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

Solution & Explanation

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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More Determinants Previous-Year Questions — Page 4

Q25 jee_main_2026_28_january_evening Higher Powers of Matrices
Let A = bmatrix 3 & -4 1 & -1 bmatrix and B be two matrices such that A¹⁰⁰ = 100B + I. Then the sum of all the elements of B¹⁰⁰ is
Numerical Answer. Answer: 0 to 0

Solution

Core Logic

To find A¹⁰⁰, express A as I + M: A = I + bmatrix 2 & -4 1 & -2 bmatrix. Let M = bmatrix 2 & -4 1 & -2 bmatrix.

Calculate M² to check for nilpotency:

M² = bmatrix 2 & -4 1 & -2 bmatrix bmatrix 2 & -4 1 & -2 bmatrix = bmatrix 4-4 & -8+8 2-2 & -4+4 bmatrix = bmatrix 0 & 0 0 & 0 bmatrix

Thus, M² = 0 (nilpotent matrix of index 2).

Execution

Expand A¹⁰⁰ = (I + M)¹⁰⁰ using Binomial Theorem (valid since I and M commute):

A¹⁰⁰ = I¹⁰⁰ + ¹⁰⁰C₁ I⁹⁹ M + ¹⁰⁰C₂ I⁹⁸ M² + …

Since M^k = 0 for k ≥ 2, all higher terms vanish.

A¹⁰⁰ = I + 100M

Given the condition A¹⁰⁰ = 100B + I, we compare the equations: I + 100M = 100B + I ⇒ B = M.

We need the sum of all elements of B¹⁰⁰. Since B = M, B² = M² = 0, which implies B¹⁰⁰ = 0. The sum of all elements of a null matrix is 0.

Pattern Recognition

When asked for extreme powers of a non-diagonal matrix, extract the identity matrix I. The residual matrix M will almost inevitably be nilpotent (M²=0 or M³=0), collapsing the binomial expansion.

Chapter Mix

Class 12 Maths: Matrices and Determinants

Q58 jee_main_2025_02_april_evening System of Linear Equations
If the system of equations aligned 2x + λ y + 3z &= 5 3x + 2y - z &= 7 4x + 5y + μ z &= 9 aligned has infinitely many solutions, then (λ² + μ²) is equal to:
  • A. 22
  • B. 18
  • C. 26
  • D. 30

Solution

Related Formula
For infinitely many solutions: Δ = 0 and Δᵢ = 0
Core Logic

For a system of 3 linear equations to have infinitely many solutions, the determinant of coefficients and all Cramer determinants must equal zero.

Step 1: Set up determinant equations

The determinant of coefficients is:

Δ = vmatrix 2 & λ & 3 3 & 2 & -1 4 & 5 & μ vmatrix = 0 2(2μ + 5) - λ(3μ + 4) + 3(15 - 8) = 0 4μ + 10 - 3λμ - 4λ + 21 = 0 4μ - 3λμ - 4λ + 31 = 0 --- (1)

Now, set Δ₃ = 0:

Δ₃ = vmatrix 2 & λ & 5 3 & 2 & 7 4 & 5 & 9 vmatrix = 0 2(18 - 35) - λ(27 - 28) + 5(15 - 8) = 0 -34 + λ + 35 = 0 λ = -1
Step 2: Solve for mu and compute the sum of squares

Substitute λ = -1 into equation (1):

4μ - 3(-1)μ - 4(-1) + 31 = 0 4μ + 3μ + 4 + 31 = 0 7μ = -35 μ = -5

Now calculate the sum of squares:

λ² + μ² = (-1)² + (-5)² = 1 + 25 = 26
Pattern Recognition

Whenever you need to solve for two variables in Cramer's theorem, identifying which determinant lacks the complex variable (like Δ₃ which lacks μ) is the fastest way to solve for one variable independently.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

Q70 jee_main_2025_02_april_evening Properties of Matrices
Let A be a 3 × 3 real matrix such that A² (A - 2I) - 4(A - I) = O, where I and O are the identity and null matrices, respectively. If A⁵ = α A² + β A + γ I, where α, β and γ are real constants, then α + β + γ is equal to:
  • A. 12
  • B. 20
  • C. 76
  • D. 4

Solution

Related Formula
Characteristic equation reduction: A³ = 2A² + 4A - 4I
Core Logic

We use the given cubic matrix equation recursively to express the fifth power of matrix A solely in terms of quadratic and linear terms.

