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205 questions with worked solutions.
Q1. If A is a square matrix, then the matrix product A A^T is a
Answer: symmetric matrix
(A A^T)^T = (A^T)^T A^T = A A^T, so A A^T equals its own transpose and is therefore symmetric.
Answer: -I₂
Each matrix is a rotation by -a, so the product is a rotation by -(alpha+beta+gamma). In a triangle alpha+beta+gamma=pi, giving rotation by -pi, which is the matrix [[-1,0],[0,-1]] = -I_2.
Q3. Let A, B, and C be n × n matrices. Which of the following statements is true?
Answer: If A³ + 2A² + 3A + 5I = 0, then A is invertible.
Rewrite as A(A²+2A+3I) = -5I, so A·[-(1/5)(A²+2A+3I)] = I, giving an explicit inverse; hence A is invertible. (AB=AC need not give B=C, and A²=0 need not give A=0, e.g. a nonzero nilpotent matrix.)
Q4. Given the matrix A_α = [cos α -sin α; sin α cos α], which of the following identities is true?
Answer: A_α · A_β = A_(α+β)
The correct option states that the product of two rotation matrices A_α and A_β results in another rotation matrix A_(α+β), which reflects the additive property of angles in trigonometric functions. This means that rotating by angle α followed by angle β is equivalent to a single rotation by the sum of those angles.
Q5. Let A be a square matrix satisfying (A−2I)(A+I)=O. Then A−1 equals:
Answer: (A−I)/2
The equation (A−2I)(A+I)=O implies that A has eigenvalues 2 and -1. The eigenvalue 2 leads to the inverse A−1 being expressed as (A−I)/2, since for eigenvalue 2, the corresponding eigenvector satisfies A*v = 2*v, leading to v = (A−I)/2*v.
Answer: 6
From P^2 = I - P, each power has the form P^k = a_k I + b_k P with (a,b): P^1=(0,1), P^2=(1,-1), P^3=(-1,2), P^4=(2,-3), P^5=(-3,5), P^6=(5,-8). Thus P^6 = 5I - 8P, so n = 6.
Answer: 7A⁸
Direct multiplication gives A^2 = I, so A^4=A^6=I and A^2+2A^4+4A^6=7I. Since 7A^8 = 7(A^2)^4 = 7I, the answer is 7A^8.
Q8. Let A = [1 2; 3 4] and B = [a 0; 0 b], where a, b are natural numbers. Then
Answer: there are infinitely many matrices B for which AB = BA
With B = diag(a,b), AB = BA requires 2b=2a and 3a=3b, hence a=b. Then B = aI commutes with A for every natural number a, so there are infinitely many such matrices B.
Q9. If [1 ω ω²; ω ω² 1; ω² 1 ω] [k 1 1; 1 1 1; 1 1 1] = [0 0 0; 0 0 0; 0 0 0], then the value of k is
Answer: 1
Multiplying the two matrices, the only non-zero entries lie in the first column; the (1,1) entry equals k-1, and the other first-column entries also reduce to 0 precisely when k=1. So k=1.
Answer: I + A
With A=[[0,-t],[t,0]], t=tan(α/2), and using cosα=(1-t²)/(1+t²), sinα=2t/(1+t²), direct multiplication of (I−A) by the rotation matrix gives [[1,-t],[t,1]] = I + A.
Answer: [6 6 7; 6 2 5; 7 5 1]
A + A^T = [[12,12,14],[12,4,10],[14,10,2]], so B = (A+A^T)/2 = [[6,6,7],[6,2,5],[7,5,1]]. (Option 1 shown is the skew-symmetric part, not B.)
Q12. If B is an idempotent matrix and A is defined by A = I − B, which of the following is true?
Answer: A² = A
An idempotent matrix B satisfies the property B² = B. Therefore, when we compute A² = (I - B)(I - B), we can expand this to I - 2B + B². Since B² = B, this simplifies to I - 2B + B = I - B, which is equal to A. Thus, A² = A.
Q13. For the matrix F(α) = [cos α −sin α 0; sin α cos α 0; 0 0 1], the product F(α)F(β) is equal to
Answer: F(α + β)
F(a) is a rotation by angle a (with a fixed third axis). The product of two rotations is rotation by the sum of the angles, so F(a)F(b) = F(a+b).
Answer: A(z) = A(x)A(y)
Computing A(x)A(y) and simplifying with z=(x+y)/(1+xy) shows A(x)A(y)=A(z). Hence A(z)=A(x)A(y).
Q15. For the matrix A = [0 −1; 1 0], what is the value of A¹⁶?
Answer: [1 0; 0 1]
A = [[0,-1],[1,0]] rotates by 90 degrees, so A^4 = I. Since 16 = 4*4, A^16 = I = [[1,0],[0,1]].
Answer: α² + βγ = 1
For M=[[a,b],[g,-a]], M^2 = (a^2+bg) I. Setting this equal to the identity requires a^2 + beta*gamma = 1.
Q17. If A and B are matrices of the same order, then the matrix expression AB^T - BA^T is a
Answer: skew symmetric matrix
Transposing: (AB^T-BA^T)^T = (AB^T)^T-(BA^T)^T = BA^T-AB^T = -(AB^T-BA^T). A matrix equal to the negative of its transpose is skew-symmetric.
Q18. If a matrix A is both symmetric and skew-symmetric, then A must be
Answer: Zero matrix
A matrix that is both symmetric and skew-symmetric must have all its entries equal to their negatives and themselves, which can only occur if all entries are zero, resulting in the zero matrix.
Q19. If the matrices AB, A, and B are all equal, then what is the value of B squared?
Answer: B
Since AB = A = B, we can substitute B for A in the equation AB = B, leading to B² = B. This indicates that B is idempotent, confirming that the value of B squared is B itself.
Q20. Let A and B be square matrices satisfying B = -A⁻¹BA. Then what is the value of (A + B)²?
Answer: A² + B²
The equation B = -A⁻¹BA implies a specific relationship between A and B that leads to the cancellation of the cross term when expanding (A + B)², resulting in A² + B².
Answer: (AB)^T = I
AB = [[2,-1],[-7,4]][[4,1],[7,2]] = I, and BA = I too, so AB = BA. Also A A^T != I and B B^T != I. Hence (AB)^T = I^T = I is the correct statement.
Q22. For any square matrix A, which of the following expressions is skew-symmetric?
Answer: A - A^T
A matrix S is skew-symmetric if S^T=-S. For S=A-A^T, S^T=A^T-A=-(A-A^T)=-S, so A-A^T is always skew-symmetric (while A+A^T and AA^T are symmetric).
Q23. If A = [ 3 7 1 2 ], then the value of the determinant | A²⁰¹³ − 3A²⁰¹² | is equal to
Answer: −7
The determinant of a matrix raised to a power can be calculated using the property that |Aⁿ| = |A|ⁿ. In this case, the determinant of A is -1, so |A²⁰¹³| = (-1)²⁰¹³ = -1 and |3A²⁰¹²| = 3²⁰¹² * |A| = 3²⁰¹² * (-1). Therefore, the determinant of the expression simplifies to -1 - 3²⁰¹², which evaluates to -7.
Answer: p = -1
When p = -1, the equations simplify in a way that allows for consistent solutions across all three equations, ensuring that the system is solvable. This value maintains the integrity of the relationships between x and y, making it the only viable option.
Q25. Let A, B, and C be n × n matrices. Which of the following statements is true?
Answer: If A³ + 2A² + 3A + 5I = 0, then A is nonsingular.
The statement is true because if A satisfies the polynomial equation A³ + 2A² + 3A + 5I = 0, it implies that the determinant of A cannot be zero, thus A must be nonsingular.
Answer: F(−α)
F(alpha) is a rotation through angle alpha about the z-axis, so its inverse is the rotation through -alpha, i.e. [F(alpha)]^-1 = F(-alpha).
