Let A be a 3 x 3 matrix such that If S = n Z : (|adj(adj A)|)((n - 1)2/2) = |A|3n2 - 5n

Year	2026	2025	2024	Total
Questions	16	27	16	59

Q61 jee_main_2025_24_jan_evening System of Linear Equations

If the system of equations

x+2y-3z=2

2x+λ y+5z=5

14x+3y+μ z=33

has infinitely many solutions, then

λ+μ

is equal to:

A. $13$ $13$
B. $10$ $10$
C. $11$ $11$
D. $12$ $12$

Solution

Related Formula

Cramer\'s rule states that a system of non-homogeneous linear equations has infinitely many solutions if the main determinant $D = 0$ $D = 0$ and the variable determinants $D₁ = D₂ = D₃ = 0$ $D_1 = D_2 = D_3 = 0$.

Core Logic

Set the main determinant $D$ $D$ to zero :

D = vmatrix 1 & 2 & -3 2 & λ & 5 14 & 3 & μ vmatrix = 0

1(λμ - 15) - 2(2μ - 70) - 3(6 - 14λ) = 0

λμ - 15 - 4μ + 140 - 18 + 42λ = 0

λμ + 42λ - 4μ + 107 = 0

Step 1: Use $D₂ = 0$ $D_2 = 0$ to solve for $μ$ $\mu$

Form determinant $D₂$ $D_2$ by replacing the second column with the constant terms vector :

D₂ = vmatrix 1 & 2 & -3 2 & 5 & 5 14 & 33 & μ vmatrix = 0

1(5μ - 165) - 2(2μ - 70) - 3(66 - 70) = 0

5μ - 165 - 4μ + 140 + 12 = 0 \Rightarrow μ - 13 = 0 \Rightarrow μ = 13

Step 2: Solve for $λ$ $\lambda$

Substitute $μ = 13$ $\mu = 13$ back into the first equation derived from $D=0$ $D=0$ :

13λ + 42λ - 4(13) + 107 = 0

55λ - 52 + 107 = 0 \Rightarrow 55λ + 55 = 0 \Rightarrow λ = -1

Thus, $λ + μ = -1 + 13 = 12$ $\lambda + \mu = -1 + 13 = 12$.

Pattern Recognition

When evaluating infinite solutions, look for columns that are easily solvable using $Dᵢ = 0$ $D_i = 0$ forms. Calculating $D₂ = 0$ $D_2 = 0$ avoids dealing with any non-linear products of $λμ$ $\lambda\mu$ directly at the start.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

Q65 jee_main_2025_24_jan_evening Evaluation of Determinants using Limits

For some a, b, let

f(x)= vmatrix a+( x)/(x) & 1 & b a & 1+( x)/(x) & b a & 1 & b+( x)/(x) vmatrix, x\neq0

xarrow0f(x)=λ+μ a+vb

Then

(λ+μ+ν)²

is equal to:

A. $25$ $25$
B. $9$ $9$
C. $36$ $36$
D. $16$ $16$

Solution

Related Formula

The fundamental trigonometric limit is given by:

x \to 0 ( x)/(x) = 1

Core Logic

Apply the limit inside each element of the matrix determinant :

x \to 0 f(x) = vmatrix a+1 & 1 & b a & 2 & b a & 1 & b+1 vmatrix

Step 1: Simplify the Determinant via Row Operations

Perform rows reductions $R₂ \to R₂ - R₁$ $R_2 \to R_2 - R_1$ and $R₃ \to R₃ - R₁$ $R_3 \to R_3 - R_1$ to create zeros:

x \to 0 f(x) = vmatrix a+1 & 1 & b -1 & 1 & 0 -1 & 0 & 1 vmatrix

Expand across the first row :

= (a+1)[1(1) - 0] - 1[-1(1) - 0] + b[0 - (-1)]

= (a+1)(1) + 1 + b = a + b + 2

Step 2: Match Coefficients

Equate this outcome with the target parameter template $λ + μ a + ν b$ $\lambda + \mu a + \nu b$:

λ = 2, μ = 1, ν = 1

Calculate $(λ + μ + ν)²$ $(\lambda + \mu + \nu)^2$:

(2 + 1 + 1)² = 4² = 16

Pattern Recognition

Standard row manipulations on identity-shifted arrays quickly eliminate complex parameter symbols, reducing determinant calculations into straightforward polynomial expansions.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants Class 11 Mathematics: Limits and Derivatives

Q jee_main_2025_24_jan_morning Properties of Matrices and Adjoints

Let

A

be a

3 \times 3

matrix such that

XTAX = 0

for all nonzero

3 \times 1

matrices

X = bmatrix x y z bmatrix

. If

A bmatrix 1 1 1 bmatrix = bmatrix 1 4 -5 bmatrix

,

A bmatrix 1 2 1 bmatrix = bmatrix 0 4 -8 bmatrix

, and

(adj(2(A + I))) = 2α3β5γ

for

α, β, γ in N

, then

α² + β² + γ²

is ________.

