About Linear Algebra Previous Year Questions (PYQs)

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

GATE DA Linear Algebra Unit-wise Weightage from Past Papers

We counted every GATE DA Linear Algebra previous year question in our bank (20 questions from 3 papers) and grouped them by unit.

Unit	Chapters	PYQs	Share of section	Avg per paper
Matrices	8	13	65%	4.3
Matrix Decompositions	4	2	10%	0.7
Vector Spaces	5	5	25%	1.7
Unit 1 — Linear Algebra	3	0	0%	0
Unit 2 — Linear Algebra	3	0	0%	0

Suggested Linear Algebra Study Order for GATE DA

Matrices: 65% of past Linear Algebra questions, about 4.3 per paper.
Vector Spaces: 25% of past Linear Algebra questions, about 1.7 per paper.
Matrix Decompositions: 10% of past Linear Algebra questions, about 0.7 per paper.

Start where the marks are. Units at the top of this list have appeared most often in past GATE DA papers.

Units in GATE DA Linear Algebra

All Linear Algebra chapters

One Solved Question from Each Linear Algebra Chapter

Question 1 · Matrix Operations, Determinants and Gaussian Elimination · 2025 MCQ

The number of additions and multiplications involved in performing Gaussian elimination on any $n \times n$ upper triangular matrix is of the order

A.
$O (n)$
B.
$O (n^{2})$
C.
$O (n^{3})$
D.
$O (n^{4})$

Correct Answer:

B

Step-by-Step Solution

Insight: An upper triangular matrix already has all the zeros that forward elimination would create, so the cubic phase collapses to zero work. The only cost left is the quadratic back-substitution phase.

Exam route:

Recognise the matrix structure: $a_{ij} = 0$ for all $i > j$ .
Forward elimination has nothing to eliminate below any pivot $\Rightarrow$ cost $= 0$ .
The task "perform Gaussian elimination" means the complete solve (elimination + back substitution).
Back substitution on an $n \times n$ upper triangular system uses $\sum_{i = 1}^{n} (n - i) = \frac{n ( n - 1 )}{2}$ multiplications and the same number of additions.
Total additions + multiplications $= n (n - 1) = O (n^{2})$ .

Learning route:

Gaussian elimination has two phases. Phase 1 (forward elimination) converts $A \to U$ by zeroing entries below each pivot. For a general dense $n \times n$ matrix this costs $\frac{2 ( n ^{3} - n )}{3}$ additions and multiplications, i.e. $O (n^{3})$ . Phase 2 (back substitution) solves $U x = c$ from the bottom row upward, costing $n (n - 1)$ additions and multiplications, i.e. $O (n^{2})$ .

When the input is already upper triangular, every entry below the diagonal is already zero. At pivot step $k$ , there are no entries in column $k$ below row $k$ to remove, so no row updates are performed. Forward elimination contributes exactly $0$ operations.

The system still must be solved, so back substitution runs in full. Variable $x_{i}$ requires $n - i$ multiplications and $n - i$ additions. Summing over $i = 1, \dots, n$ gives $\frac{n ( n - 1 )}{2}$ of each, for a combined total of $n (n - 1)$ , which is $O (n^{2})$ .

The common wrong path is to memorise "Gaussian elimination is $O (n^{3})$ " and answer $O (n^{3})$ without inspecting the matrix structure. That ignores the fact that the cubic cost comes entirely from the row updates that this matrix does not need.

Verification: For $n = 3$ , back substitution does $2 + 1 + 0 = 3$ multiplications and $3$ additions, total $6 = 3 (2)$ , matching $n (n - 1)$ . Growth is clearly quadratic, not cubic.

Question 2 · Singular Value Decomposition and Principal Component Analysis · 2025 NAT

Let

D = {x^{(1)}, ..., x^{(n)}}

be a dataset of

n

observations where each

x^{(i)} \in R^{100}

. It is given that

\sum_{i = 1}^{n} x^{(i)} = 0

. The covariance matrix computed from

D

has eigenvalues

λ_{i} = 100 2^{- i}

,

1 \leq i \leq 100

. Let

u \in R^{100}

be the direction of maximum variance with

u^{T} u = 1

.

