Matrix Operations, Determinants and Gaussian Elimination PYQs for GATE DA

Solve 2+ Matrix Operations, Determinants and Gaussian Elimination previous year questions for GATE DA with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Matrix Operations, Determinants and Gaussian Elimination

Chapter Journey

Gaussian Elimination & Complexity

Counting additions/multiplications and seeing how matrix structure changes cost.

CURRENT TOPIC

Determinants of Matrix Expressions

Using determinant properties to simplify matrix expressions.

Target: Build the operation-counting habit first. The structural thinking here supports determinant shortcuts later.

Topic Hero: Gaussian Elimination as a Costed Algorithm

Two-Phase Process for $A x = b$

Phase 1: Forward Elimination

Create zeros below pivots
Convert to $U x = c$

Phase 2: Back Substitution

Solve from last variable upward
Uses upper triangular form

Why count operations? The cost matters as much as the answer. We track additions, subtractions, multiplications, and divisions to understand scaling with

n

Core intuition: Gaussian elimination is cheap when the matrix already has the zeros that elimination would otherwise create.

Matrix Operations, Determinants and Gaussian Elimination: Solved Questions with Step-by-Step Explanations (2 Problems)

Question 1 · Linear Algebra · 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:

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 · Linear Algebra · 2024 NAT

Consider the

3 \times 3

matrix

M = 134213336

.
The determinant of

(M^{2} + 12 M)

is ______.

Correct Answer:

0.00

Step-by-Step Solution

Insight: The matrix polynomial $M^{2} + 12 M$ factors as $M (M + 12 I)$ . If $det (M) = 0$ , the whole product has determinant $0$ without any need to square the matrix or compute the second factor.

Exam route:

Factor the expression: $M^{2} + 12 M = M (M + 12 I)$ .
Use multiplicativity: $det (M^{2} + 12 M) = det (M) \cdot det (M + 12 I)$ .
Inspect $M$ for linear dependence. Row 3 $= (4, 3, 6) = (1, 2, 3) + (3, 1, 3) = R_{1} + R_{2}$ .
Therefore $det (M) = 0$ .
The product is $0 \times det (M + 12 I) = 0$ .

Learning route:

The determinant is multiplicative: $det (A B) = det (A) det (B)$ . Since $M$ and $M + 12 I$ are both polynomials in the same matrix $M$ , they commute, and the factorisation $M^{2} + 12 M = M (M + 12 I)$ is valid at the matrix level. Hence

$det (M^{2} + 12 M) = det (M) \cdot det (M + 12 I) .$

Compute $det (M)$ by inspection. The rows of $M$ are

$R_{1} = (1, 2, 3), R_{2} = (3, 1, 3), R_{3} = (4, 3, 6) .$

Observe that $R_{1} + R_{2} = (1 + 3, 2 + 1, 3 + 3) = (4, 3, 6) = R_{3}$ . The rows are linearly dependent, so $det (M) = 0$ .

(Equivalently, expanding along the first row: $det (M) = 1 (1 \cdot 6 - 3 \cdot 3) - 2 (3 \cdot 6 - 3 \cdot 4) + 3 (3 \cdot 3 - 1 \cdot 4) = 1 (- 3) - 2 (6) + 3 (5) = - 3 - 12 + 15 = 0$ .)

Since one factor has determinant $0$ , the entire product has determinant $0$ , regardless of the value of $det (M + 12 I)$ .

The common wrong path is to compute $M^{2}$ explicitly (a $3 \times 3$ matrix multiplication), then add $12 M$ , then expand a $3 \times 3$ determinant. That is legal but wasteful and invites arithmetic errors. The factorisation shortcut collapses the work to a single dependency check.

Verification: $det (M) = 0$ is confirmed by two independent routes (row dependence and cofactor expansion). Any product containing a singular factor is singular, so $det (M (M + 12 I)) = 0$ is exact.

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Matrix Operations, Determinants and Gaussian Elimination PYQs for GATE DA

Chapter Roadmap: Matrix Operations, Determinants and Gaussian Elimination

Chapter Journey

Topic Hero: Gaussian Elimination as a Costed Algorithm

Two-Phase Process for $A x = b$

Matrix Operations, Determinants and Gaussian Elimination: Solved Questions with Step-by-Step Explanations (2 Problems)

Step-by-Step Solution

Step-by-Step Solution

More previous year questions (pyqs) in this unit

Try a question

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

Matrix Operations, Determinants and Gaussian Elimination PYQs for GATE DA

Chapter Roadmap: Matrix Operations, Determinants and Gaussian Elimination

Chapter Journey

Topic Hero: Gaussian Elimination as a Costed Algorithm

Two-Phase Process for Ax=b

Matrix Operations, Determinants and Gaussian Elimination: Solved Questions with Step-by-Step Explanations (2 Problems)

Step-by-Step Solution

Step-by-Step Solution

More previous year questions (pyqs) in this unit

Two-Phase Process for $A x = b$