Step 1: Simplify the cubic matrix equation

The given equation is:

A² (A - 2I) - 4(A - I) = O A³ - 2A² - 4A + 4I = O A³ = 2A² + 4A - 4I

Multiply by matrix A to find the fourth power:

A⁴ = 2A³ + 4A² - 4A
Step 2: Reduce the fourth power term

Substitute the expression for A³ into our formula for A⁴:

A⁴ = 2( 2A² + 4A - 4I ) + 4A² - 4A A⁴ = 4A² + 8A - 8I + 4A² - 4A = 8A² + 4A - 8I

Multiply by matrix A to find the fifth power:

A⁵ = 8A³ + 4A² - 8A
Step 3: Reduce the fifth power term and solve

Substitute the expression for A³ again:

A⁵ = 8( 2A² + 4A - 4I ) + 4A² - 8A A⁵ = 16A² + 32A - 32I + 4A² - 8A = 20A² + 24A - 32I

Comparing this with A⁵ = α A² + β A + γ I, we find:

  • α = 20
  • β = 24
  • γ = -32
  • Sum the coefficients:

α + β + γ = 20 + 24 - 32 = 12
Pattern Recognition

Cayley-Hamilton reduction: For any polynomial equation of a matrix, higher powers A^k can always be reduced down to polynomials of order less than the degree of the characteristic equation by recursive substitution.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

Q jee_main_2025_02_april_morning Idempotent Matrices
Let A = bmatrix α & -1 6 & β bmatrix, α > 0, such that (A) = 0 and α + β = 1. If I denotes the 2 × 2 identity matrix, then the matrix (I + A)⁸ is:
  • A. bmatrix 4 & -1 6 & -1 bmatrix
  • B. bmatrix 257 & -64 514 & -127 bmatrix
  • C. bmatrix 1025 & -511 2024 & -1024 bmatrix
  • D. bmatrix 766 & -255 1530 & -509 bmatrix

Solution

Related Formula

For a matrix satisfying A² = A (Idempotent Matrix):

(I+A)ⁿ = I + (2ⁿ - 1)A
Core Logic

Given (A) = αβ + 6 = 0 αβ = -6 and α + β = 1. Solving these gives α = 3, β = -2 (since α > 0).

Step 1: Check Powers of A

Substitute values into A:

A = bmatrix 3 & -1 6 & -2 bmatrix

Compute A²:

A² = bmatrix 3 & -1 6 & -2 bmatrix bmatrix 3 & -1 6 & -2 bmatrix = bmatrix 9-6 & -3+2 18-12 & -6+4 bmatrix = bmatrix 3 & -1 6 & -2 bmatrix = A
Step 2: Expand Matrix Expression

Since A² = A, it follows that Aⁿ = A for all integers n ≥ 1.

(I + A)⁸ = I + Σk=1⁸ 8k A^k = I + A Σk=1⁸ 8k = I + (2⁸ - 1)A = I + 255A
Step 3: Construct the Final Matrix
(I + A)⁸ = bmatrix 1 & 0 0 & 1 bmatrix + 255 bmatrix 3 & -1 6 & -2 bmatrix = bmatrix 1 + 765 & -255 1530 & 1 - 510 bmatrix = bmatrix 766 & -255 1530 & -509 bmatrix
Pattern Recognition

Whenever tr(A) = 1 and (A) = 0 for a 2 × 2 matrix, Cayley-Hamilton theorem gives A² - tr(A)A + (A)I = 0 A² = A. Thus A is idempotent, simplifying polynomial expansions exponentially.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

Q jee_main_2025_02_april_morning System of Linear Equations
If the system of linear equations 3x + y + β z = 3 2x + α y - z = -3 x + 2y + z = 4 has infinitely many solutions, then the value of 22β - 9α is:
  • A. 49
  • B. 31
  • C. 43
  • D. 37

Solution

Related Formula

Cramer's Rule for infinite solutions specifies that the main determinant and all component determinants must vanish:

Δ = 0 and Δ₁ = Δ₂ = Δ₃ = 0
Core Logic

Set the key system determinants to zero to form equations linking α and β, then isolate the constants.

Step 1: Set Main Determinant to Zero
Δ = vmatrix 3 & 1 & β 2 & α & -1 1 & 2 & 1 vmatrix = 0

Expand along the first row:

3(α + 2) - 1(2 + 1) + β(4 - α) = 0 3α + 6 - 3 + 4β - αβ = 0 3α + 4β - αβ + 3 = 0 (1)
Step 2: Set Subsidiary Determinant to Zero

Using Δ₃ = 0 by substituting the constants vector into the third column:

Δ₃ = vmatrix 3 & 1 & 3 2 & α & -3 1 & 2 & 4 vmatrix = 0

Expand along the first row:

3(4α + 6) - 1(8 + 3) + 3(4 - α) = 0 12α + 18 - 11 + 12 - 3α = 0 9α + 19 = 0 α = -(19)/(9)
Step 3: Solve for Beta and Final Expression

Substitute α = -(19)/(9) into equation (1):

3(-(19)/(9)) + 4β - (-(19)/(9))β + 3 = 0 -(19)/(3) + 3 + β(4 + (19)/(9)) = 0 -(10)/(3) + β((55)/(9)) = 0 (55)/(9)β = (10)/(3) β = (10)/(3) · (9)/(55) = (6)/(11)

Now compute 22β - 9α:

22((6)/(11)) - 9(-(19)/(9)) = 12 + 19 = 31
Pattern Recognition

Choosing Δ₃ over Δ₁ or Δ₂ eliminates β entirely because the variable parameters are localized in specific positions. This yields α directly without requiring a coupled system solution.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

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