Q27. Let A be the 2×2 matrix [[1, 1], [1, 1]]. What is A raised to the 100th power?
Answer: 2⁹⁹ [[1, 1], [1, 1]]
Since A=[[1,1],[1,1]] gives A^2=[[2,2],[2,2]]=2A, by induction A^n = 2^(n-1) A. Hence A^100 = 2^99 [[1,1],[1,1]].
Q28. Let A = [[cos θ, -sin θ], [sin θ, cos θ]]. If A^T + A = I₂, then the value of θ is
Answer: 2nπ + π/3, n ∈ Z
For A=[[cos,-sin],[sin,cos]], A^T+A has diagonal 2cos(theta) and off-diagonals cancel, giving 2cos(theta)*I = I. Hence cos(theta)=1/2, so theta = 2n*pi +/- pi/3; the matching option is 2n*pi + pi/3.
Q29. For a matrix A satisfying A² = 0, what is the expansion of (aI + bA)²?
Answer: (aI + bA)² = a² I + b² A² + 2ab A
Expand: (aI+bA)^2 = a^2 I + ab(IA) + ab(AI) + b^2 A^2 = a^2 I + 2ab A + b^2 A^2. Since A^2 = 0, this equals a^2 I + 2ab A, i.e. a^2 I + b^2 A^2 + 2ab A.
Answer: nilpotent
det(P) = -1 and det(Q) = -1; computing PQ gives the identity matrix I = [[1,0],[0,1]], so (PQ)^-1 = I. The identity is idempotent (I^2 = I), involutory (I^2 = I), and symmetric, but it is never nilpotent. The property that does NOT hold is nilpotent.
Answer: T₃(A²) = [T₃(A)]²
The statement T₃(A²) = [T₃(A)]² is not necessarily true because the trace of the product of two matrices does not equal the product of their traces. In general, T₃(A²) equals T₃(A) when A is a diagonal matrix, but this does not hold for all matrices.
Answer: k = 3, m ≠ 4/5
Determinant = 24 - 8k, which is 0 only at k=3. With k=3 the equations force 4y=1 from eqs 1&2 but 4y=5m-3 from eqs 2&3, consistent only if 5m-3=1 i.e. m=4/5. So no solution when k=3 and m != 4/5.
Q33. If A = [a b; b a] and A² = [α β; β α], then
Answer: α = a² + b², β = 2ab
The correct option states that α equals the sum of the squares of the elements on the diagonal of matrix A, while β represents twice the product of the off-diagonal elements. This is derived from the matrix multiplication of A with itself, which results in the specified forms for α and β.
Answer: Aⁿ = nA − (n − 1)I
The correct option is valid because it can be proven using mathematical induction. By establishing a base case and then assuming it holds for n, we can show that it also holds for n+1, confirming that the relationship between Aⁿ, nA, and I is consistent for all n ≥ 1.
Answer: AB = BA
The equation A² − B² = (A − B)(A + B) holds true for any square matrices A and B, indicating that the matrices commute (AB = BA) under this condition, which is a fundamental property of matrix multiplication.
Q36. Let A = [1 2; 3 4] and B = [a 0; 0 b], a, b ∈ N. Then
Answer: there exist infinitely many B's such that AB = BA
The matrices A and B commute if their product is the same regardless of the order of multiplication. Since B is a diagonal matrix, it can take on infinitely many combinations of the natural numbers for a and b, allowing for an infinite number of matrices B that satisfy the commutation relation with A.
Q37. The number of 3 × 3 non-singular matrices, with four entries as 1 and all other entries as 0, is
Answer: at least 7
A 3 × 3 non-singular matrix with four entries as 1 and one entry as 0 must have a unique arrangement that maintains linear independence among its rows and columns. The configurations that satisfy this condition can be achieved through various placements of the 1s, leading to at least 7 valid matrices.
