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Matrix Algebra and Determinants — JEE Main Questions

Questions & explanations

1. For what value of k is the matrix [[k, 2, 3], [2, k, 3], [3, 3, 6]] singular?

k = 1
k = 3
k = 2
k = 4

Answer: k = 3

A matrix is singular if its determinant is zero. Compute det = k(k*6 - 3*3) - 2(2*6 - 3*3) + 3(2*3 - k*3) = k(6k - 9) - 2(12 - 9) + 3(6 - 3k) = 6k^2 - 9k - 6 + 18 - 9k = 6k^2 - 18k + 12 = 6(k^2 - 3k + 2) = 6(k-1)(k-2). Set to zero gives k = 1 or 2. Among options, k = 2 is correct.

2. For matrices A = [[1,2],[0,1]] and B = [[1,0],[1,1]], which of the following is correct?

AB = BA = [[3,2],[1,3]]
AB = BA = [[1,2],[1,3]]
AB = [[1,2],[1,3]] and BA = [[3,2],[1,1]]
AB = [[3,2],[1,1]] and BA = [[1,2],[1,3]]

Answer: AB = [[1,2],[1,3]] and BA = [[3,2],[1,1]]

Matrix multiplication is not commutative. Compute AB: (1,1)=1*1+2*1=3, (1,2)=1*0+2*1=2, (2,1)=0*1+1*1=1, (2,2)=0*0+1*1=1 → AB=[[3,2],[1,1]]. Compute BA: (1,1)=1*1+0*0=1, (1,2)=1*2+0*1=2, (2,1)=1*1+1*0=1, (2,2)=1*2+1*1=3 → BA=[[1,2],[1,3]]. Thus AB≠BA, and option d matches.

3. For matrix A = [[1,2,3],[4,5,6],[7,8,10]], what is the determinant when expanded along row 2?

-3
3
0
-1

Answer: -3

Expanding along row 2: det = a₂₁C₂₁ + a₂₂C₂₂ + a₂₃C₂₃. C₂₁ = (-1)^(2+1) M₂₁ = -det[[2,3],[8,10]] = -(20-24)=4. C₂₂ = (-1)^(2+2) M₂₂ = det[[1,3],[7,10]] = 10-21=-11. C₂₃ = (-1)^(2+3) M₂₃ = -det[[1,2],[7,8]] = -(8-14)=6. So det = 4×4 + 5×(-11) + 6×6 = 16 - 55 + 36 = -3.

4. Which of the following is NOT a valid elementary row operation?

R_i → R_i + k R_j
R_i → k R_i (k ≠ 0)
R_i → R_i + R_j
R_i → k R_i + R_j (k ≠ 0)

Answer: R_i → k R_i + R_j (k ≠ 0)

Elementary row operations are: swap, multiply a row by a non-zero scalar, and add a multiple of one row to another. Option d replaces a row with a linear combination that includes scaling the same row, which is not allowed as a single elementary operation.

5. The system 2x - y + z = 4, x + 3y - 2z = -1, 3x + y + z = 5 is written as AX = B. What is matrix A?

[[2,-1,1],[1,3,-2],[3,1,-1]]
[[2,-1,1],[1,3,2],[3,1,1]]
[[2,-1,1],[1,3,-2],[3,1,1]]
[[2,1,1],[1,3,-2],[3,1,1]]

Answer: [[2,-1,1],[1,3,-2],[3,1,1]]

The coefficient matrix A is formed by taking the coefficients of x, y, z from each equation in order. For the given system, the first row is [2, -1, 1], second row [1, 3, -2], third row [3, 1, 1]. Thus A = [[2,-1,1],[1,3,-2],[3,1,1]].

6. If Δ = det [[1, 1, 1], [a, b, c], [a^2, b^2, c^2]], then Δ equals

(a - b)(b - c)(a - c)
(a - b)(b - c)(c - a)
(a + b)(b + c)(c + a)
(a - b)(b + c)(c - a)

Answer: (a - b)(b - c)(c - a)

This is the Vandermonde determinant. Setting a = b makes columns 2 and 3 identical, so (a - b) is a factor. Similarly, (b - c) and (c - a) are factors. The product (a - b)(b - c)(c - a) matches the degree and leading coefficient.

7. If D(x) = det [[1, x, x^2], [1, 2, 4], [1, 3, 9]], then which of the following is a factor of D(x)?

(x - 2)(x - 3)
(x - 2)(x + 3)
(x + 2)(x - 3)
(x + 2)(x + 3)

Answer: (x - 2)(x - 3)

Using the factor theorem for determinants: if x = 2, rows 1 and 2 become identical, so D(2) = 0, thus (x - 2) is a factor. Similarly, x = 3 makes rows 1 and 3 identical, so (x - 3) is a factor. Hence (x - 2)(x - 3) is a factor.

8. For the system x + y = 3, 2x - y = 0, what is the solution using X = A^{-1}B?

x = -1, y = 2
x = 2, y = 1
x = 1, y = -2
x = 1, y = 2

Answer: x = 1, y = 2

Using matrix inversion method: A = [[1,1],[2,-1]], B = [[3],[0]]. det A = -3, so A^{-1} = (-1/3)[[-1,-1],[-2,1]] = [[1/3,1/3],[2/3,-1/3]]. Then X = A^{-1}B = [[1/3*3+1/3*0],[2/3*3+(-1/3)*0]] = [[1],[2]]. Hence x=1, y=2.

9. If A is a 2×2 matrix such that Aᵀ = A and AᵀA = I, then which of the following is true?

A² = I and det(A) = ±1
A² = I and det(A) = 1
A² = A and det(A) = 1
A² = A and det(A) = ±1

Answer: A² = I and det(A) = ±1

Given Aᵀ = A (symmetric) and AᵀA = I (orthogonal). Substituting Aᵀ = A into AᵀA = I gives A² = I, so A is involutory. Taking determinant: det(A²) = det(I) ⇒ (det A)² = 1 ⇒ det A = ±1. Hence option a is correct.

10. Using Sarrus rule, what is det of [[1, 2, 3], [4, 5, 6], [7, 8, 10]]?

3
0
-3
6

Answer: -3

Using Sarrus rule: write matrix and repeat first two columns. Downward products: 1×5×10=50, 2×6×7=84, 3×4×8=96; sum=230. Upward products: 3×5×7=105, 1×6×8=48, 2×4×10=80; sum=233. Determinant = 230 – 233 = –3.

11. If A is an orthogonal matrix of order 3, what is the value of det(A)?

0
1
-1
±1

Answer: ±1

For an orthogonal matrix A, we have A A^T = I. Taking determinant on both sides: det(A A^T) = det(I) = 1. Since det(A A^T) = det(A) det(A^T) = det(A) det(A) = (det A)^2, we get (det A)^2 = 1, so det A = ±1.

12. Let A(k) = [[k,1,1],[1,k,1],[1,1,k]]. For what value(s) of k is |adj A| = 81?

k = 4 or k = -3
k = 3 or k = -4
k = 1 or k = -2
k = 2 or k = -5

Answer: k = 2 or k = -5

For a 3×3 matrix, |adj A| = |A|^2. Compute det A = (k-1)^2(k+2). Then |adj A| = (k-1)^4(k+2)^2 = 81 ⇒ (k-1)^2|k+2| = 9. Solve: (k-1)^2(k+2)=9 gives k=2; (k-1)^2(k+2)=-9 gives k=-5. Hence k=2 or -5.