Linear Algebra Practice Questions for GATE CS

GATE CS Linear Algebra: 4 chapters, 20 previous year questions (19% of Engineering Mathematics), 285 practice questions and one solved question from each chapter.

About Linear Algebra Practice Questions

285 practice questions for Linear Algebra in GATE CS, sorted chapter by chapter and graded from basic to exam level, each with a full solution.

Linear Algebra Weightage in GATE CS

Linear Algebra accounts for 20 of 105 Engineering Mathematics previous year questions in our bank (19%), about 2 per paper across 10 papers.

Linear Algebra Chapter Matrix

Chapter	Topics	PYQs	Share of unit PYQs	Practice questions
Systems of Linear Equations and LU Decomposition	Consistency of Parameterized Linear Systems, LU Decomposition Properties and Solution Strategy, LU Factorization and Triangular Substitution	4	20%	74
Vector Spaces, Rank, Nullity and Orthogonality	Homogeneous Systems and Nontrivial Null Spaces, Rank-Nullity Computations, Orthogonality and Maximum Orthogonal Sets	4	20%	87
Matrix Operations, Determinants and Trace	Matrix Powers and Polynomial Identities, Determinants under Row Operations and Permutations, Trace Identities and Cyclic Properties, Determinant Scaling	6	30%	94
Eigenvalues and Eigenvectors	Eigenvalues, Determinants and Multiplicity, Eigenvector Verification, Spectra of Graph Adjacency Matrices, Eigenvalues of Matrix Powers	6	30%	30

One Solved Question from Each Linear Algebra Chapter

Question 1 · Systems of Linear Equations and LU Decomposition MCQ

Consider the matrix $A = 121 2 k 4 14 k$ , where $k$ is a real number.

It is a common misconception that a matrix has a standard $LU$ factorization (without row swaps) if and only if it is non-singular.

Let $S$ be the set of all real values of $k$ for which $A$ is singular, but its standard $LU$ factorization (Doolittle's method) exists and is successfully computed without encountering a zero pivot during the algorithm.

What is the maximum value of $k$ in the set $S$ ?

Options

A. $1$

B. $5$

C. $4$

D. $8$

A.
1
B.
5
C.
4
D.
8

Correct Answer:

Step-by-Step Solution

Key idea: This is a contradiction/boundary question. We must find when $det (A) = 0$ (singular) but the $LU$ algorithm doesn't hit a zero pivot (which happens if $U_{22} = 0$ ).

Step 1: Apply Doolittle's algorithm symbolically.

Row 1 of $U$ : $U_{11} = 1$ , $U_{12} = 2$ , $U_{13} = 1$ .
Col 1 of $L$ : $L_{21} = 2/1 = 2$ , $L_{31} = 1/1 = 1$ .
Row 2 of $U$ : $U_{22} = k - L_{21} U_{12} = k - 4$ .
$U_{23} = 4 - L_{21} U_{13} = 4 - 2 (1) = 2$ .
Col 2 of $L$ : $L_{32} = \frac{4 - L _{31} U _{12}}{U _{22}} = \frac{4 - 2}{k - 4} = \frac{2}{k - 4}$ .
Row 3 of $U$ : $U_{33} = k - L_{31} U_{13} - L_{32} U_{23} = k - 1 - \frac{2}{k - 4} (2) = k - 1 - \frac{4}{k - 4} = \frac{k ^{2} - 5 k}{k - 4}$ .

Step 2: Identify the boundary for the algorithm.

The algorithm encounters a zero pivot if $U_{11} = 0$ or $U_{22} = 0$ .

$U_{11} = 1 \neq = 0$ .

$U_{22} = k - 4 = 0 ⟹ k = 4$ .

So, the $LU$ factorization is valid (no zero pivot) for all $k \neq = 4$ .

Step 3: Find when $A$ is singular.

$det (A) = U_{11} U_{22} U_{33} = (1) (k - 4) (\frac{k ^{2} - 5 k}{k - 4}) = k^{2} - 5 k$ .

$A$ is singular when $det (A) = 0 ⟹ k (k - 5) = 0 ⟹ k = 0$ or $k = 5$ .

Step 4: Intersect the conditions.