Numerical Answer. Answer: 44

Solution

Related Formula

The quadratic form condition $XTAX = 0$ holds for all non-zero vectors $X$ $\mathbf{X}$ if and only if $A$ $A$ is a skew-symmetric matrix. For a $3 \times 3$ $3 \times 3$ skew-symmetric matrix, the components satisfy:

A = bmatrix 0 & x₁ & x₂ -x₁ & 0 & x₃ -x₂ & -x₃ & 0 bmatrix

Core Logic

Let's define the matrix $A$ $A$ using the parameters of a standard skew-symmetric form:

A = bmatrix 0 & a & b -a & 0 & c -b & -c & 0 bmatrix

Apply the first given matrix multiplication vector condition:

A bmatrix 1 1 1 bmatrix = bmatrix a+b -a+c -b-c bmatrix = bmatrix 1 4 -5 bmatrix

This gives the system of linear equations:

a + b = 1 (1)

-a + c = 4 (2)

-b - c = -5 b + c = 5 (3)

Apply the second given matrix multiplication vector condition:

A bmatrix 1 2 1 bmatrix = bmatrix 2a+b -a+c -b-2c bmatrix = bmatrix 0 4 -8 bmatrix

This gives the equation:

2a + b = 0 (4)

Step 1: Solve for Matrix Elements

Subtract equation (1) from equation (4):

(2a + b) - (a + b) = 0 - 1 a = -1

Substitute $a = -1$ $a = -1$ back into equation (1):

-1 + b = 1 b = 2

Substitute $a = -1$ $a = -1$ into equation (2):

-(-1) + c = 4 1 + c = 4 c = 3

Thus, the explicit matrix $A$ $A$ is:

A = bmatrix 0 & -1 & 2 1 & 0 & 3 -2 & -3 & 0 bmatrix

Step 2: Compute Target Matrix Determinant

Construct the modified target matrix $2(A+I)$ $2(A+I)$:

A + I = bmatrix 1 & -1 & 2 1 & 1 & 3 -2 & -3 & 1 bmatrix

2(A + I) = bmatrix 2 & -2 & 4 2 & 2 & 6 -4 & -6 & 2 bmatrix

Calculate its determinant value:

(2(A+I)) = 2[4 - (-36)] - (-2)[4 - (-24)] + 4[-12 - (-8)]

(2(A+I)) = 2[40] + 2[28] + 4[-4] = 80 + 56 - 16 = 120

Step 3: Analyze Adjoint Power and Prime Factors

Using the standard determinant identity for adjoints, $(adj(M)) = ( M)ⁿ⁻¹$ where $n=3$ $n=3$:

(adj(2(A+I))) = (120)³⁻¹ = 120²

Find the prime factorization of the result:

120 = 2³ \cdot 3¹ \cdot 5¹ 120² = (2³ \cdot 3¹ \cdot 5¹)² = 2⁶ \cdot 3² \cdot 5²

This maps the exponents directly to our target variables:

α = 6, β = 2, γ = 2

Finally, calculate the sum of their squares:

α² + β² + γ² = 6² + 2² + 2² = 36 + 4 + 4 = 44

Pattern Recognition

The condition $X^T A X = 0$ $\mathbf{X}^T A \mathbf{X} = 0$ always implies that $A$ $A$ is a skew-symmetric matrix, which instantly forces the diagonal entries to be zero, reducing the number of unknown parameters from 9 down to 3.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

Q70 jee_main_2025_24_jan_morning System of Linear Equations

If the system of equations

2x - y + z = 4

5x + λ y + 3z = 12

100x - 47y + μ z = 212

has infinitely many solutions, then

μ - 2λ

is equal to :

A. $56$ $56$
B. $59$ $59$
C. $55$ $55$
D. $57$ $57$

Solution

Related Formula

According to Cramer's Rule, for a system of linear equations to have infinitely many solutions, the main determinant $Δ$ $\Delta$ and all directional determinants $Δ₁, Δ₂, Δ₃$ $\Delta_1, \Delta_2, \Delta_3$ must equal zero simultaneously.

Core Logic

Set up the directional determinant equation $Δ₃ = 0$ $\Delta_3 = 0$ by replacing the third column with the constant vector:

Δ₃ = | matrix 2 & -1 & 4 5 & λ & 12 100 & -47 & 212 matrix | = 0

Expand the determinant along the first row:

2[212λ - 12(-47)] - (-1)[5(212) - 12(100)] + 4[5(-47) - 100λ] = 0

2[212λ + 564] + 1[1060 - 1200] + 4[-235 - 100λ] = 0

424λ + 1128 - 140 - 940 - 400λ = 0

24λ + 48 = 0 λ = -2

Step 1: Solve for Mu using the main determinant

Set the primary coefficient matrix determinant $Δ = 0$ $\Delta = 0$ and substitute $λ = -2$ $\lambda = -2$:

Δ = | matrix 2 & -1 & 1 5 & -2 & 3 100 & -47 & μ matrix | = 0

Expand the determinant along the first row:

2[-2μ - 3(-47)] - (-1)[5μ - 3(100)] + 1[5(-47) - (-2)(100)] = 0

2[-2μ + 141] + [5μ - 300] + [-235 + 200] = 0

-4μ + 282 + 5μ - 300 - 35 = 0

μ - 53 = 0 μ = 53

Step 2: Calculate the Target Value

Substitute the values of $μ$ $\mu$ and $λ$ $\lambda$ into the expression:

μ - 2λ = 53 - 2(-2) = 53 + 4 = 57

Pattern Recognition

When solving systems of equations for infinite solution parameters, choosing a directional determinant that excludes one of the variables simplifies the problem into two separate single-variable equations.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

Q jee_main_2025_28_jan_evening Properties of Matrices and Powers

Let

A = bmatrix 1\sqrt(2) & -2 0 & 1 bmatrix

and

P = bmatrix θ & - θ θ & θ bmatrix

,

θ > 0

. If

B = PAPT

,

C = PTB¹⁰P

and the sum of the diagonal elements of

C

is

(m)/(n)

where

(m, n) = 1

, then

m+n

is:

A. $65$ $65$
B. $127$ $127$
C. $258$ $258$
D. $2049$ $2049$

Solution

Related Formula

For orthogonal matrix $P$ $P$, $P^T P = P P^T = I$ $P^T P = P P^T = I$.

If $B = PAP^T$ $B = PAP^T$, then:

B^k = (PAP^T)(PAP^T) (PAP^T) = PA^kP^T

Core Logic

Given $C = P^T B¹⁰ P$ $C = P^T B^{10} P$. Substitute $B¹⁰ = P A¹⁰ P^T$ $B^{10} = P A^{10} P^T$ into the expression:

C = P^T (P A¹⁰ P^T) P

C = (P^T P) A¹⁰ (P^T P)

Since $P$ $P$ is an orthogonal rotation matrix, $P^T P = I$ $P^T P = I$, meaning:

C = I \cdot A¹⁰ \cdot I = A¹⁰

Therefore, the sum of diagonal elements of $C$ $C$ is simply the trace of $A¹⁰$ $A^{10}$.

Step 1: Analyze Powers of Upper Triangular Matrix A

Matrix $A$ $A$ is upper triangular:

A = bmatrix 1\sqrt(2) & -2 0 & 1 bmatrix

For any upper triangular matrix, any integer power $k$ $k$ preserves the main diagonal entries as simply the powers of the individual diagonal elements:

A¹⁰ = bmatrix ( 1\sqrt(2))¹⁰ & * 0 & 1¹⁰ bmatrix = bmatrix (1)/(32) & * 0 & 1 bmatrix

Step 2: Calculate the Trace and sum m+n

Sum of diagonal elements = Trace(C) = Trace(A¹⁰) = (1)/(32) + 1 = (33)/(32)

Given $(m)/(n) = (33)/(32)$ $\frac{m}{n} = \frac{33}{32}$ with $(33, 32) = 1$ $\gcd(33, 32) = 1$:

m = 33, n = 32

m + n = 33 + 32 = 65

Pattern Recognition

Traces of matrices are invariant under cyclic permutations, so $Tr(P^T B¹⁰ P) = Tr(P P^T B¹⁰) = Tr(B¹⁰)$ . Furthermore, $Tr(P A¹⁰ P^T) = Tr(A¹⁰)$ . This identity bypasses the need to evaluate any outer matrix multiplication.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

Solution & Explanation

Related Formula

Core Logic

Step 1: Solve for Exponent Parameter

Step 2: Calculate the Target Summation

Pattern Recognition

Chapter Mix

More Matrices and Determinants Previous-Year Questions — Page 8

Solution

Related Formula

Core Logic

Step 1: Use D₂ = 0$D_2 = 0$ to solve for μ$\mu$

Step 2: Solve for λ$\lambda$

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Simplify the Determinant via Row Operations

Step 2: Match Coefficients

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve for Matrix Elements

Step 2: Compute Target Matrix Determinant

Step 3: Analyze Adjoint Power and Prime Factors

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve for Mu using the main determinant

Step 2: Calculate the Target Value

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Analyze Powers of Upper Triangular Matrix A

Step 2: Calculate the Trace and sum m+n

Pattern Recognition

Chapter Mix

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back

Step 1: Use $D₂ = 0$ $D_2 = 0$ to solve for $μ$ $\mu$

Step 2: Solve for $λ$ $\lambda$