JEE Main Maths: Determinants questions with solutions

Q1. If [ ] represents the greatest integer less than or equal to the given real number, and -1 ≤ x < 0, 0 ≤ y < 1, 1 ≤ z < 2, then the determinant [x] [y] [z] [x+1] [y+1] [z] [x] [y] [z]+1 is equal to:

1. [z]
2. [y]
3. [x]
4. None of these

Answer: [x]

With -1<=x<0, 0<=y<1, 1<=z<2: [x]=-1, [y]=0, [z]=1, [x+1]=0, [y+1]=1. The determinant of [[-1,0,1],[0,1,1],[-1,0,2]] expands to -1, which equals [x].

Q2. Let x, y and z be complex numbers. If Δ denotes the determinant of the matrix [0, -y, -z; y, 0, -x; z, x, 0], then Δ is

1. purely real
2. purely imaginary
3. complex
4. zero

Answer: zero

The matrix is skew-symmetric (M^T = -M) and of odd order 3. For an odd-order skew-symmetric matrix det = 0; direct expansion of [[0,-y,-z],[y,0,-x],[z,x,0]] also gives 0.

Q3. For the linear system x1 + 2x2 + 3x3 = 6 x1 + 3x2 + 5x3 = 9 2x1 + 5x2 + ax3 = b if it is consistent and admits infinitely many solutions, then which statement must be true?

1. a = 8, and b may be any real number
2. b = 15, and a may be any real number
3. a belongs to R {8} and b belongs to R {15}
4. a = 8, b = 15

Answer: a = 8, b = 15

Subtracting R1 and R2 from R3 gives (0, 0, a-8 | b-15). For the system to be consistent with infinitely many solutions, this row must vanish: a - 8 = 0 and b - 15 = 0, so a = 8 and b = 15.

Q4. For which value(s) of x does the determinant of the matrix | 1 (x−3) (x−3)² | | 1 (x−4) (x−4)² | | 1 (x−5) (x−5)² | become zero?

1. 3 values of x
2. 2 values of x
3. 1 value of x
4. no value of x

Answer: no value of x

Subtracting rows turns this into a Vandermonde-type determinant in the entries (x-3),(x-4),(x-5), which differ by fixed constants. The determinant evaluates to -2 for every x, so it is never zero.

Q5. Let Δ_r denote the determinant [2r-1, {^mC_r}, 1; m²-1, 2^m, m+1; sin²(m²), sin²(m), sin²(m+1)]. Then the value of ∑_(r=0)^(m) Δ_r is

1. 0
2. 4
3. 3
4. 1

Answer: 0

The determinant Delta_r is structured such that its rows exhibit linear dependencies when summed over the range of r from 0 to m, leading to a total sum of zero. This is due to the properties of determinants where linear combinations of rows can result in a determinant of zero.

Q6. If the determinant expression [a, b, ax+by; b, c, bx+cy; ax+by, bx+cy, 0] is considered under the conditions b²-ac<0 and a<0, what is its sign?

1. Zero
2. Positive
3. Negative
4. b²+ac

Answer: Positive

The determinant is positive because the conditions given, specifically that the quadratic form defined by the matrix is negative definite due to the inequality, imply that the overall determinant must be positive when evaluated.

Q7. Let α, β, γ be real numbers with sin α, sin β, sin γ all nonzero. Define Δ = | sin²α sinα cosα cos²α | | sin²β sinβ cosβ cos²β | | sin²γ sinγ cosγ cos²γ |. What is the greatest value Δ can attain?

1. 1
2. 0
3. −1/2
4. None of these

Answer: None of these

Using sin²=(1-cos2θ)/2, sin·cos=sin2θ/2, cos²=(1+cos2θ)/2, the three columns are independent combinations of 1, sin2θ, cos2θ, so Δ is generally nonzero. Its maximum value is about 0.6495 (=3√3/8), which is none of 1, 0, or -1/2; hence None of these.

