Matrices PYQs for GATE DA

GATE DA Matrices: 8 chapters, 13 previous year questions (65% of Linear Algebra), 685 practice questions and one solved question from each chapter.

About Matrices Previous Year Questions (PYQs)

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

Matrices Weightage in GATE DA

Matrices accounts for 13 of 20 Linear Algebra previous year questions in our bank (65%), about 4.3 per paper across 3 papers.

Matrices Chapter Matrix

Chapter	Topics	PYQs	Share of unit PYQs	Practice questions
Chapter 1 — Matrices		0	0%	0
Matrix Operations, Determinants and Gaussian Elimination	Gaussian Elimination and Matrix Operation Complexity, Determinants of Matrix Expressions	2	15%	104
Chapter 2 — Matrices		0	0%	0
Rank, Invertibility and Linear Systems	Rank, Nullity and Matrix Polynomial Equations, Invertibility and Eigenvalues of Rank-One Updates	3	23%	158
Chapter 3 — Matrices		0	0%	0
Eigenvalues, Eigenvectors and Matrix Powers	Rotation Matrices, Matrix Powers and Trace-Eigenvalue Relations, Characteristic Polynomial and Nature of Eigenvalues	3	23%	154
Orthogonal, Projection and Special Matrices	Projection Matrices and Quadratic Forms, Orthogonal and Involutory Matrices	3	23%	157
Singular Values and Gram Matrices	Gram Matrices and Positive Definiteness, Singular Values and Spectral Properties of Special Matrices	2	15%	112

One Solved Question from Each Matrices 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:

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 · Rank, Invertibility and Linear Systems · 2025 MCQ

Let $A \in R^{n \times n}$ be such that $A^{3} = A$ . Which one of the following statements is ALWAYS correct?

A.
$A$ is invertible
B.
Determinant of $A$ is $0$
C.
The sum of the diagonal elements of $A$ is $1$
D.
$A$ and $A^{2}$ have the same rank

Correct Answer:

Step-by-Step Solution

Key idea: The matrix equation $A^{3} = A$ implies that the minimal polynomial of $A$ has distinct linear factors, making $A$ diagonalizable, and forcing a specific relationship between the null spaces of $A$ and $A^{2}$ .

Step 1: Rewrite the given equation as $A^{3} - A = 0$ , which factors to $A (A - I) (A + I) = 0$ .

Step 2: This means the minimal polynomial of $A$ must divide $p (x) = x (x - 1) (x + 1)$ . Since $p (x)$ has distinct linear roots ( $0, 1, - 1$ ), the matrix $A$ is diagonalizable.

Step 3: Evaluate the options using counterexamples or structural properties.

Option A: " $A$ is invertible". Counterexample: $A = 0$ (the zero matrix) satisfies $0^{3} = 0$ , but is not invertible.
Option B: "Determinant of $A$ is $0$ ". Counterexample: $A = I$ (identity matrix) satisfies $I^{3} = I$ , but $det (I) = 1 \neq = 0$ .
Option C: "The sum of diagonal elements is $1$ ". Counterexample: $A = 0$ has trace $0$ ; $A = I$ (for $n = 2$ ) has trace $2$ . Neither is always $1$ .
Option D: " $A$ and $A^{2}$ have the same rank". Let's prove this structurally.

Step 4: Prove $ker (A) = ker (A^{2})$ .

Clearly, if $A x = 0$ , then $A^{2} x = A (A x) = A (0) = 0$ . So $ker (A) \subseteq ker (A^{2})$ .
Conversely, if $A^{2} x = 0$ , multiply both sides by $A$ to get $A^{3} x = 0$ . Since $A^{3} = A$ , this means $A x = 0$ . So $ker (A^{2}) \subseteq ker (A)$ .

Step 5: Since $ker (A) = ker (A^{2})$ , their dimensions (nullities) are equal: $nullity (A) = nullity (A^{2})$ .

Step 6: By the Rank-Nullity Theorem, $rank (A) = n - nullity (A) = n - nullity (A^{2}) = rank (A^{2})$ . This is ALWAYS correct.

Answer: $A$ and $A^{2}$ have the same rank.

Question 3 · Eigenvalues, Eigenvectors and Matrix Powers · 2026 MCQ

Let

M = (cos θ s in θ - s in θ cos θ)

be a

2 \times 2

matrix, where

θ = \frac{2 π}{5}

, and

I_{2} = (1001)

.
Which of the following options is equal to

M^{2026}

A.
$M^{2}$
B.
$M$
C.
$M^{- 1}$
D.
$I_{2}$

Correct Answer:

Step-by-Step Solution

Insight: $M = R (2 π /5)$ is a rotation by $2 π /5$ , so $M^{5} = I$ . Reduce the exponent $2026$ modulo $5$ .

Exam route: $2026 = 5 \times 405 + 1$ , so $M^{2026} = (M^{5})^{405} \cdot M^{1} = I^{405} \cdot M = M$ .

Learning route:

Identify $M$ as the standard counter-clockwise rotation matrix $R (θ)$ with $θ = \frac{2 π}{5}$ .
A rotation by $θ$ raised to the $n$ -th power is a rotation by $n θ$ : $M^{n} = R (n θ)$ .
Compute $n θ = 2026 \times \frac{2 π}{5} = \frac{4052 π}{5}$ .
Reduce modulo $2 π$ : $\frac{4052}{5} = 810 + \frac{2}{5}$ , so $\frac{4052 π}{5} = 810 π + \frac{2 π}{5} = 405 (2 π) + \frac{2 π}{5}$ .
The effective angle is $\frac{2 π}{5} = θ$ , so $M^{2026} = R (θ) = M$ .
Alternatively, $M^{5} = R (5 \cdot 2 π /5) = R (2 π) = I$ . Then $2026 mod 5 = 1$ , so $M^{2026} = M^{1} = M$ .

