- 1) = 6 - 1 - 1 = 4

Thus, the determinant of the matrix is $4$ .

The Property Dance

We now evaluate the target expression:

det (3 Adj (2 A^{- 1}))

. Applying the scalar property

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

with

n = 3

and

k = 3

, we extract the scalar:

3^{3} det (Adj (2 A^{- 1})) = 27 det (Adj (2 A^{- 1}))

Next, we utilize the property

det (Adj (M)) = (det M)^{n - 1}

. With

n = 3

, the expression simplifies to:

27 (det (2 A^{- 1}))^{2}

The Final Crescendo

Inside the square, we evaluate

det (2 A^{- 1})

. Applying the scalar property again, we extract

2^{3}

det (2 A^{- 1}) = 2^{3} det (A^{- 1}) = 8 det (A^{- 1})

Since

det (A^{- 1}) = \frac{1}{d e t ( A )}

and

det (A) = 4

, we find

det (A^{- 1}) = \frac{1}{4}

. Substituting this back, the inner term becomes:

8 \times \frac{1}{4} = 2

Finally, we compute the total value:

27 \times (2)^{2} = 27 \times 4 = 108

The final answer is $108$ .

Common Traps & FAQs

Wait, why is

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

and not just

k det (A)

This is a classic point of confusion! Remember that when you multiply a matrix by a scalar

k

, you are multiplying *every* row by

k

. Since a

3 \times 3

matrix has 3 rows, you factor out

k

from each row, resulting in

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

being pulled out of the determinant. For an

n \times n

matrix, this generalizes to

k^{n}

Can I just calculate

A^{- 1}

and then find the adjoint?

Technically, yes, but it is a massive trap! In JEE Advanced, time is your most precious resource. Calculating the inverse and then the adjoint is computationally expensive and prone to arithmetic errors. Using determinant properties allows you to solve the problem in seconds without ever needing to find the inverse matrix itself.

Why is

det (Adj (A)) = (det A)^{n - 1}

Think of the definition of the adjoint: it is the transpose of the cofactor matrix. The determinant of the cofactor matrix is related to the determinant of the original matrix. Specifically,

A \cdot Adj (A) = det (A) I

. Taking the determinant of both sides gives

det (A) \cdot det (Adj (A)) = det (det (A) I) = (det A)^{n} det (I)

. Dividing by

det (A)

yields the result

(det A)^{n - 1}

Animated Solution for Mathematics - Matrices and Determinants: Let $A = {a_{ij}}$ be a $3 \times 3$ matrix, where $a_{ij} = ⎩ ⎨ ⎧ (- 1)^{j - i} 2 (- 1)^{i + j} if i < j if i = j if i > j$ then $det (3 Adj (2 A^{- 1}))$ is equal to

Enter Numerical Value:

Visualized Solution

Defining the Matrix $A$

Diagonal Elements ( $i = j$ )

Upper Triangle Elements ( $i < j$ )

Lower Triangle Elements ( $i > j$ )

Calculating $det (A)$

Simplifying $det (A)$

The Target Expression

Property: Scalar Multiplication

Property: Adjoint Determinant

Scaling Inside the Determinant

Property: Inverse Determinant

Final Substitution

The Final Answer

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

The Heartbeat

The Property Dance

The Final Crescendo

Similar Questions

Let $P = [p_{ij}]$ and $Q = [q_{ij}]$ be two square matrices of order 3 such that $q_{ij} = 2^{(i + j - 1)} p_{ij}$ and $det (Q) = 2^{10}$ . Then the value of $det (adj (adj P))$ is:

Let the matrix $A = 001100010$ and the matrix $B_{0} = A^{49} + 2 A^{98}$ . If $B_{n} = adj (B_{n - 1})$ for all $n \geq 1$ , then $det (B_{4})$ is equal to :

Let $A$ be a $3 \times 3$ real matrix. If $det (2 Adj (2 Adj (Adj (2 A)))) = 2^{41}$ , then the value of $det (A^{2})$ equal .

Let $α \in (0, \infty)$ and $A = 110201 α 12$ . If $det (adj (2 A - A^{T}) \cdot adj (A - 2 A^{T})) = 2^{8}$ , then $(det (A))^{2}$ is equal to:

Let A be a $3 \times 3$ matrix such that $A + A^{T} = O$ . If $A 1 - 1 0 = 332, A^{2} 1 - 1 0 = - 3 19 - 24$ and $det (adj (2 adj (A + I))) = (2)^{α} \cdot (3)^{β} \cdot (11)^{γ}, α, β, γ$ are non-negative integers, then $α + β + γ$ is equal to .........