Q8. How many different real values of x in the interval −π/4 ≤ x ≤ π/4 satisfy | sin x cos x cos x | | cos x sin x cos x | = 0 | cos x cos x sin x | ?

1. 0
2. 2
3. 1
4. 3

Answer: 1

This is a matrix with diagonal sin x and all off-diagonals cos x, so its determinant is (sin x - cos x)^2 (sin x + 2 cos x) = 0. In [-pi/4, pi/4], sin x = cos x gives x = pi/4, and tan x = -2 lies outside the interval. So exactly 1 solution.

Q9. Find the value of the determinant | (a^x + a^-x)² (a^x - a^-x)² 1 | | (b^x + b^-x)² (b^x - b^-x)² 1 | | (c^x + c^-x)² (c^x - c^-x)² 1 |

1. 0
2. 2abc
3. a²b²c²
4. None of these

Answer: 0

For each row, (a^x + a^-x)^2 - (a^x - a^-x)^2 = 4*(a^x)*(a^-x) = 4, a constant. So column1 - column2 = 4 in every row, making it proportional to the column of 1's. Hence the determinant is 0.

Q10. For n ∈ N, evaluate the determinant D = | n! (n+1)! (n+2)! | | (n+1)! (n+2)! (n+3)! | | (n+2)! (n+3)! (n+4)! |.

1. (n!)² (2n³ − 8n²)
2. (2n!)³ (3n² + 4n − 5)
3. (n!)³ (2n³ + 8n² + 10n + 4)
4. None of these

Answer: (n!)³ (2n³ + 8n² + 10n + 4)

Taking out n!, (n+1)!, (n+2)! from the rows and reducing the resulting determinant gives D = (n!)^3 (2n^3 + 8n^2 + 10n + 4) (verified for n=1,2,3).

Q11. If Sₖ = a^k + b^k + c^k, then the determinant Δ = [S₀, S₁, S₂; S₁, S₂, S₃; S₂, S₃, S₄] is equal to:

1. S₆
2. S₅ - S₃
3. S₆ - S₄
4. None of these

Answer: None of these

This is the symmetric-function (Hankel) determinant, which evaluates to ((a-b)(b-c)(c-a))^2. This equals none of S6, S5-S3, or S6-S4, so the answer is None of these.

Q12. Let f(x) = ax⁶ + bx⁵ + cx⁴ + dx³ + ex² + fx + g, where | x² − 2x + 3 7x + 2 x + 4 | | 2x + 7 x² − x + 2 3x | | 3 2x − 1 x² − 4x + 7 | Then the value of g is:

1. −200
2. 100
3. 112
4. −108

Answer: −108

The constant term g equals f(0), the determinant evaluated at x=0: |3,2,4; 7,2,0; 3,-1,7|. Expanding gives 3(14-0) - 2(49-0) + 4(-7-6) = 42 - 98 - 52 = -108.

Q13. A homogeneous system of equations −ax + y + z = 0 x − by + z = 0 x + y − cz = 0 where a, b, c ≠ −1, admits a non-trivial solution. Then the value of 1/(1+a) + 1/(1+b) + 1/(1+c) is:

1. 0
2. 1
3. 2
4. 3

Answer: 1

The system of equations has a non-trivial solution if the determinant of the coefficient matrix is zero. Given the conditions on a, b, and c, the relationship derived from the determinant leads to the conclusion that the sum of the fractions equals 1.

Q14. Let a, b, and c be the side lengths of a triangle. If | a² b² c² | | (a+1)² (b+1)² (c+1)² | | (a−1)² (b−1)² (c−1)² | = 0, then which of the following must be true?

1. The triangle ABC cannot be equilateral.
2. The triangle ABC is a right-angled isosceles triangle.
3. The triangle ABC is isosceles.
4. None of these.

Answer: The triangle ABC is isosceles.

The determinant equals 4(a-b)(a-c)(b-c). It is zero precisely when at least two of a,b,c are equal, i.e. the triangle is isosceles (which also includes the equilateral case, so 'cannot be equilateral' is false).

Q15. For the variables x, y and z, consider the homogeneous linear system x sin 3θ − y + z = 0, x cos 2θ + 4y + 3z = 0, 2x + 7y + 7z = 0. If a non-zero solution exists, then for an integer n, the possible values of θ are

1. π(n + (−1)ⁿ/3)
2. π(n + (−1)ⁿ/4)
3. π(n + (−1)ⁿ/6)
4. nπ/2

Answer: π(n + (−1)ⁿ/6)

For a non-trivial solution the determinant must vanish: -28 sin^2(theta) + 7 sin(3theta) = 0. Using sin(3theta)=3 sin-4 sin^3 gives 7 sin(theta)(3 - 4 sin^2(theta) - 4 sin(theta)) = 0, so sin(theta)=1/2 (the other root -3/2 is invalid). Thus theta = n*pi + (-1)^n * pi/6 = pi(n + (-1)^n/6).

Q16. If the adjugate of B is A and |P| = |Q| = 1, then the adjugate of Q⁻¹BP⁻¹ equals

1. PQ
2. QAP
3. PAQ
4. PA⁻¹Q

Answer: PAQ

adj(Q^-1 B P^-1) = adj(P^-1) adj(B) adj(Q^-1). Since |P|=|Q|=1, adj(P^-1)=det(P^-1)(P^-1)^-1=P and adj(Q^-1)=Q, and adj(B)=A, giving P A Q.

Q17. For real parameters α and β, consider the homogeneous system λ x + (sinα)y + (cosα)z = 0, x + (cosα)y + (sinα)z = 0, -x + (sinα)y - (cosα)z = 0. The collection of all x-values for which this system admits a non-zero solution is:

1. [0,√(2)]
2. [-√(2),0]
3. [-√(2),√(2)]
4. None of these

Answer: [-√(2),√(2)]

For a nonzero solution the coefficient determinant must vanish: λ = √2·sin(2α+π/4). As α varies this expression takes every value in [-√2, √2], so that is the required set.

Q18. Given that a² + b² + c² = -2 and f(x) = |[1+a²x, (1+b²)x, (1+c²)x; (1+a²)x, 1+b²x, (1+c²)x; (1+a²)x, (1+b²)x, 1+c²x] |, what is the degree of the polynomial f(x)?

1. 1
2. 0
3. 3
4. 2

Answer: 2

Doing R1->R1-R2 and R2->R2-R3 factors out (1-x)^2, and the remaining 3x3 expands to 1+(2+a^2+b^2+c^2)x = 1 (since a^2+b^2+c^2=-2). So f(x)=(1-x)^2, degree 2.

Q19. Let D be the determinant of the matrix | 1 1 1 | | 1 1+x 1 | | 1 1 1+y | where x ≠ 0 and y ≠ 0. Then D is

1. divisible by x, but not by y
2. divisible by y, but not by x
3. divisible by neither x nor y
4. divisible by both x and y

Answer: divisible by both x and y

The determinant D can be computed using properties of determinants and shows that it contains terms involving both x and y, indicating that it is divisible by both variables. This is due to the structure of the matrix, where the presence of x and y in the second and third rows leads to terms that include these variables when expanded.

Q20. Let A be the matrix [5, 5α, α; 0, α, 5α; 0, 0, 5]. If |A²| = 25, then the value of |α| is

1. (1)/(5)
2. 5
3. 5²
4. 1

Answer: (1)/(5)

A is upper triangular with diagonal 5, alpha, 5, so det(A) = 25*alpha. Then |A^2| = (det A)^2 = 625*alpha^2 = 25 -> alpha^2 = 1/25 -> |alpha| = 1/5.

Q21. Let a, b, and c be real numbers. Assume there exist real numbers x, y, z, not all zero, satisfying x = cy + bz, y = az + cx, and z = bx + ay. Then the value of a² + b² + c² + 2abc is

1. 2
2. -1
3. 0
4. 1

Answer: 1

Writing the three relations as a homogeneous system in x,y,z, a non-trivial solution requires the coefficient determinant to vanish. Expanding |[-1,c,b],[c,-1,a],[b,a,-1]|=0 gives 1-(a^2+b^2+c^2)-2abc=0, so a^2+b^2+c^2+2abc=1.

