Linear Algebra PYQs 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 Previous Year Questions (PYQs)

20 previous year questions from Linear Algebra in GATE CS, grouped by chapter with the exam year, answer key and step-by-step solution for each.

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 · 2026_Set2 NAT

Consider the system of linear equations given below.

a x + y = b

16 x + a y = 24

Suppose the values of

a

and

b

are chosen such that the system of linear equations produce multiple solutions. Then the product of

a

and

b

is __________. (answer in integer)

Correct Answer:

Step-by-Step Solution

Key idea: This is a parameterized system question asking for infinite solutions. It is recognisable because it uses parameters $a$ and $b$ in the coefficients and constants, and explicitly states the system has "multiple solutions".

Step 1: Write down the condition for infinite solutions in a 2x2 system.

For the system $a_{1} x + b_{1} y = c_{1}$ and $a_{2} x + b_{2} y = c_{2}$ to have infinitely many solutions, the two equations must represent the same line. This means their coefficients and constants must be strictly proportional:

$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ .

Step 2: Apply the proportionality condition to the given system.

The system is:

$a x + y = b$

$16 x + a y = 24$

So, $\frac{a}{16} = \frac{1}{a} = \frac{b}{24}$ .

Step 3: Solve for $a$ .

From the first equality: $\frac{a}{16} = \frac{1}{a} ⟹ a^{2} = 16 ⟹ a = 4$ or $a = - 4$ .

Step 4: Solve for $b$ in both cases.

Case 1: $a = 4$ .

$\frac{1}{4} = \frac{b}{24} ⟹ b = 6$ .

The product $ab = 4 \times 6 = 24$ .

Case 2: $a = - 4$ .

$\frac{1}{- 4} = \frac{b}{24} ⟹ b = - 6$ .

The product $ab = (- 4) \times (- 6) = 24$ .

Step 5: Conclude the final answer.

In both valid scenarios, the product $ab$ is exactly 24.

Answer: 24

Question 2 · Vector Spaces, Rank, Nullity and Orthogonality · 2026_Set1 MSQ

Let

n > 1

. Consider an

n \times n

matrix

M

with its elements from

R

. Let the vector

(0, 1, 0, 0, \dots, 0) \in R^{n}

be in the null space of

M

.

Which of the following options is/are always correct?

A.
Determinant of $M$ is 1
B.
Determinant of $M$ is 0
C.
Rank of $M$ is 1
D.
There are at least two non-zero vectors in the null space of $M$

Correct Answer:

["B","D"]

Step-by-Step Solution

Key idea: This is a "null space membership implies matrix properties" question, recognisable because a specific non-zero vector is given as belonging to the null space, and we must deduce which matrix properties are guaranteed.

Step 1: Interpret the given information.

The vector $v = (0, 1, 0, 0, \dots, 0)^{T} \in R^{n}$ is in the null space of $M$ . This means:

$M v = 0$

Since $v \neq = 0$ , the homogeneous system $M x = 0$ has a non-trivial solution.

Step 2: Evaluate Option A (det $M = 1$ ).

A non-trivial null space means $M$ is singular. For a singular matrix, $det (M) = 0 \neq = 1$ . Option A is FALSE.

Step 3: Evaluate Option B (det $M = 0$ ).

Since $M v = 0$ with $v \neq = 0$ , the columns of $M$ are linearly dependent (specifically, the second column must be the zero vector, since $M v$ picks out the second column). A matrix with linearly dependent columns has determinant zero. Option B is TRUE.

Step 4: Evaluate Option C (rank $M = 1$ ).

We know $det (M) = 0$ , so rank $(M) < n$ . But the rank could be any value from $0$ to $n - 1$ . For example:

If $M = 0$ (zero matrix), rank = 0.
If $M = diag (1, 0, 1, 1, \dots, 1)$ , rank = $n - 1$ .

Neither is necessarily 1. Option C is FALSE.

Step 5: Evaluate Option D (at least two non-zero vectors in null space).

The null space is a vector subspace. If $v$ is in the null space, then every scalar multiple $c v$ is also in the null space. Since $v \neq = 0$ , the vectors $v$ and $2 v$ are both non-zero and distinct. In fact, there are infinitely many non-zero vectors in the null space. Option D is TRUE.

Answer: B, D

Question 3 · Matrix Operations, Determinants and Trace · 2026_Set2 NAT

The determinant of a $4 \times 4$ matrix $A$ is 3. The value of the determinant of $2 A$ is ____________. <i>(answer in integer)</i>

Correct Answer:

48.00

Step-by-Step Solution

Insight: This is a determinant scaling question, recognizable because it asks for the determinant of a scalar multiple of a matrix, $det (k A)$ .