Question 4 · Orthogonal, Projection and Special Matrices · 2026 MSQ

Let

M = (I_{n} - \frac{1}{n} 1 1^{T})

be a matrix, where

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

and

I_{n}

is the identity matrix of order

n

.

Which of the following options is/are correct?

A.
$M^{T} = M$
B.
$M^{2} = I_{n}$
C.
$T r a ce (M) = n$
D.
$M$ is a projection matrix

Correct Answer:

["A","D"]

Step-by-Step Solution

Key idea: This question tests the algebraic properties of the centering matrix $M = I_{n} - \frac{1}{n} 1 1^{T}$ .

Step 1: Analyze Option A ( $M^{T} = M$ ). The transpose of a sum is the sum of transposes, and $(1 1^{T})^{T} = (1^{T})^{T} 1^{T} = 1 1^{T}$ . Therefore, $M^{T} = (I_{n} - \frac{1}{n} 1 1^{T})^{T} = I_{n} - \frac{1}{n} 1 1^{T} = M$ . Option A is correct.

Step 2: Analyze Option B ( $M^{2} = I_{n}$ ). Compute $M^{2} = (I_{n} - \frac{1}{n} 1 1^{T}) (I_{n} - \frac{1}{n} 1 1^{T}) = I_{n} - \frac{2}{n} 1 1^{T} + \frac{1}{n ^{2}} 1 (1^{T} 1) 1^{T}$ . Notice that $1^{T} 1$ is the dot product of the all-ones vector with itself, which equals $n$ . Substituting this gives $I_{n} - \frac{2}{n} 1 1^{T} + \frac{1}{n ^{2}} (n) 1 1^{T} = I_{n} - \frac{1}{n} 1 1^{T} = M$ . Since $M^{2} = M$ (and $M \neq = I_{n}$ for $n > 1$ ), Option B is incorrect.

Step 3: Analyze Option C ( $T r a ce (M) = n$ ). Using the linearity of trace, $T r a ce (M) = T r a ce (I_{n}) - \frac{1}{n} T r a ce (1 1^{T})$ . We know $T r a ce (I_{n}) = n$ and $T r a ce (1 1^{T}) = 1^{T} 1 = n$ . Thus, $T r a ce (M) = n - \frac{1}{n} (n) = n - 1$ . Option C is incorrect.

Step 4: Analyze Option D ( $M$ is a projection matrix). A matrix is an orthogonal projection matrix if and only if it is symmetric ( $M^{T} = M$ ) and idempotent ( $M^{2} = M$ ). We proved both in Steps 1 and 2. Therefore, Option D is correct.

Answer: ["A", "D"]

Question 5 · Singular Values and Gram Matrices · 2025 MCQ

Let ${x_{1}, x_{2}, ..., x_{n}}$ be a set of linearly independent vectors in $R^{n}$ . Let the $(i, j)$ -th element of matrix $A \in R^{n \times n}$ be given by $A_{ij} = x_{i}^{T} x_{j}$ , $1 \leq i, j \leq n$ . Which one of the following statements is correct?

A.
$A$ is invertible
B.
$0$ is a singular value of $A$
C.
Determinant of $A$ is $0$
D.
$z^{T} A z = 0$ for some non-zero $z \in R^{n}$

Correct Answer:

Step-by-Step Solution

Key idea: This is a Gram matrix and positive definiteness question. We need to recognize that $A$ is formed by inner products of linearly independent vectors.

Step 1: Express $A$ in matrix form. Let $X$ be the $n \times n$ matrix whose columns are the vectors $x_{1}, x_{2}, \dots, x_{n}$ .

Step 2: Relate $X$ to $A$ . The $(i, j)$ -th element of $X^{T} X$ is the dot product of the $i$ -th row of $X^{T}$ (which is $x_{i}^{T}$ ) and the $j$ -th column of $X$ (which is $x_{j}$ ). Thus $(X^{T} X)_{ij} = x_{i}^{T} x_{j} = A_{ij}$ . So $A = X^{T} X$ .

Step 3: Since ${x_{1}, \dots, x_{n}}$ are linearly independent vectors in $R^{n}$ , the matrix $X$ is full rank (rank $n$ ) and therefore invertible.

Step 4: For any non-zero vector $z \in R^{n}$ : $z^{T} A z = z^{T} X^{T} X z = (X z)^{T} (X z) = ∥ X z ∥^{2}$ . Since $X$ is invertible, $X z \neq = 0$ for $z \neq = 0$ . Thus $∥ X z ∥^{2} > 0$ , which means $A$ is strictly positive definite.

Step 5: Evaluate options. Since $A$ is positive definite, all eigenvalues are strictly positive. Thus $det (A) > 0$ (not 0), 0 is not a singular value, and $z^{T} A z > 0$ for all non-zero $z$ . A positive definite matrix is always invertible.

Answer: Option A is correct.

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Matrices PYQs for GATE DA

About Matrices Previous Year Questions (PYQs)

Matrices Weightage in GATE DA

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One Solved Question from Each Matrices Chapter

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