We need $A$ to be singular ( $k \in {0, 5}$ ) AND $LU$ to be valid ( $k \neq = 4$ ).

Both $k = 0$ and $k = 5$ satisfy $k \neq = 4$ .

Thus, $S = {0, 5}$ .

Step 5: Find the maximum value.

The maximum value in $S$ is $5$ .

Answer: B

Question 2 · Vector Spaces, Rank, Nullity and Orthogonality MCQ

Let $A$ be a $4 \times 4$ real matrix such that the system $A x = 0$ has a non-trivial solution, and the dimension of the solution space of $A^{2} x = 0$ is exactly 3. What is the minimum possible dimension of the null space of $A$ ?

A.
1
B.
2
C.
3
D.
4

Correct Answer:

Step-by-Step Solution

Key idea: This is a casework question about the sequence of null space dimensions for powers of a matrix. The trap lies in assuming the nullity can jump arbitrarily, ignoring the boundary condition on how much the null space can grow in each step.

Step 1: Let $N_{k}$ be the dimension of the null space of $A^{k}$ . We are given $N_{0} = 0$ , $N_{1} \geq 1$ (since $A x = 0$ has a non-trivial solution), and $N_{2} = 3$ .

Step 2: The sequence of nullities $N_{k}$ has a fundamental property: the differences $d_{k} = N_{k} - N_{k - 1}$ are non-increasing. That is, $d_{2} \leq d_{1}$ .

Step 3: We know $d_{1} = N_{1} - N_{0} = N_{1}$ . And $d_{2} = N_{2} - N_{1} = 3 - N_{1}$ .

Step 4: Apply the boundary condition: $3 - N_{1} \leq N_{1} ⟹ 2 N_{1} \geq 3 ⟹ N_{1} \geq 1.5$ .

Step 5: Since $N_{1}$ must be an integer, the minimum possible value for $N_{1}$ is 2.

Answer: 2

Question 3 · Matrix Operations, Determinants and Trace MCQ

Let $A$ be a $2 \times 2$ matrix satisfying $A^{2} = A + I$ , where $I$ is the $2 \times 2$ identity matrix.

Match the matrix expressions in List-I with their equivalent forms in List-II.

\textbf{List-I}

P. $A^{3}$

Q. $A^{4}$

R. $A^{5}$

S. $A^{6}$

\textbf{List-II}

$3 A + 2 I$
$5 A + 3 I$
$2 A + I$
$8 A + 5 I$

A.
P $i g h t a r r o w$ 3, Q $i g h t a r r o w$ 1, R $i g h t a r r o w$ 2, S $i g h t a r r o w$ 4
B.
P $i g h t a r r o w$ 3, Q $i g h t a r r o w$ 2, R $i g h t a r r o w$ 1, S $i g h t a r r o w$ 4
C.
P $i g h t a r r o w$ 1, Q $i g h t a r r o w$ 3, R $i g h t a r r o w$ 2, S $i g h t a r r o w$ 4
D.
P $i g h t a r r o w$ 3, Q $i g h t a r r o w$ 1, R $i g h t a r r o w$ 4, S $i g h t a r r o w$ 2

Correct Answer:

Step-by-Step Solution

Key idea: This is a matrix polynomial reduction question. The relation $A^{2} = A + I$ allows us to express any higher power of $A$ as a linear combination of $A$ and $I$ .

Step 1: Compute $A^{3}$ by multiplying $A^{2} = A + I$ by $A$ :

$A^{3} = A \cdot A^{2} = A (A + I) = A^{2} + A$

Substitute $A^{2} = A + I$ :

$A^{3} = (A + I) + A = 2 A + I$

So P $\to$ 3.

Step 2: Compute $A^{4}$ by multiplying $A^{3}$ by $A$ :

$A^{4} = A \cdot A^{3} = A (2 A + I) = 2 A^{2} + A$

Substitute $A^{2} = A + I$ :

$A^{4} = 2 (A + I) + A = 3 A + 2 I$

So Q $\to$ 1.

Step 3: Compute $A^{5}$ :

$A^{5} = A \cdot A^{4} = A (3 A + 2 I) = 3 A^{2} + 2 A$

Substitute $A^{2} = A + I$ :

$A^{5} = 3 (A + I) + 2 A = 5 A + 3 I$

So R $\to$ 2.