Q22. For how many values of k do the homogeneous equations 4x + ky + 2z = 0, kx + 4y + z = 0, and 2x + 2y + z = 0 admit a non-trivial solution?

1. 2
2. 1
3. zero
4. 3

Answer: 2

The homogeneous system of equations admits a non-trivial solution when the determinant of the coefficient matrix is zero. By calculating the determinant and setting it to zero, we find that there are two specific values of k that satisfy this condition.

Q23. For the system x-ky+z=0 kx+3y-kz=0 3x+y-z=0 if the zero solution is the only solution, then the possible values of k are:

1. R∖{2,-3}
2. R∖{2}
3. R∖{-3}
4. {2,-3}

Answer: R∖{2,-3}

The zero solution being the only solution indicates that the system of equations must be consistent and have a unique solution, which occurs when the determinant of the coefficient matrix is non-zero. The values of k that make the determinant zero, specifically k = 2 and k = -3, must be excluded, leading to the conclusion that k can take any real number except these two.

Q24. The number of values of k, for which the system of equations (k + 1)x + 8y = 4k kx + (k + 3)y = 3k − 1 has no solution, is

1. infinite
2. 1
3. 2
4. 3

Answer: 1

Det = (k+1)(k+3)-8k = (k-1)(k-3) = 0 gives k=1 or k=3. At k=1 both equations become x+4y=2 (infinitely many solutions). At k=3 they give x+2y=3 and x+2y=8/3, which are inconsistent (no solution). So exactly 1 value of k.

Q25. If P = [ [1, α, 3], [1, 3, 3], [2, 4, 4] ] is the adjoint of a 3 × 3 matrix A and |A| = 4, then α is equal to:

1. 4
2. 11
3. 5
4. 0

Answer: 11

Since P = adj(A) for a 3x3 matrix, |P| = |A|^2 = 4^2 = 16. Expanding det(P) gives 2*alpha - 6 = 16, so alpha = 11.

Q26. If α, β ≠ 0, and f(n) = αⁿ + βⁿ and | 3 1 + f(1) 1 + f(2) | | 1 + f(1) 1 + f(2) 1 + f(3) | | 1 + f(2) 1 + f(3) 1 + f(4) | = K(1 − α)²(1 − β)²(α − β)², then K is equal to:

1. 1
2. −1
3. αβ
4. 1/(αβ)

Answer: 1

With D = [[1,1,1],[1,a,b],[1,a^2,b^2]], the given matrix equals D D^T, so its determinant = (det D)^2 = ((a-1)(b-1)(b-a))^2 = (1-a)^2(1-b)^2(a-b)^2. Hence K=1.

Q27. The set of all values of λ for which the system of linear equations: 2x − 2x + z = λx, 2x1 − 3x2 + 2x3 = λx2, −x1 + 2x2 = λx3 has a non-trivial solution,

1. contains two elements.
2. contains more than two elements
3. is an empty set.
4. is a singleton

Answer: contains two elements.

The characteristic equation factors as (lambda-1)^2(lambda+3)=0, giving lambda=1 and lambda=-3, so the set has exactly two elements.

Q28. The system of linear equations x + λy − z = 0 λx − y − z = 0 x + y − λz = 0 has a non-trivial solution for:

1. exactly two values of λ.
2. exactly three values of λ.
3. infinitely many values of λ.
4. exactly one value of λ.

Answer: exactly three values of λ.

Non-trivial solutions need det=0. Expanding the coefficient determinant gives lambda^3 - lambda = lambda(lambda-1)(lambda+1) = 0, so lambda = 0, 1, -1 -> exactly three values.

Q29. If S is the set of distinct values of 'b' for which the following system of linear equations x + y + z = 1 x + ay + z = 1 ax + by + z = 0 has no solution, then S is:

1. a singleton
2. an empty set
3. an infinite set
4. a finite set containing two or more elements

Answer: a singleton

The system of equations has no solution when the determinant of the coefficient matrix is zero, leading to a unique value of 'b' that causes this condition. Thus, the set of distinct values of 'b' is a singleton, indicating only one specific value leads to no solution.

Q30. If the determinant of the matrix with rows (a, a², 1+a³) and (b, b², 1+b³) is zero, and the vectors (1,a,a²), (1,b,b²), and (1,c,c²) are not coplanar, then what is the value of abc?

1. 0
2. 2
3. −1
4. 1

Answer: −1

The determinant being zero indicates that the rows are linearly dependent, which implies that the values of a and b must be related in a specific way. Given that the vectors are not coplanar, the only way for this to hold true while satisfying the conditions is if the product abc equals -1, indicating a specific relationship among the values.

Q31. If α, β ≠ 0, and f(n) = αⁿ + βⁿ and | 3 1+f(1) 1+f(2) | | 1+f(1) 1+f(2) 1+f(3) | | 1+f(2) 1+f(3) 1+f(4) | = K(1−α)²(1−β)²(α−β)², then K is equal to

1. −1
2. αβ
3. 1/(αβ)
4. 1

Answer: 1

The determinant of the given matrix simplifies to a form that reveals K as a constant factor, specifically 1, when considering the structure of the functions involved and their relationships. This is consistent with the properties of determinants and the specific forms of f(n) in the context of the problem.

Q32. The system of linear equations x + λy - z = 0 λx - y - z = 0 x + y - λz = 0 has a non-trivial solution for

1. infinitely many values of λ
2. exactly one value of λ
3. exactly two values of λ
4. exactly three values of λ

Answer: exactly three values of λ

The system has a non-trivial solution when the determinant of the coefficient matrix is zero, which occurs for specific values of λ. In this case, the determinant is a polynomial in λ that can be factored to find three distinct roots, indicating that there are exactly three values of λ for which the system has non-trivial solutions.

Q33. If the system of linear equations x + ay + z = 3, x + 2y + 2z = 0, x + 5y + 3z = b has no solution, then

1. (A) a = 1, b ≠ 9
2. (B) a ≠ −1, b = 9
3. (C) a = −1, b = 9
4. (D) a = −1, b ≠ 9

Answer: (D) a = −1, b ≠ 9

For the system to have no solution, the equations must be inconsistent. Setting a = -1 leads to a situation where the first equation becomes dependent on the others, while b must not equal 9 to ensure that the third equation does not align with the others, thus creating inconsistency.

Q34. If f(x) = | cos x x 1 | | 2 sin x x² 2x | | tan x x 1 |, then lim x→0 f'(x)/x

1. exists and is equal to −2
2. does not exist
3. exist and is equal to 0
4. exists and is equal to 2

Answer: exists and is equal to −2

The limit of f'(x)/x as x approaches 0 exists and equals -2 because the derivative of the matrix function f(x) can be computed using the properties of derivatives and the behavior of trigonometric functions near zero, leading to a consistent result.

Q35. Let S be the set of all real values of k for which the system of linear equations x + y + z = 2 2x + y − z = 3 3x + 2y + kz = 4 has a unique solution. Then S is

1. an empty set
2. equal to R − {0}
3. equal to {0}
4. equal to R

Answer: equal to R − {0}

The system of equations has a unique solution when the determinant of the coefficient matrix is non-zero. For the given equations, this condition is satisfied for all values of k except k = 0, leading to the conclusion that the set S consists of all real numbers except zero.

Q36. If [x-4, 2x, 2x; 2x, x-4, 2x; 2x, 2x, x-4] = (A+Bx)(x-A)², then the ordered pair (A, B) is equal to:

1. (-4, -5)
2. (-4, 3)
3. (-4, 5)
4. (4, 5)

Answer: (-4, 5)

The determinant simplifies to a polynomial in terms of x, and by comparing coefficients with the given form, we find that A must be -4 and B must be 5 to satisfy the equation.

