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Matrices and Determinants appeared 59 times across 3 years — 6.8% of Mathematics. This question is from Properties of Adjoint.

Year 2026 2025 2024 Total
Questions 16 27 16 59

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

Solution & Explanation

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

More Matrices and Determinants Previous-Year Questions — Page 10

Q1 jee_main_2024_29_january_evening Properties of Determinants
Let A = bmatrix 2 & 1 & 2 6 & 2 & 11 3 & 3 & 2 bmatrix and P = bmatrix 1 & 2 & 0 5 & 0 & 2 7 & 1 & 5 bmatrix. The sum of the prime factors of |P⁻¹AP - 2I| is equal to
  • A. 26
  • B. 27
  • C. 66
  • D. 23

Solution

Related Formula
|P⁻¹AP - 2I| = |P⁻¹(A - 2I)P| = |P⁻¹| · |A - 2I| · |P| = |A - 2I|
Core Logic

Since |P⁻¹| · |P| = 1, the expression simplifies completely to the determinant of A - 2I.

First, let us construct the matrix A - 2I:

A - 2I = bmatrix 2-2 & 1 & 2 6 & 2-2 & 11 3 & 3 & 2-2 bmatrix = bmatrix 0 & 1 & 2 6 & 0 & 11 3 & 3 & 0 bmatrix

Now, evaluating the determinant:

|A - 2I| = 0(0 - 33) - 1(0 - 33) + 2(18 - 0) = 33 + 36 = 69
Step 1: Finding Prime Factors

The number obtained is 69. Let us find its prime factorization:

69 = 3 × 23

Both 3 and 23 are prime numbers. Their sum is:

Sum = 3 + 23 = 26
Pattern Recognition

Whenever you encounter a matrix expression of the form P⁻¹AP - kI, always factor out P⁻¹ and P to simplify it to |A - kI|. This saves tremendous time over computing matrix multiplications.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

Q22 jee_main_2024_29_january_evening System of Linear Equations
Let for any three distinct consecutive terms a, b, c of an A.P, the lines ax + by + c = 0 be concurrent at the point P and Q (α, β) be a point such that the system of equations x + y + z = 6, 2 x + 5 y + α z = β and x + 2y + 3z = 4 has infinitely many solutions. Then (PQ)² is equal to
Numerical Answer. Answer: 113 to 113

Solution

Related Formula

For infinite solutions of a system of equations, the main determinant D and auxiliary determinants D₁, D₂, D₃ must all equal 0.

Core Logic

Since a, b, c are in A.P., we have 2b = a + c a - 2b + c = 0. Comparing this identity with the line equation ax + by + c = 0, we immediately find that the lines always pass through the fixed point configuration (1, -2). Thus, P = (1, -2).

Step 1: Evaluation of the Matrix for Infinite Solutions

Let us set the system determinant D = 0:

D = bmatrix 1 & 1 & 1 2 & 5 & α 1 & 2 & 3 bmatrix = 0 1(15 - 2α) - 1(6 - α) + 1(4 - 5) = 0 15 - 2α - 6 + α - 1 = 0 8 - α = 0 α = 8

Now set D₁ = 0 by substituting columns:

D₁ = bmatrix 6 & 1 & 1 β & 5 & 8 4 & 2 & 3 bmatrix = 0 6(15 - 16) - 1(3β - 32) + 1(2β - 20) = 0 -6 - 3β + 32 + 2β - 20 = 0 6 - β = 0 β = 6

Thus, Q = (8, 6).

Step 2: Distance Metric Calculation

Using the distance formula between P(1, -2) and Q(8, 6):

(PQ)² = (8 - 1)² + (6 - (-2))² = 7² + 8² = 49 + 64 = 113
Pattern Recognition

A.P. coefficients inside a standard linear equation reveal a fixed point of concurrency by mapping matching coefficient components (1, -2, 1).

Chapter Mix

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

Q19 jee_main_2024_27_jan_morning Matrix Multiplication
Consider the matrix f(x) = bmatrix x & - x & 0 x & x & 0 0 & 0 & 1 bmatrix Given below are two statements: Statement I: f(-x) is the inverse of the matrix f(x) Statement II: f(x)f(y) = f(x+y). In the light of the above statements, choose the correct answer from the options given below
  • A. Statement I is false but Statement II is true
  • B. Both Statement I and Statement II are false
  • C. Statement I is true but Statement II is false
  • D. Both Statement I and Statement II are true

Solution

Related Formula
(-x) = x (-x) = - x (x+y) = x y + x y (x+y) = x y - x y
Core Logic

Evaluate f(-x):

f(-x) = bmatrix (-x) & - (-x) & 0 (-x) & (-x) & 0 0 & 0 & 1 bmatrix = bmatrix x & x & 0 - x & x & 0 0 & 0 & 1 bmatrix

Checking if f(-x) is the inverse by evaluating f(x) · f(-x):

f(x) f(-x) = bmatrix ² x + ² x & x x - x x & 0 x x - x x & ² x + ² x & 0 0 & 0 & 1 bmatrix f(x) f(-x) = bmatrix 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 bmatrix = I

Thus, Statement I is true.

Step 1: Checking Statement II

Evaluate the matrix multiplication f(x) · f(y):

f(x) f(y) = bmatrix x & - x & 0 x & x & 0 0 & 0 & 1 bmatrix bmatrix y & - y & 0 y & y & 0 0 & 0 & 1 bmatrix = bmatrix x y - x y & - x y - x y & 0 x y + x y & - x y + x y & 0 0 & 0 & 1 bmatrix

Apply standard trigonometric compound angle formulas:

= bmatrix (x+y) & - (x+y) & 0 (x+y) & (x+y) & 0 0 & 0 & 1 bmatrix = f(x+y)

Thus, Statement II is also true.