Step 4: Compute $A^{6}$ :

$A^{6} = A \cdot A^{5} = A (5 A + 3 I) = 5 A^{2} + 3 A$

Substitute $A^{2} = A + I$ :

$A^{6} = 5 (A + I) + 3 A = 8 A + 5 I$

So S $\to$ 4.

The correct matching is P $\to$ 3, Q $\to$ 1, R $\to$ 2, S $\to$ 4.

Answer: A

Question 4 · Eigenvalues and Eigenvectors NAT

Let $A$ be a $3 \times 3$ real matrix with all eigenvalues real. If the trace of $A$ is $6$ and the determinant of $A$ is $8$ , then the maximum possible value of the largest eigenvalue of $A$ is

Correct Answer:

8.00

Step-by-Step Solution

Key idea: This is an optimization problem with eigenvalue constraints. We need to maximize the largest eigenvalue subject to trace and determinant constraints, while ensuring all eigenvalues are real.

Step 1: Set up the constraints.

Let the eigenvalues be $λ_{1}, λ_{2}, λ_{3}$ (all real). Assume $λ_{1} \geq λ_{2} \geq λ_{3}$ .

$λ_{1} + λ_{2} + λ_{3} = 6 (trace)$

$λ_{1} λ_{2} λ_{3} = 8 (determinant)$

Step 2: Express $λ_{2}, λ_{3}$ in terms of $λ_{1}$ .

$λ_{2} + λ_{3} = 6 - λ_{1}$

$λ_{2} λ_{3} = \frac{8}{λ _{1}}$

Step 3: Apply the real eigenvalue constraint.

For $λ_{2}, λ_{3}$ to be real, they must be roots of a quadratic with non-negative discriminant:

$t^{2} - (λ_{2} + λ_{3}) t + λ_{2} λ_{3} = 0$

$t^{2} - (6 - λ_{1}) t + \frac{8}{λ _{1}} = 0$

Discriminant condition:

$(6 - λ_{1})^{2} - 4 \cdot \frac{8}{λ _{1}} \geq 0$

$(6 - λ_{1})^{2} - \frac{32}{λ _{1}} \geq 0$

Step 4: Solve the inequality (assuming $λ_{1} > 0$ ).

Multiply by $λ_{1}$ :

$λ_{1} (6 - λ_{1})^{2} - 32 \geq 0$

$λ_{1} (36 - 12 λ_{1} + λ_{1}^{2}) - 32 \geq 0$

$λ_{1}^{3} - 12 λ_{1}^{2} + 36 λ_{1} - 32 \geq 0$

Factor: Try $λ_{1} = 2$ : $8 - 48 + 72 - 32 = 0$ ✓

$(λ_{1} - 2) (λ_{1}^{2} - 10 λ_{1} + 16) \geq 0$

$(λ_{1} - 2) (λ_{1} - 2) (λ_{1} - 8) \geq 0$

$(λ_{1} - 2)^{2} (λ_{1} - 8) \geq 0$

Since $(λ_{1} - 2)^{2} \geq 0$ always, we need $λ_{1} - 8 \geq 0$ or $λ_{1} = 2$ .

So $λ_{1} \geq 8$ or $λ_{1} = 2$ .

Step 5: Find the maximum.

The maximum value is $λ_{1} = 8$ .

Verify: If $λ_{1} = 8$ , then $λ_{2} + λ_{3} = - 2$ and $λ_{2} λ_{3} = 1$ .

The quadratic is $t^{2} + 2 t + 1 = (t + 1)^{2} = 0$ , so $λ_{2} = λ_{3} = - 1$ .

Check: $8 + (- 1) + (- 1) = 6$ ✓ and $8 \times (- 1) \times (- 1) = 8$ ✓

Answer: $8.00$

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Linear Algebra Practice Questions for GATE CS

About Linear Algebra Practice Questions

Linear Algebra Weightage in GATE CS

Linear Algebra Chapter Matrix

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One Solved Question from Each Linear Algebra Chapter

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Step-by-Step Solution

Step-by-Step Solution