Q37. If the system of linear equations x + ky + 3z = 0, 3x + ky - 2z = 0, 2x + 4y - 3z = 0 has a non-zero solution (x, y, z) then (xz)/(y²) is equal to

1. -10
2. 10
3. -30
4. 30

Answer: 10

The system of equations has a non-zero solution if the determinant of the coefficient matrix is zero, which leads to a specific relationship between the variables. Solving for the ratio (xz)/(y²) using the values derived from the equations shows that it simplifies to 10.

Q38. If | a − b − c 2a 2a | | 2b b − c − a 2b | = (a + b + c)(x + a + b + c)², x ≠ 0 and a + b + c ≠ 0, then x is equal to: | 2c 2c c − a − b |

1. −2(a + b + c)
2. 2(a + b + c)
3. abc
4. −(a + b + c)

Answer: −2(a + b + c)

Adding all rows factors out (a+b+c); reducing gives the determinant = (a+b+c)^3 = (a+b+c)(x+a+b+c)^2. So (x+a+b+c)^2 = (a+b+c)^2; since x != 0, x+a+b+c = -(a+b+c), giving x = -2(a+b+c).

Q39. Let A and B be two invertible matrices of order 3 × 3. If det(ABA^T) = 8 and det(AB⁻¹) = 8, then det(BA⁻¹B^T) is equal to:

1. 1/4
2. 16
3. 1/16
4. 1

Answer: 1/16

To find det(BA⁻¹B^T), we can use the properties of determinants. We know that det(ABA^T) = det(A)det(B)det(A^T) = det(A)²det(B) and det(AB⁻¹) = det(A)det(B⁻¹) = det(A)/det(B). Given that both determinants equal 8, we can derive that det(B) = 2det(A). Using these relationships, we can compute det(BA⁻¹B^T) and find that it equals 1/16.

Q40. The number of values of θ ∈ (0, π) for which the system of linear equations x + 3y + 7z = 0, -x + 4y + 7z = 0, (sin 3θ)x + (cos 2θ)y + 2z = 0 has a non-trivial solution, is -

1. (1) two
2. (2) one
3. (3) four
4. (4) three

Answer: (1) two

The system of equations has a non-trivial solution when the determinant of the coefficient matrix is zero. By analyzing the conditions under which the determinant vanishes, we find that there are two specific values of θ ∈ (0, π) that satisfy this condition.

Q41. If A = [[e^t, e^(-t) cos t, e^(-t) sin t], [e^t, -e^(-t) cos t, -e^(-t) sin t], [e^t, 2e^(-t) sin t, -2e^(-t) cos t]] then A is:

1. invertible only if t = π
2. invertible for all t ∈ R
3. invertible only if t = π/2
4. not invertible for any t ∈ R

Answer: invertible for all t ∈ R

Computing the determinant gives det(A) = 4e^(-t), which is nonzero for every real t. Hence A is invertible for all t in R.

Q42. Let λ be a real number for which the system of linear equations x + y + z = 6, 4x + λy - λz = λ - 2, 3x + 2y - 4z = 5 has infinitely many solutions. Then λ is a root of the quadratic equation:

1. λ² + λ - 6 = 0
2. λ² - λ - 6 = 0
3. λ² - 3λ - 4 = 0
4. λ² + 3λ - 4 = 0

Answer: λ² - λ - 6 = 0

The correct option is right because for the system to have infinitely many solutions, the equations must be dependent, leading to a specific relationship between the coefficients of the variables, which results in the quadratic equation λ² - λ - 6 = 0.

Q43. If the system of equations 2x + 3y - z = 0, x + ky - 2z = 0 and 2x - y + z = 0 has a non-trivial solution (x, y, z), then x/y + y/z + z/x + k is equal to:

1. 1/2
2. 3/4
3. -1/4
4. -4

Answer: 1/2

For the system of equations to have a non-trivial solution, the determinant of the coefficient matrix must be zero. Solving for k under this condition reveals that the expression x/y + y/z + z/x + k simplifies to 1/2.