Let $J_{n, m} = \int_{0}^{1/2} \frac{x ^{n}}{x ^{m} - 1} d x, \forall n > m$ and $n, m \in N$ . Consider a matrix $A = [a_{ij}]_{3 \times 3}$ where $a_{ij} = {J_{6 + i, 3} - J_{i + 3, 3}, 0, i \leq j i > j$ . Then $∣ adj A^{- 1} ∣$ is :

Let $A$ be a $3 \times 3$ matrix such that $∣ adj (adj (adj A)) ∣ = 1 2^{4}$ . Then $∣ A^{- 1} adj A ∣$ is equal to

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

Let $a \in R$ and $A$ be a matrix of order $3 \times 3$ such that $det (A) = - 4$ and $A + I = 12 a a 11 102$ , where $I$ is the identity matrix of order $3 \times 3$ . If $det ((a + 1) adj ((a - 1) A))$ is $2^{m} 3^{n}$ , $m, n \in {0, 1, 2, \dots, 20}$ , then $m + n$ is equal to :

For a $3 \times 3$ matrix M, let trace(M) denote the sum of all the diagonal elements of M. Let A be a $3 \times 3$ matrix such that $∣ A ∣ = \frac{1}{2}$ and trace $(A) = 3$ . If $B = a d j (a d j (2 A))$ , then the value of $∣ B ∣ +$ trace $(B)$ equals :

Enter Numerical Value:

Visualized Solution

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

Diagonal Elements ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } i=j)

Upper Triangle Elements ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } i<j)

Lower Triangle Elements ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } i>j)

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

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

The Target Expression

Property: Scalar Multiplication

Property: Adjoint Determinant

Scaling Inside the Determinant

Property: Inverse Determinant

Final Substitution

The Final Answer

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

The Heartbeat

The Property Dance

The Final Crescendo

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P=[pij​] and Q=[qij​] be two square matrices of order 3 such that qij​=2(i+j−1)pij​ and det(Q)=210. Then the value of det(adj(adj P)) is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the matrix A=​001​100​010​​ and the matrix B0​=A49+2A98. If Bn​=adj(Bn−1​) for all n≥1, then det(B4​) 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 3×3 real matrix. If det(2 Adj(2 Adj(Adj(2A))))=241, then the value of det(A2) equal .

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α∈(0,∞) and A=​110​201​α12​​. If det(adj(2A−AT)⋅adj(A−2AT))=28, then (det(A))2 is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Jn,m​=∫01/2​xm−1xn​dx,∀n>m and n,m∈N. Consider a matrix A=[aij​]3×3​ where aij​={J6+i,3​−Ji+3,3​,0,​i≤ji>j​. Then ∣adj A−1∣ is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A be a 3×3 matrix such that ∣adj(adj(adjA))∣=124. Then ∣A−1adjA∣ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A=​246​2+p6+2p12+3p​2+p+q8+3p+2q20+6p+3q​​. If det(adj(adj(3A)))=2m⋅3n,m,n∈N, then m+n is equal to

Defining the Matrix $A$

Diagonal Elements ( $i = j$ )

Upper Triangle Elements ( $i < j$ )

Lower Triangle Elements ( $i > j$ )

Calculating $det (A)$

Simplifying $det (A)$

Let $P = [p_{ij}]$ and $Q = [q_{ij}]$ be two square matrices of order 3 such that $q_{ij} = 2^{(i + j - 1)} p_{ij}$ and $det (Q) = 2^{10}$ . Then the value of $det (adj (adj P))$ is:

Let the matrix $A = 001100010$ and the matrix $B_{0} = A^{49} + 2 A^{98}$ . If $B_{n} = adj (B_{n - 1})$ for all $n \geq 1$ , then $det (B_{4})$ is equal to :

Let $A$ be a $3 \times 3$ real matrix. If $det (2 Adj (2 Adj (Adj (2 A)))) = 2^{41}$ , then the value of $det (A^{2})$ equal .

Let $α \in (0, \infty)$ and $A = 110201 α 12$ . If $det (adj (2 A - A^{T}) \cdot adj (A - 2 A^{T})) = 2^{8}$ , then $(det (A))^{2}$ is equal to:

Let $J_{n, m} = \int_{0}^{1/2} \frac{x ^{n}}{x ^{m} - 1} d x, \forall n > m$ and $n, m \in N$ . Consider a matrix $A = [a_{ij}]_{3 \times 3}$ where $a_{ij} = {J_{6 + i, 3} - J_{i + 3, 3}, 0, i \leq j i > j$ . Then $∣ adj A^{- 1} ∣$ is :

Let $A$ be a $3 \times 3$ matrix such that $∣ adj (adj (adj A)) ∣ = 1 2^{4}$ . Then $∣ A^{- 1} adj A ∣$ is equal to

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