The value of

\frac{1}{n} i = 1 \sum n (u^{T} x^{(i)})^{2} =

(Answer in integer)

Correct Answer:

50

Step-by-Step Solution

Key idea: This is a PCA maximum variance direction problem. The expression given is the definition of variance along a direction, which is maximized by the largest eigenvalue of the covariance matrix.

Step 1: Recognize that the dataset has mean zero ( $\sum_{i = 1}^{n} x^{(i)} = 0$ ). Thus, the covariance matrix $C$ is given by $C = \frac{1}{n} \sum_{i = 1}^{n} x^{(i)} (x^{(i)})^{T}$ .

Step 2: The expression to evaluate is $\frac{1}{n} \sum_{i = 1}^{n} (u^{T} x^{(i)})^{2}$ . Notice that $(u^{T} x^{(i)})^{2} = (u^{T} x^{(i)}) (x^{(i) T} u) = u^{T} (x^{(i)} (x^{(i)})^{T}) u$ .

Step 3: Substitute this into the sum: $\frac{1}{n} \sum_{i = 1}^{n} u^{T} (x^{(i)} (x^{(i)})^{T}) u = u^{T} (\frac{1}{n} \sum_{i = 1}^{n} x^{(i)} (x^{(i)})^{T}) u = u^{T} C u$ .

Step 4: The problem states that $u$ is the direction of maximum variance with $u^{T} u = 1$ . By the properties of PCA, the maximum variance along any unit vector is the largest eigenvalue of the covariance matrix $C$ .

Step 5: The eigenvalues are given as $λ_{i} = 100 \cdot 2^{- i}$ for $1 \leq i \leq 100$ . The largest eigenvalue occurs at $i = 1$ , which is $λ_{1} = 100 \cdot 2^{- 1} = 50$ .

Step 6: Therefore, $u^{T} C u = λ_{1} = 50$ .

Answer: 50

Question 3 · Vector Spaces, Subspaces and Bases · 2026 MCQ

Consider a set

S_{1} = {x = (x_{1}, x_{2}, x_{3})^{T} \in R^{3} ∣ x^{T} x \leq 16}

. Let

S_{2}

be another set which is a subspace of

R^{3}

with dimension two.
Which of the following gives the area of

S_{1} \cap S_{2}

?

A.
$16 π$
B.
$4 π$
C.
$4 π^{2}$
D.
$16 π^{2}$

Correct Answer:

A

Step-by-Step Solution

Insight: The intersection of a solid n-dimensional ball with a k-dimensional subspace passing through the origin is a solid k-dimensional ball of the exact same radius.

Exam route: $x^{T} x \leq 16$ means $∥ x ∥_{2} \leq 4$ , so it's a solid 3D ball of radius $R = 4$ . $S_{2}$ is a 2D subspace (a plane through the origin). The intersection is a 2D solid disk of radius 4. Area = $π R^{2} = 16 π$ .

Learning route:

Identify $S_{1}$ : The condition $x^{T} x \leq 16$ is equivalent to $∥ x ∥_{2}^{2} \leq 16$ , which means $∥ x ∥_{2} \leq 4$ . This defines a solid 3-dimensional $L_{2}$ ball centered at the origin with radius $R = 4$ .
Identify $S_{2}$ : A subspace of $R^{3}$ with dimension 2 is a flat plane that passes exactly through the origin.
Find the intersection $S_{1} \cap S_{2}$ : Slicing a solid 3D sphere with a plane that passes through its exact center (the origin) yields a solid 2D disk (a "great disk") with the same radius $R = 4$ .
Calculate the area: The area of a 2D disk of radius $R$ is $π R^{2}$ . Substituting $R = 4$ , we get Area = $π (4)^{2} = 16 π$ .

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 DA

About Linear Algebra Previous Year Questions (PYQs)

GATE DA Linear Algebra Unit-wise Weightage from Past Papers

Suggested Linear Algebra Study Order for GATE DA

Units in GATE DA Linear Algebra

All Linear Algebra chapters

One Solved Question from Each Linear Algebra Chapter

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution