GATE CS Linear Algebra Previous Year Questions

GATE CS 2025 Set 1 | Question 23

MSQ · 1 mark

Consider the given system of linear equations for variables \(x\) and \(y\), where \(k\) is a real-valued constant. Which of the following option(s) is/are CORRECT?

\(x + ky = 1\)
\(kx + y = -1\)

There is exactly one value of \(k\) for which the above system of equations has no solution. There exist an infinite number of values of \(k\) for which the system of equations has no solution. There exists exactly one value of \(k\) for which the system of equations has exactly one solution. There exists exactly one value of \(k\) for which the system of equations has an infinite number of solutions.

Explanation

Answer: (A), (D)

The determinant of the coefficient matrix is \(1 \cdot 1 - k \cdot k = 1 - k^{2}\). For every \(k \ne \pm 1\) it is non-zero, so the system has exactly one solution.

For \(k = 1\): \(x + y = 1\) and \(x + y = -1\) contradict each other, so there is no solution. For \(k = -1\): \(x - y = 1\) and \(-x + y = -1\) are the same equation, so there are infinitely many solutions.

(A) True: only \(k = 1\) gives no solution.
(B) False: no solution happens for \(k = 1\) alone.
(C) False: infinitely many values (every \(k \ne \pm 1\)) give exactly one solution.
(D) True: only \(k = -1\) gives infinitely many solutions.

Answer: A, D

GATE CS 2025 Set 1 | Question 41

MCQ · 2 marks · −0.66

Let \(A\) be a \(2 \times 2\) matrix as given.

\[ A = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \]

What are the eigenvalues of the matrix \(A^{13}\)?

\(1, -1\) \(2\sqrt{2}, -2\sqrt{2}\) \(4\sqrt{2}, -4\sqrt{2}\) \(64\sqrt{2}, -64\sqrt{2}\)

Explanation

Answer: (D)

Squaring \(A\) gives a shortcut:

\[ A^2 = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = 2I, \qquad A^{13} = (A^2)^6 A = 2^6 A = 64A. \]

The eigenvalues of \(A\) satisfy \(\det(A - \lambda I) = (1 - \lambda)(-1 - \lambda) - 1 = \lambda^2 - 2 = 0\), so \(\lambda = \pm\sqrt{2}\). Hence \(A^{13} = 64A\) has eigenvalues \(\pm 64\sqrt{2}\), the same as \((\pm\sqrt{2})^{13} = \pm 2^6\sqrt{2}\).

Answer: D

GATE CS 2025 Set 2 | Question 11

MCQ · 1 mark · −0.33

If \(A = \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix}\), then which ONE of the following is \(A^{8}\) ?

\(\begin{pmatrix} 25 & 0 \\ 0 & 25 \end{pmatrix}\) \(\begin{pmatrix} 125 & 0 \\ 0 & 125 \end{pmatrix}\) \(\begin{pmatrix} 625 & 0 \\ 0 & 625 \end{pmatrix}\) \(\begin{pmatrix} 3125 & 0 \\ 0 & 3125 \end{pmatrix}\)

Explanation

Answer: (C)

Square the matrix first. The diagonal entries are \(1 \cdot 1 + 2 \cdot 2 = 5\) and \(2 \cdot 2 + (-1)(-1) = 5\), and both off-diagonal entries are \(1 \cdot 2 + 2 \cdot (-1) = 0\), so

\[A^{2} = \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} = 5I.\]

Then \(A^{8} = (A^{2})^{4} = 5^{4}I = 625I\). The other options are \(5^{2}I\), \(5^{3}I\) and \(5^{5}I\), which are \(A^{4}\), \(A^{6}\) and \(A^{10}\).

Answer: C

GATE CS 2025 Set 2 | Question 14

MCQ · 1 mark · −0.33

Let \(L, M\), and \(N\) be non-singular matrices of order 3 satisfying the equations

\(L^{2} = L^{-1}\) , \(M = L^{8}\) and \(N = L^{2}\).

Which ONE of the following is the value of the determinant of \((M - N)\)?

0 1 2 3

Explanation

Answer: (A)

Multiply \(L^{2} = L^{-1}\) on the right by \(L\) to get \(L^{3} = I\). Powers of \(L\) therefore repeat every three steps:

\[M = L^{8} = (L^{3})^{2}L^{2} = L^{2} = N.\]

So \(M - N\) is the zero matrix, and its determinant is 0. Non-singularity only guarantees that \(L^{-1}\) exists; it says nothing about the difference \(M - N\), so the non-zero options cannot occur.

Answer: A

GATE CS 2025 Set 2 | Question 44

MSQ · 2 marks

Consider a system of linear equations \(PX = Q\) where \(P \in \mathbb{R}^{3 \times 3}\) and \(Q \in \mathbb{R}^{3 \times 1}\). Suppose \(P\) has an LU decomposition, \(P = LU\), where

\[ L = \begin{bmatrix} 1 & 0 & 0 \\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1 \end{bmatrix} \text{ and } U = \begin{bmatrix} u_{11} & u_{12} & u_{13} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & u_{33} \end{bmatrix}. \]

Which of the following statement(s) is/are TRUE?

The system \(PX = Q\) can be solved by first solving \(LY = Q\) and then \(UX = Y\). If \(P\) is invertible, then both \(L\) and \(U\) are invertible. If \(P\) is singular, then at least one of the diagonal elements of \(U\) is zero. If \(P\) is symmetric, then both \(L\) and \(U\) are symmetric.

Explanation

Answer: (A), (B), (C)

A true. \(PX = Q\) is \(L(UX) = Q\). Put \(Y = UX\): solve \(LY = Q\) by forward substitution, then \(UX = Y\) by back substitution.
B true. \(L\) has a unit diagonal, so \(\det L = 1\) and \(\det P = \det L \cdot \det U = u_{11}u_{22}u_{33}\). If \(P\) is invertible this is non-zero, so \(U\) is invertible, and \(L\) always is.
C true. If \(P\) is singular, \(u_{11}u_{22}u_{33} = \det P = 0\), so some \(u_{ii} = 0\).
D false. A triangular matrix is symmetric only if it is diagonal. The symmetric \(P = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 1 \end{bmatrix}\) has \(l_{21} = 1\) and \(u_{12} = 1\) with \(l_{12} = u_{21} = 0\), so neither factor is symmetric.

Answer: A, B, C

GATE CS 2024 Set 1 | Question 12

MCQ · 1 mark · −0.33

The product of all eigenvalues of the matrix \(\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}\) is

\(-1\) 0 1 2

Explanation

Answer: (B)

The product of all eigenvalues of a square matrix equals its determinant.

Here the rows are linearly dependent: each row exceeds the previous one by \((3, 3, 3)\), so \(R_{1} + R_{3} = 2R_{2}\). A matrix with dependent rows has determinant 0:

\[\det\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} = 0.\]

So 0 is one of the eigenvalues, and the product of all the eigenvalues is 0, whatever the other two are.

Answer: B

GATE CS 2024 Set 1 | Question 49

MSQ · 2 marks

Let \(A\) be any \(n \times m\) matrix, where \(m > n\). Which of the following statements is/are TRUE about the system of linear equations \(Ax = \mathbf{0}\) ?

There exist at least \(m - n\) linearly independent solutions to this system There exist \(m - n\) linearly independent vectors such that every solution is a linear combination of these vectors There exists a non-zero solution in which at least \(m - n\) variables are 0 There exists a solution in which at least \(n\) variables are non-zero

Explanation

Answer: (A)

Since \(A\) has only \(n\) rows, \(\operatorname{rank}(A) \le n\). By rank–nullity, the solutions of \(Ax = \mathbf{0}\) form a subspace of dimension \(m - \operatorname{rank}(A) \ge m - n \ge 1\).

(A) True: a basis of this subspace contains at least \(m - n\) linearly independent solutions.
(B) False: it needs the dimension to be exactly \(m - n\). For the \(1 \times 3\) zero matrix every vector in \(\mathbb{R}^{3}\) is a solution, and \(m - n = 2\) vectors cannot span a 3-dimensional space.
(C) False: for \(A = \begin{pmatrix} 1 & 1 \end{pmatrix}\) the solutions are \((t, -t)\), so no non-zero solution has a zero entry.
(D) False: for \(A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}\) every solution is \((0, 0, t)\), with at most one non-zero entry rather than \(n = 2\).

Answer: A

GATE CS 2024 Set 2 | Question 47

MSQ · 2 marks

Let \(A\) be an \(n \times n\) matrix over the set of all real numbers \(\mathbb{R}\). Let \(B\) be a matrix obtained from \(A\) by swapping two rows. Which of the following statements is/are TRUE?

The determinant of \(B\) is the negative of the determinant of \(A\) If \(A\) is invertible, then \(B\) is also invertible If \(A\) is symmetric, then \(B\) is also symmetric If the trace of \(A\) is zero, then the trace of \(B\) is also zero

Explanation

Answer: (A), (B)

(A) True: swapping two rows multiplies the determinant by \(-1\), so \(\det B = -\det A\).
(B) True: if \(A\) is invertible, \(\det B = -\det A \ne 0\), so \(B\) is invertible too.
(C) False: \(\begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix}\) is symmetric, but swapping its rows gives \(\begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}\), which is not.
(D) False: \(\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\) has trace 0, but swapping its rows gives the identity matrix, whose trace is 2. A row swap moves entries onto and off the diagonal, so the trace can change.

Answer: A, B