Answer: Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
Statement-1 is true because the product of symmetric matrices is symmetric under certain conditions, and both A(BA) and (AB)A maintain symmetry. Statement-2 is also true since if A and B commute, their product AB is symmetric, but it does not explain the symmetry of the products in Statement-1.
Answer: H
Since H is a diagonal matrix with the same eigenvalue ω on the diagonal, raising H to any power will yield the same matrix scaled by ω raised to that power. Since ω is a cube root of unity, ω³ = 1, and thus ω⁷⁰ = ω^(3*23 + 1) = ω, leading to H⁷⁰ = H.
Answer: (-2, -1)
Row1.Row3=a+4+2b=0 and Row2.Row3=2a+2-2b=0 give b=a+1 and 3a+6=0, so a=-2, b=-1. Thus (a,b)=(-2,-1).
Q41. If A = [[5a, -b], [3, 2]] and A adj A = A A^T, then 5a + b is equal to:
Answer: 5
A adj A = det(A) I = (10a+3b)I. A A^T = [[25a^2+b^2, 15a-2b],[15a-2b,13]]. Off-diagonal: 15a-2b=0. Diagonal: 13=10a+3b. Solving gives a=2/5, b=3, so 5a+b = 2+3 = 5.
Q42. If A = [[2, -3], [-4, 1]], then adj(3A² + 12A) is equal to:
Answer: [[51, 63], [84, 72]]
A^2 = [[16,-9],[-12,13]], so 3A^2+12A = [[48,-27],[-36,39]] + [[24,-36],[-48,12]] = [[72,-63],[-84,51]]. adj of [[a,b],[c,d]] = [[d,-b],[-c,a]] = [[51,63],[84,72]].
Q43. If A = [[cos θ, -sin θ], [sin θ, cos θ]], then the matrix A⁻⁵⁰ when θ = π/12, is equal to:
Answer: [[√3/2, 1/2], [-1/2, √3/2]]
A is a rotation matrix, so A^n rotates by n*theta. With theta=pi/12, -50*theta = -25pi/6 = -pi/6 (mod 2pi). Thus A^(-50) = [[cos(-pi/6), -sin(-pi/6)],[sin(-pi/6), cos(-pi/6)]] = [[sqrt(3)/2, 1/2],[-1/2, sqrt(3)/2]].
Answer: [[1, -13], [0, 1]]
Multiplying the upper-triangular matrices adds the off-diagonal entries: 1+2+...+(n-1)=n(n-1)/2=78 -> n(n-1)=156 -> n=13. The inverse of [[1,13],[0,1]] is [[1,-13],[0,1]].
Answer: A³
The matrix A is constructed using the roots of the polynomial, which are related to the cube roots of unity. Since the roots satisfy the relation α³ = 1, raising A to the power of 3 results in a matrix that cycles back to its original form, thus A³ is equivalent to A.
Q46. If a square matrix A satisfies A² - A + I = 0, what is A⁻¹ equal to?
Answer: I - A
From A^2-A+I=0, A(I-A)=I, so A^-1=I-A.
Answer: Statement 1 is true, Statement 2 is false
Statement 1 is true because if A² = I and A is neither I nor -I, it must have eigenvalues of 1 and -1, leading to a determinant of -1. Statement 2 is false because the trace can be zero if the eigenvalues are 1 and -1, which contradicts the claim.
Answer: If det A = ±1, then A⁻¹ exists and all of its entries are integers
For an integer matrix A, A^(-1) = adj(A)/det(A). The adjugate (cofactor) entries are integers, so when det A = +/-1 the inverse exists and every entry is an integer.
Answer: Statement I is true, Statement II is true, but Statement II does not correctly explain Statement I.
For order n=2: adj(adj A)=|A|^(n-2) A = A, so I is true; and |adj A|=|A|^(n-1)=|A|, so II is true. But II does not explain I, so the correct choice is 'both true, II not a correct explanation'.
Answer: The equations are inconsistent, so no solution exists.
The equations are inconsistent because they do not intersect at any point in space, indicating that there is no set of values for the variables that satisfies all three equations simultaneously.