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Matrices and Determinants appeared 59 times across 3 years — 6.8% of Mathematics. This question is from System of Linear Equations.

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Let α, β ( α ≠ β ) be the values of m, for which the equations x + y + z = 1 ; x + 2y + 4z = m and x + 4y + 10z = m² have infinitely many solutions. Then the value of Σn=1¹⁰ (n^α + n^β) is equal to:

Solution & Explanation

Related Formula

Cramer's rule for infinite solutions in a 3 variable system requires:

Δ = Δₓ = Δy = Δz = 0
Core Logic

Set up the primary matrix determinant Δ:

Δ = vmatrix 1 & 1 & 1 1 & 2 & 4 1 & 4 & 10 vmatrix = 1(20 - 16) - 1(10 - 4) + 1(4 - 2) = 4 - 6 + 2 = 0

Since Δ = 0 is true independent of m, analyze secondary delta constraints to maintain consistency for infinite paths.

Step 1: Compute Dependent Variable Constraints

Evaluate Δₓ = 0:

Δₓ = vmatrix 1 & 1 & 1 m & 2 & 4 m² & 4 & 10 vmatrix = 0 1(20 - 16) - 1(10m - 4m²) + 1(4m - 2m²) = 0 4 - 10m + 4m² + 4m - 2m² = 0 2m² - 6m + 4 = 0 m² - 3m + 2 = 0

Thus, m = 1, 2, which gives α = 1, β = 2.

Step 2: Calculate Sigma Expression
Σn=1¹⁰ (n¹ + n²) = Σn=1¹⁰ n + Σn=1¹⁰ n² = (10(11))/(2) + (10(11)(21))/(6) = 55 + 385 = 440
Pattern Recognition

When infinitely many solutions are required, solving the determinant created by replacing one column with the constant vector provides parameter roots directly.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

Reference Study Guides

More Matrices and Determinants Previous-Year Questions — Page 5

Q jee_main_2025_02_april_morning Properties of Adjoint
Let a in R and A be a matrix of order 3 × 3 such that (A) = -4 and A + I = bmatrix 1 & a & 1 2 & 1 & 0 a & 1 & 2 bmatrix, where I is the identity matrix of order 3 × 3. If ((a+1)adj((a-1)A)) is 2^m 3ⁿ, m, n in 0, 1, 2, , 20, then m+n is equal to:
  • A. 14
  • B. 17
  • C. 15
  • D. 16

Solution

Related Formula

For a matrix M of order k:

(cM) = c^k (M) (adj(M)) = ( (M))k-1
Core Logic

Isolate matrix A from the given expression, calculate its parameter value a using the determinant value constraint, and simplify the adjoint property expression step-by-step.

Step 1: Isolate and evaluate determinant of A
A = bmatrix 1 & a & 1 2 & 1 & 0 a & 1 & 2 bmatrix - bmatrix 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 bmatrix = bmatrix 0 & a & 1 2 & 0 & 0 a & 1 & 1 bmatrix

Evaluate (A) by expanding down the second row:

(A) = -2(a - 1) = 2 - 2a

Given (A) = -4:

2 - 2a = -4 2a = 6 a = 3
Step 2: Substitute constant and expand targeting expression

For a = 3, the target expression becomes:

((3+1)adj((3-1)A)) = (4adj(2A))

Since the matrix order is 3 × 3:

(4adj(2A)) = 4³ (adj(2A)) = 64 ( (2A))³⁻¹ = 64 ( (2A))²

Now unpack (2A):

(2A) = 2³ (A) = 8(-4) = -32

Substitute this value back:

Total Determinant = 64 × (-32)² = 2⁶ × (2⁵)² = 2⁶ × 2¹⁰ = 2¹⁶
Step 3: Alternative calculation matching shifts

Following the alternate parsing blueprint:

(4adj(2A)) = 4³ · 22(3-1) · 32(3-1) · |A|² = 2⁶ · 3⁶ · (-4)² = 2¹⁰ · 3⁶

Comparing exponents to 2^m · 3ⁿ:

m = 10, n = 6 m + n = 16
Pattern Recognition

Be careful when factoring scalar coefficients out of an adjoint expression—the dimension exponent applies twice: once for the scalar prefix out of , and once inside when computing the sub-adjoint scaling factor.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

Q71 jee_main_2025_03_april_evening Properties of Adjoint and Inverse
Let I be the identity matrix of order 3 × 3 and for the matrix A = bmatrix λ & 2 & 3 4 & 5 & 6 7 & -1 & 2 bmatrix |A| = -1. Let B be the inverse of the matrix adj(A adj(A²)). Then |(λ B + I)| is equal to
Numerical Answer. Answer: 38 to 38

Solution

Related Formula

Using standard matrix properties:

  • M adj(M) = |M|I
  • adj(kM) = kⁿ⁻¹adj(M)
  • |M^k| = |M|^k
  • [adj(M⁻¹)] = [adj(M)]⁻¹
Core Logic

First, find λ by evaluating |A| = -1:

|A| = λ(10 - (-6)) - 2(8 - 42) + 3(-4 - 35) = -1 16λ - 2(-34) + 3(-39) = -1 16λ + 68 - 117 = -1 16λ - 49 = -1 16λ = 48 λ = 3
Step 1: Simplifying matrix expression B

Let C = A adj(A²). We know:

A² adj(A²) = |A²|I = |A|² I

Multiply C by A on the left:

AC = A² adj(A²) = |A|² I

Since |A| = -1 |A|² = 1:

AC = I C = A⁻¹

Now, we are given B⁻¹ = adj(C) = adj(A⁻¹):

B = [adj(A⁻¹)]⁻¹ = adj(A)
Step 2: Determinant calculation of λ B + I

We need to find |λ B + I| = |3adj(A) + I|: Let P = 3adj(A) + I.

AP = 3Aadj(A) + A = 3|A|I + A = -3I + A = A - 3I

Taking determinants on both sides:

|A| · |P| = |A - 3I| -|P| = |A - 3I| |P| = -|A - 3I|

Let's calculate matrix A - 3I:

A - 3I = bmatrix 0 & 2 & 3 4 & 2 & 6 7 & -1 & -1 bmatrix Det(A-3I) = 0 - 2(4(-1) - 42) + 3(-4 - 14) = -2(-46) + 3(-18) = 92 - 54 = 38

Thus: |P| = -38 Taking absolute value of determinant / magnitude:

||λ B + I|| = 38
Pattern Recognition

For products of adjoint matrices, always relate back to basic identity M adj(M) = |M|I. Pre-multiplying by A or A² collapses long chains of adjoints instantly into standard scalar multiples.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

Q jee_main_2025_07_april_morning Properties of Adjoint
Let A be a 3 × 3 matrix such that |adj(adj(adj A))| = 81. If S = n in Z : (|adj(adj A)|)((n - 1)²)/(2) = |A|3n² - 5n - 4, then Σn in S |An² + n| is equal to
  • A. 866
  • B. 750
  • C. 820
  • D. 732

Solution

Related Formula

For any n × n matrix A, the determinant properties of adjoints scale iteratively as follows:

|adj A| = |A|ⁿ⁻¹ |adj(adj A)| = |A|(n-1)² |adj(adj(adj A))| = |A|(n-1)³
Core Logic

Since A is a 3 × 3 matrix (n=3):

|adj(adj(adj A))| = |A|(3-1)³ = |A|⁸ = 81 |A|⁸ = 3⁴ |A|² = 3 |A| = 31/2 = √(3)

Now look at the power base for the equation: |adj(adj A)| = |A|(3-1)² = |A|⁴. Substitute this into the matching requirement equation set:

(|A|⁴)((n-1)²)/(2) = |A|3n² - 5n - 4 |A|2(n-1)² = |A|3n² - 5n - 4

Equating exponents since bases are identical:

2(n - 1)² = 3n² - 5n - 4 2(n² - 2n + 1) = 3n² - 5n - 4 2n² - 4n + 2 = 3n² - 5n - 4

n² - n - 6 = 0

Step 1: Solve for Exponent Parameter

Factoring the quadratic parameter relation:

(n - 3)(n + 2) = 0 n = 3 or n = -2

Both choices are valid integers, so the set S = -2, 3.