Q44. An ordered pair (α, β) for which the system of linear equations (1 + α)x + βy + z = 2 αx + (1 + β)y + z = 3 αx + βy + 2z = 2 has a unique solution is:

1. (-3, 1)
2. (1, -3)
3. (-4, 2)
4. (2, 4)

Answer: (2, 4)

The ordered pair (2, 4) ensures that the determinant of the coefficient matrix of the system is non-zero, which is a requirement for the system to have a unique solution. This means that the equations are linearly independent and intersect at a single point.

Q45. The set of all values of λ for which the system of linear equations x - 2y - 2z = λx x + 2y + z = λy -x - y = λz has a non-trivial solutions:

1. is an empty set
2. contains more than two elements
3. is a singleton
4. contains exactly two elements

Answer: is a singleton

The system of equations can be rewritten in matrix form, and for non-trivial solutions to exist, the determinant of the coefficient matrix must be zero. Solving for λ reveals that there is only one specific value that satisfies this condition, indicating that the set of values for which non-trivial solutions exist is a singleton.

Q46. Let S be the set of all λ ∈ R for which the system of linear equations 2x - y + 2z = 2, x - 2y + λz = -4, x + λy + z = 4 has no solution. Then the set S

1. contains more than two elements.
2. is a singleton.
3. contains exactly two elements.
4. is an empty set.

Answer: contains exactly two elements.

The system of linear equations has no solution when the determinant of the coefficient matrix is zero, leading to a specific condition on λ that results in exactly two values where the system becomes inconsistent.

Q47. Suppose the vectors x1, x2 and x3 are the solutions of the system of linear equations, Ax = b when the vector b on the right side is equal to b1, b2 and b3 respectively. If x1 = [1, 1, 1]^T, x2 = [0, 2, 1]^T, x3 = [0, 0, 1]^T, b1 = [1, 0, 0]^T, b2 = [0, 2, 0]^T and b3 = [0, 0, 2]^T, then the determinant of A is equal to:

1. 1/2
2. 4
3. 2
4. 3/2

Answer: 2

The determinant of matrix A can be calculated using the vectors x1, x2, and x3 as they represent the solutions to the system of equations. The volume of the parallelepiped formed by these vectors in three-dimensional space is given by the absolute value of the determinant, which in this case is 2, indicating that the vectors are linearly independent and span a volume of 2.

Q48. If the minimum and the maximum values of the function f: [π/4, π/2] → R, defined by: f(θ) = | -sin² θ -1 - sin² θ 1 | | -cos² θ -1 - cos² θ 1 | | 12 10 -2 | are m and M respectively, then the ordered pair (m, M) is equal to:

1. (0, 4)
2. (-4, 4)
3. (0, 2√2)
4. (-4, 0)

Answer: (-4, 0)

Subtracting rows reduces the determinant to 4(cos^2 theta - sin^2 theta) = 4cos(2theta). For theta in [pi/4, pi/2], 2theta in [pi/2, pi], so cos(2theta) in [-1,0] and the value lies in [-4,0]. Hence (m,M) = (-4, 0).

Q49. If a + x = b + y = c + z + 1, where a, b, c, x, y, z are non-zero distinct real numbers, then | x a + y x + a / y b + y y + b / z c + y z + c | is equal to:

1. (1) 0
2. (2) y(a - b)
3. (3) y(b - a)
4. (4) y(a - c)

Answer: (1) 0

The expression simplifies to zero because the equalities imply that the terms involving a, b, and c balance out, leading to a cancellation in the overall expression.

Q50. [2020] Let m and M be respectively the minimum and maximum values of the determinant [cos² x, 1+sin² x, sin 2x; 1+cos² x, sin² x, sin 2x; cos² x, sin² x, 1+sin 2x]. Then the ordered pair (m, M) is equal to -

1. (-3, -1)
2. (1, 3)
3. (-3, 3)
4. (-4, -1)

Answer: (-3, -1)

Expanding the determinant gives a value ranging over [-3,-1] as x varies, so (m,M)=(-3,-1).