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 Type:
Enter a numerical value Answer: 28 to 28 +4 marks

Solution & Explanation

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

Reference Study Guides

More Matrices Previous-Year Questions — Page 9

Q jee_main_2025_29_jan_morning Properties of Determinants
Let M and m respectively be the maximum and the minimum values of f (x) = | arrayc c c 1 + ^ 2 x & ^ 2 x & 4 4 x ^ 2 x & 1 + ^ 2 x & 4 4 x ^ 2 x & ^ 2 x & 1 + 4 4 x array |, x in R Then M⁴ -m⁴ is equal to :
  • A. 1280
  • B. 1295
  • C. 1040
  • D. 1215

Solution

Related Formula
² x + ² x = 1 -1 ≤ 4x ≤ 1
Core Logic

Apply the row operations R₂ → R₂ - R₁ and R₃ → R₃ - R₁ to simplify the determinant:

f(x) = | arrayc c c 1 + ^ 2 x & ^ 2 x & 4 4 x -1 & 1 & 0 -1 & 0 & 1 array |
Step 1: Expand the Determinant

Expanding along the first row:

f(x) = (1 + ² x)(1 - 0) - ² x(-1 - 0) + 4 4x(0 - (-1)) f(x) = 1 + ² x + ² x + 4 4x

Since ² x + ² x = 1, we get:

f(x) = 2 + 4 4x
Step 2: Find Maximum and Minimum Values

The range of 4x is [-1, 1].

M = 2 + 4(1) = 6 m = 2 + 4(-1) = -2
Step 3: Calculate M⁴ - m⁴
M⁴ - m⁴ = 6⁴ - (-2)⁴ = 1296 - 16 = 1280
Pattern Recognition

Look for repeated structures or cyclic additions in rows. Subtracting rows quickly creates zeros, reducing complex trigonometric matrices into elementary algebraic expressions.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants Class 11 Mathematics: Trigonometric Functions

Q jee_main_2025_29_jan_morning Cofactors and Determinant Value
Let A = [aᵢⱼ] = bmatrix ₅ 128 & ₄ 5 ₅ 8 & ₄ 25 bmatrix . If Aᵢⱼ is the cofactor of aᵢⱼ , Cᵢⱼ = Σk=1² aik Ajk , 1 ≤ i, j ≤ 2 , and C = [Cᵢⱼ] , then 8|C| is equal to:
  • A. 262
  • B. 288
  • C. 242
  • D. 222

Solution

Related Formula
Σk aik Ajk = δᵢⱼ |A| C = bmatrix |A| & 0 0 & |A| bmatrix |C| = |A|²
Core Logic

Evaluate the determinant of matrix A:

|A| = ( ₅ 128)( ₄ 25) - ( ₄ 5)( ₅ 8)

Using change of base rules:

|A| = (7 ₅ 2)(2 ₄ 5) - ((1)/(2) ₂ 5)(3 ₅ 2) |A| = 14( ₅ 2 · (1)/(2) ₂ 5) - (3)/(2) = 7 - 1.5 = 5.5 = (11)/(2)
Step 1: Compute |C| and evaluate response target

Since matrix properties dictate |C| = |A|²:

|C| = ((11)/(2))² = (121)/(4)

Evaluate targeted multiplier:

8|C| = 8 × (121)/(4) = 2 × 121 = 242
Pattern Recognition

Recognize the core cofactor theorem identity instantly: multiplying rows by cofactors of other rows creates zero elements, yielding basic diagonal scalar structures matching matrix attributes.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

Q jee_main_2025_29_jan_morning Matrix Powers
Let S = | m in Z : Am² + A^m = 3I - A⁻⁶ | , where A = bmatrix 2 & -1 1 & 0 bmatrix . Then n(S) is equal to
Numerical Answer. Answer: 2

Solution

Related Formula
Inductive formulation for exponent powers of a pattern matrix
Core Logic

Evaluate lower power forms of A to establish inductive sequence patterns:

A = bmatrix 2 & -1 1 & 0 bmatrix, A² = bmatrix 3 & -2 2 & -1 bmatrix, A³ = bmatrix 4 & -3 3 & -2 bmatrix

This cleanly establishes general power state rule configuration expression:

A^m = bmatrix m+1 & -m m & -m+1 bmatrix
Step 1: Setup Matrix Power Equation

Using the pattern, find expressions for terms:

A⁶ = bmatrix 7 & -6 6 & -5 bmatrix, A⁻⁶ = (A⁶)⁻¹ = bmatrix -5 & 6 -6 & 7 bmatrix

Substitute into the targeted equation block:

Am² + A^m = 3 bmatrix 1 & 0 0 & 1 bmatrix - bmatrix -5 & 6 -6 & 7 bmatrix = bmatrix 8 & -6 6 & -4 bmatrix
Step 2: Equate Corresponding Elements

From the bottom-left entry [2,1]:

m² + m = 6 m² + m - 6 = 0 (m+3)(m-2) = 0 m = -3, 2

Both integer states satisfy all matrix components seamlessly. Therefore, the number of elements n(S) = 2.

Pattern Recognition

Always calculate first 2–3 matrix powers to spot linear arithmetic trends across specific cell values (m+1, -m), avoiding tedious full Cayley-Hamilton character expansions.

Chapter Mix

Class 12 Mathematics: Matrices

Q3 jee_main_2024_01_february_morning Properties of Determinants
If A= bmatrix√(2) & 1 -1 & √(2) bmatrix, B= bmatrix1 & 0 1 & 1 bmatrix, C=ABAT and X=ATC²A, then X is equal to:
  • A. 243
  • B. 729
  • C. 27
  • D. 891

Solution

Related Formula

Properties of Determinants:

  • (AB) = (A) · (B)
  • (A^T) = (A)
  • (Aⁿ) = ( (A))ⁿ
Core Logic

First, find the determinant of matrix A and matrix B:

(A) = |A| = vmatrix√(2) & 1 -1 & √(2) vmatrix = (√(2))(√(2)) - (1)(-1) = 2 + 1 = 3 (B) = |B| = vmatrix1 & 0 1 & 1 vmatrix = (1)(1) - (0)(1) = 1
Step 1: Calculate determinant of C

Given C = ABAT, we calculate its determinant:

(C) = (ABAT) = (A) · (B) · (AT)

Using (A^T) = (A) = 3:

(C) = 3 · 1 · 3 = 9
Step 2: Calculate determinant of X

Given X = ATC²A, we determine |X|:

(X) = |ATC²A| = |AT| · |C²| · |A| (X) = |A| · |C|² · |A| = |A|² · |C|²

Substitute the values |A| = 3 and |C| = 9:

(X) = (3)² · (9)² = 9 · 81 = 729
Pattern Recognition

Sees: Transpose and multiplication operations nested inside a determinant statement. Shortcut: Never explicitly compute product matrices like ABA^T or A^TC²A. Apply the distributive identity of determinants |XYZ| = |X||Y||Z| entirely to work strictly with scalar multiplication.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

Q12 jee_main_2024_01_february_morning System of Linear Equations
If the system of equations 2x+3y-z=5 x+α y+3z=-4 3x-y+β z=7 has infinitely many solutions, then 13 α β is equal to:
  • A. 1110
  • B. 1120
  • C. 1210
  • D. 1220

Solution

Related Formula

For a system of linear equations representing planes to have infinitely many solutions, the planes must belong to a single family sharing a common line of intersection:

P₁ = k₁ P₂ + k₂ P₃
Core Logic

Let the planes be defined as:

  • P₁: 2x + 3y - z - 5 = 0
  • P₂: x + α y + 3z + 4 = 0
  • P₃: 3x - y + β z - 7 = 0
  • Expressing P₁ as a linear combination of P₂ and P₃:

2x + 3y - z - 5 = k₁(x + α y + 3z + 4) + k₂(3x - y + β z - 7)
Step 1: Evaluate parameters k1 and k2

Comparing the coefficients of x and the constant terms on both sides:

  • For x:
k₁ + 3k₂ = 2 (1)
  • For the constants:
4k₁ - 7k₂ = -5 (2)

Multiplying equation (1) by 4 gives 4k₁ + 12k₂ = 8. Subtracting equation (2) from this result:

(4k₁ + 12k₂) - (4k₁ - 7k₂) = 8 - (-5) 19k₂ = 13 k₂ = (13)/(19)

Substituting k₂ back into equation (1):

k₁ + 3((13)/(19)) = 2 k₁ = 2 - (39)/(19) = -(1)/(19)
Step 2: Calculate alpha, beta and final product

Comparing coefficients for y and z:

  • For y:
k₁α - k₂ = 3 -(1)/(19)α - (13)/(19) = 3 -α - 13 = 57 α = -70
  • For z:
3k₁ + k₂β = -1 3(-(1)/(19)) + (13)/(19)β = -1 -3 + 13β = -19 13β = -16 β = -(16)/(13)

Now, compute 13 α β:

13 α β = 13 × (-70) × (-(16)/(13)) = 70 × 16 = 1120
Pattern Recognition

Sees: Infinite solution framework for three linear planes. Shortcut: Using the family of planes equation is significantly less prone to fractional algebraic mistakes compared to establishing Cramer's rule determinants (D = Dₓ = Dy = Dz = 0).

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

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