Step 2: Final Conclusion

Both Statement I and Statement II are true.

Pattern Recognition

This specific matrix represents a standard 2D rotation matrix embedded in 3D space. Rotation matrices naturally follow R(x)R(y) = R(x+y) (additive property of angles) and their inverse is always obtained by negating the angle R(-x) = R(x)⁻¹.

Chapter Mix

Class 12 Maths: Matrices Class 11 Maths: Trigonometric Functions

Q29 jee_main_2024_27_jan_morning Matrix Inverse and Determinant
Let A= bmatrix 2 & 0 & 1 1 & 1 & 0 1 & 0 & 1 bmatrix, B=[B₁, B₂, B₃], where B₁, B₂, B₃ are column matrices, and AB₁= bmatrix 1 0 0 bmatrix, AB₂= bmatrix 2 3 0 bmatrix, AB₃= bmatrix 3 2 1 bmatrix. If α=|B| and β is the sum of all the diagonal elements of B, then α³+β³ is equal to:
Numerical Answer. Answer: 28 to 28

Solution

Related Formula

|AB| = |A||B| Trace(B) = Σ bᵢᵢ

Core Logic

Since B = [B₁, B₂, B₃], the matrix multiplication AB effectively applies A to each column of B:

AB = [AB₁, AB₂, AB₃] = bmatrix 1 & 2 & 3 0 & 3 & 2 0 & 0 & 1 bmatrix

Let this target matrix be C. We know AB = C. Thus, taking determinants on both sides: |A| |B| = |C|

Step 1: Finding Determinants

Calculate determinant of A:

|A| = 2(1 - 0) - 0 + 1(0 - 1) = 2 - 1 = 1

Calculate determinant of C. Since C is an upper triangular matrix, its determinant is simply the product of its main diagonal:

|C| = 1 × 3 × 1 = 3

Therefore, 1 × |B| = 3 ⇒ |B| = 3. So α = 3.

Step 2: Finding Matrix B

To find β (the trace of B), we must explicitly find B = A⁻¹C. Rather than finding the full inverse, manually solve A Bᵢ = Cᵢ: For B₁ = [x, y, z]^T: 2x+z=1, x+y=0, x+z=0 ⇒ x=1, z=-1, y=-1. Thus B₁ = [1, -1, -1]^T. For B₂: 2x+z=2, x+y=3, x+z=0 ⇒ x=2, z=-2, y=1. Thus B₂ = [2, 1, -2]^T. For B₃: 2x+z=3, x+y=2, x+z=1 ⇒ x=2, z=-1, y=0. Thus B₃ = [3, 0, -1]^T.

Step 3: Calculating Trace and Final Output

Constructing Matrix B:

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

The diagonal elements are 1, 1, -1. Trace β = 1 + 1 - 1 = 1.

Finally compute α³ + β³:

3³ + 1³ = 27 + 1 = 28
Pattern Recognition

When given AXᵢ = Yᵢ for multiple columns, they collectively form A X = Y. Using |A||X| = |Y| bypasses full matrix inversion if you strictly need determinants. To grab the trace, solving equations systematically column-by-column is generally less error-prone than forming the full adjoint inverse matrix.

Chapter Mix

Class 12 Maths: Matrices

Q15 jee_main_2024_29_jan_morning Properties of Determinants
Let A= bmatrix1&0&0 0&α&β 0&β&α bmatrix and |2A|³=2²¹ where α, βin Z, Then a value of α is
  • A. 3
  • B. 5
  • C. 17
  • D. 9

Solution

Related Formula

|kA| = kⁿ |A| Where A is an n × n matrix, and k is a scalar.

Core Logic

Find the determinant of the 3 × 3 matrix A:

|A| = 1(α · α - β · β) - 0 + 0 |A| = α² - β²

Given the condition |2A|³ = 2²¹. Since A is a 3 × 3 matrix, applying the scalar property |kA| = k³|A|:

|2A| = 2³|A| = 8|A|

Substitute this back into the original condition:

(2³|A|)³ = 2²¹ 2⁹ |A|³ = 2²¹ |A|³ = 2²¹2⁹ = 2¹²

Taking the cube root of both sides: |A| = 2⁴ = 16

Step 1: Solve the Diophantine Equation

We have:

α² - β² = 16 (α - β)(α + β) = 16

Since α and β are integers, their sum and difference must also be integers. Also, (α + β) and (α - β) must share the same parity (both even or both odd) because their sum is 2α (an even number).

Since their product is 16, the only valid integer factor pairs of 16 that share the same parity are (8, 2) and (-8, -2) and (4, 4) and (-4, -4).

Case 1: (α + β) = 8 and (α - β) = 2 Adding them gives 2α = 10 ⇒ α = 5. Thus β = 3.

Case 2: (α + β) = 4 and (α - β) = 4 Adding them gives 2α = 8 ⇒ α = 4. Thus β = 0.

Looking at the options provided (3, 5, 17, 9), the value α = 5 is listed.

Pattern Recognition

Extracting scalar multipliers from determinants always depends on the dimension n of the matrix. For Diophantine equations like x² - y² = k, factoring into (x-y)(x+y) and analyzing parity constraints restricts the solution space instantly.

Chapter Mix

Class 12 Mathematics: Determinants

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