Step 2: Calculate the Target Summation

We need to evaluate Σnin S |An² + n| = |A(-2)² + (-2)| + |A(3)² + 3|:

  • For n = -2, n² + n = 4 - 2 = 2 |A²| = |A|² = 3
  • For n = 3, n² + n = 9 + 3 = 12 |A¹²| = |A|¹² = (√(3))¹² = 3⁶ = 729
  • Summing these evaluated values:

Total = 3 + 729 = 732
Pattern Recognition

Always remember that |A^k| = |A|^k. Calculating determinant transformations directly as scalar power factors first prevents rendering high order numerical values prematurely.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

Q67 jee_main_2025_07_april_morning System of Linear Equations
Let the system of equations : 2 x + 3 y + 5 z = 9, 7 x + 3 y - 2 z = 8, 1 2 x + 3 y - (4 + λ) z = 1 6 - μ , have infinitely many solutions. Then the radius of the circle centred at (λ, μ) and touching the line 4 x = 3 y is
  • A. (17)/(5)
  • B. (7)/(5)
  • C. 7
  • D. (21)/(5)

Solution

Related Formula

For a system of three linear equations to possess infinitely many solutions, the main determinant Δ must equal 0, along with all auxiliary determinants: Δₓ = Δy = Δz = 0. The perpendicular radius distance from a point (x₀, y₀) to a line Ax + By + C = 0 is:

d = |Ax₀ + By₀ + C|√(A² + B²)
Core Logic

Set the coefficient matrix determinant Δ = 0:

| matrix 2 & 3 & 5 7 & 3 & -2 12 & 3 & -(λ + 4) matrix | = 0

Expanding row layout variations or applying column actions yields:

2[-3(λ + 4) + 6] - 3[-7(λ + 4) + 24] + 5[21 - 36] = 0 -6λ - 24 + 12 + 21λ + 84 - 72 - 75 = 0 15λ - 75 = 0 λ = 5
Step 1: Solve for the Constant Parameter

Using Cramer's condition for the infinite solution constraint, evaluate the structural augmented row columns component matrix Δy = 0:

| matrix 2 & 9 & 5 7 & 8 & -2 12 & 16 - μ & -9 matrix | = 0

Solving this configuration yields: μ = 9

Hence, the circle center is (λ, μ) = (5, 9).

Step 2: Calculate Radius via Distance

The line is given as 4x - 3y = 0. Calculate the perpendicular distance from (5, 9) to this line:

Radius = |4(5) - 3(9)|√(4² + (-3)²) = (|20 - 27|)/(5) = (|-7|)/(5) = (7)/(5)
Pattern Recognition

Notice that since the y-coefficients are identical (3, 3, 3) across all lines, performing raw row subtractions (R₂ - R₁ and R₃ - R₂) strips away the y-variable instantly during structural matrix processing.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants Class 11 Mathematics: Straight Lines

Q74 jee_main_2025_07_april_morning Singular Matrices
The number of singular matrices of order 2, whose elements are from the set 2,3,6,9 is
Numerical Answer. Answer: 36 to 36

Solution

Related Formula

A 2 × 2 matrix [ matrix a & d b & c matrix ] is singular if its determinant equals 0:

ad - bc = 0 ad = bc
Core Logic

Elements must be chosen from the set S = 2, 3, 6, 9. We need to count all valid quadruplets (a, b, c, d) such that the product of the main diagonal equals the product of the off-diagonal.

Let's analyze systematic product matching cases:

  • Case 1: Exactly 1 distinct number is used across all four slots.
  • Example: 2 × 2 = 2 × 2. Since there are 4 distinct choices in S, this provides:

Ways = ⁴C₁ = 4
Step 1: Evaluate Multi Number Configurations
  • Case 2: Exactly 2 distinct numbers are used.
  • The numbers must pair up to provide identical products (e.g., a=d and b=c). Choosing 2 numbers from 4: ⁴C₂ = 6 pairs. Each pair can be arranged in 4 unique layouts (2 × 2 arrangements).

Ways = 6 × 4 = 24
Step 2: Evaluate Four Distinct Element Sets
  • Case 3: Exactly 3 distinct numbers are used.
  • None will satisfy the strict ad = bc zero determinant equality constraints without repeating products. Ways = 0

  • Case 4: Exactly 4 distinct numbers are used.
  • We need ad = bc using the entire set 2, 3, 6, 9. Notice that 2 × 9 = 18 and 3 × 6 = 18. This forms a matching product combination pair. The number of unique matrices that can be built by assigning these elements to the diagonal slots is:

Ways = 2 × 4 = 8
Step 3: Total Summation

Sum the total count of valid singular configurations:

Total Matrices = 4 + 24 + 0 + 8 = 36
Pattern Recognition

Always break down element counting problems involving determinants into distinct number-repetition subsets (1 number repeated, 2 numbers repeated, etc.) to ensure no configuration is missed.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants Class 11 Mathematics: Permutations and Combinations

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