Exam route: Use the master scaling formula $det (k A) = k^{n} det (A)$ , where $n$ is the order of the matrix. Here $n = 4$ , $k = 2$ , $det (A) = 3$ . So $det (2 A) = 2^{4} \times 3 = 16 \times 3 = 48$ .

Learning route:

Recall that multiplying a matrix $A$ by a scalar $k$ means multiplying every single row of $A$ by $k$ .
The determinant is a multilinear function of the rows. Pulling out the scalar $k$ from one row multiplies the determinant by $k$ .
Since an $n \times n$ matrix has $n$ rows, pulling out $k$ from every row multiplies the determinant by $k$ exactly $n$ times.
Therefore, $det (k A) = k^{n} det (A)$ .
Substitute the given values: $n = 4$ (since $A$ is a $4 \times 4$ matrix), $k = 2$ , and $det (A) = 3$ .
Calculate: $det (2 A) = 2^{4} \times 3 = 16 \times 3 = 48$ .

Common trap: Students often use the total number of elements ( $n^{2} = 16$ ) as the exponent, calculating $2^{16} \times 3$ , or they forget to raise $k$ to the power $n$ and just calculate $2 \times 3 = 6$ . The exponent must always be the number of rows, $n$ .

Verification: If $A = diag (3^{1/4}, 3^{1/4}, 3^{1/4}, 3^{1/4})$ , then $det (A) = 3$ . $2 A = diag (2 \cdot 3^{1/4}, \dots)$ , $det (2 A) = (2 \cdot 3^{1/4})^{4} = 16 \times 3 = 48$ .

Question 4 · Eigenvalues and Eigenvectors · 2026_Set1 MCQ

For $n > 1$ , the maximum multiplicity of any eigenvalue of an $n \times n$ matrix with elements from $R$ is

A.
$n$
B.
$n - 1$
C.
1
D.
$n + 1$

Correct Answer:

Step-by-Step Solution

Insight: The characteristic polynomial of an $n \times n$ matrix has degree exactly $n$ , so no eigenvalue can repeat more than $n$ times — and the identity matrix achieves this bound.

Exam route: The characteristic equation $det (A - λ I) = 0$ is a polynomial of degree $n$ . The maximum multiplicity of any root of a degree- $n$ polynomial is $n$ . The identity matrix $I_{n}$ has characteristic polynomial $(1 - λ)^{n} = 0$ , so eigenvalue $1$ has multiplicity $n$ . Answer: $n$ .

Learning route: This is a theoretical maximum-multiplicity question, recognisable because no specific matrix is given and the answer is in terms of $n$ .

Step 1: For any $n \times n$ matrix $A$ , the characteristic equation $det (A - λ I) = 0$ expands to a polynomial in $λ$ of degree exactly $n$ .

Step 2: By the Fundamental Theorem of Algebra, this polynomial has exactly $n$ roots counting multiplicity. The algebraic multiplicity of a single eigenvalue is the number of times it appears as a root.

Step 3: Since the total count of all roots (with multiplicity) is $n$ , no single eigenvalue can have algebraic multiplicity exceeding $n$ .

Step 4: To confirm $n$ is achievable, consider $A = I_{n}$ . Its characteristic polynomial is $(1 - λ)^{n} = 0$ , giving $λ = 1$ with algebraic multiplicity exactly $n$ .

Wrong path — Option B ( $n - 1$ ): A student confuses this with the rank-nullity theorem or thinks "at least one eigenvalue must differ." This produces $n - 1$ . It breaks at Step 4: the identity matrix is a direct counterexample where all $n$ eigenvalues are identical.

Wrong path — Option C ( $1$ ): A student assumes all eigenvalues must be distinct, or confuses algebraic multiplicity with the minimum geometric multiplicity. This produces $1$ . It breaks at Step 3: nothing prevents all $n$ roots from coinciding.

Wrong path — Option D ( $n + 1$ ): A student does not realise the characteristic polynomial has degree exactly $n$ and thinks multiplicity can exceed the matrix size. This produces $n + 1$ . It breaks at Step 1: a degree- $n$ polynomial cannot have a root of multiplicity $n + 1$ .

Generalization: The sum of algebraic multiplicities of all eigenvalues of an $n \times n$ matrix always equals $n$ , so the maximum any single eigenvalue can claim is the entire sum.

Verification: For $n = 3$ , $I_{3}$ has characteristic polynomial $(1 - λ)^{3} = 0$ , giving eigenvalue $1$ with multiplicity $3 = n$ . Confirmed.

A question from this chapter

Login to view the complete previous-year questions and solutions

Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

One place for everything

Honest progress

Unlock the whole course

Linear Algebra PYQs for GATE CS

About Linear Algebra Previous Year Questions (PYQs)

Linear Algebra Weightage in GATE CS

Linear Algebra Chapter Matrix

More from Engineering Mathematics

One Solved Question from Each Linear Algebra Chapter

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution