Matrix Operations, Determinants and Trace PYQs for GATE CS

Solve 6+ Matrix Operations, Determinants and Trace previous year questions for GATE CS with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Matrix Operations, Determinants and Trace

Chapter Roadmap

Matrix Operations, Determinants and Trace

Master the structural properties of matrices, moving from high-power computations to elegant invariants.

Matrix Powers and Polynomial Identities

Compute massive powers using diagonalization and characteristic equations.

Determinants under Row Operations

Predict determinant changes under swaps, scaling, and permutations.

Trace Identities and Cyclic Properties

Master

tr (A B) = tr (B A)

and eigenvalue sums.

Determinant Scaling

Calculate the exact effect of scalar multiplication on determinants.

Why this matters: These concepts are the bedrock of Linear Algebra, essential for solving systems of equations, eigenvalue problems, and quadratic forms.

Matrix Powers and Polynomial Identities

A^{k}

Matrix Powers and Polynomial Identities

Calculating $A^{100}$ by repeated multiplication is a trap. Structural properties and polynomial identities allow you to solve these problems in seconds.

What you'll learn

The diagonalization shortcut for computing $A^{k}$ .
How to identify periodic and nilpotent matrices.
Using the Cayley-Hamilton theorem to reduce high exponents.

Matrix Operations, Determinants and Trace: Solved Questions with Step-by-Step Explanations (6 Problems)

Question 1 · Engineering Mathematics · 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 2 · Engineering Mathematics · 2025_Set2 MCQ

Let

L

M

, and

N

be non-singular matrices of order 3 satisfying the equations

L^{2} = L^{- 1}

M = L^{8}

and

N = L^{2}

.

Which ONE of the following is the value of the determinant of

(M - N)

A.
0
B.
1
C.
2
D.
3

Correct Answer:

Step-by-Step Solution

Insight: This is a matrix polynomial identity question, recognizable because it gives a relation between powers of a matrix ( $L^{2} = L^{- 1}$ ) and asks for a property of high powers ( $L^{8}$ ).

Exam route:

From $L^{2} = L^{- 1}$ , multiply both sides by $L$ to get $L^{3} = I$ .
Reduce $M = L^{8}$ using $L^{3} = I$ : $L^{8} = (L^{3})^{2} \cdot L^{2} = I^{2} \cdot L^{2} = L^{2}$ .
Since $N = L^{2}$ , we have $M - N = L^{2} - L^{2} = 0$ (the zero matrix).
The determinant of the zero matrix is 0.

Learning route:

Analyze the given equation: $L^{2} = L^{- 1}$ . Since $L$ is non-singular, $L^{- 1}$ exists. Multiply both sides on the right by $L$ : $L^{2} \cdot L = L^{- 1} \cdot L ⟹ L^{3} = I$ .
This means $L$ is a periodic matrix with period 3. Any power of $L$ can be reduced modulo 3.
Evaluate $M = L^{8}$ : Divide the exponent 8 by the period 3. $8 = 3 \times 2 + 2$ . So $L^{8} = (L^{3})^{2} \cdot L^{2} = I^{2} \cdot L^{2} = I \cdot L^{2} = L^{2}$ .
We are given $N = L^{2}$ .
Compute the difference: $M - N = L^{2} - L^{2} = 0$ , where 0 is the $3 \times 3$ zero matrix.
The determinant of any zero matrix is 0. Thus, $det (M - N) = 0$ .

Common trap: Students might assume $L^{3} = I$ implies $L = I$ , which is false (e.g., rotation matrices). Even if they did, $M = I, N = I ⟹ M - N = 0 ⟹ det = 0$ , so it accidentally gives the right answer, but the reasoning is flawed. Another trap is trying to compute $det (M) - det (N)$ , which is invalid since $det (A - B) \neq = det (A) - det (B)$ .

Verification: Let $L$ be a $3 \times 3$ rotation matrix of $12 0^{\circ}$ around some axis. Then $L^{3} = I$ . $L^{2}$ is a $24 0^{\circ}$ rotation. $M = L^{8} = L^{2}$ . $N = L^{2}$ . $M - N = 0$ . $det (0) = 0$ .

Question 3 · Engineering Mathematics · 2025_Set2 MCQ

If $A = (12 2 - 1)$ , then which ONE of the following is $A^{8}$ ?

A.
$(250025)$
B.
$(12500125)$
C.
$(62500625)$
D.
$(3125003125)$

Correct Answer:

Step-by-Step Solution

Key idea: This is a matrix powers question, recognizable because it asks for a high power ( $A^{8}$ ) of a $2 \times 2$ matrix. The trigger is the large exponent, which signals that brute-force multiplication is a trap and a structural shortcut (like diagonalization or finding a minimal polynomial) is required.

Step 1: Compute $A^{2}$ to look for a pattern.

$A = (12 2 - 1)$

$A^{2} = (12 2 - 1) (12 2 - 1) = (1 (1) + 2 (2) 2 (1) + (- 1) (2) 1 (2) + 2 (- 1) 2 (2) + (- 1) (- 1))$

$A^{2} = (5005) = 5 I$ .

Step 2: Use the pattern to find $A^{8}$ .

Since $A^{2} = 5 I$ , we can raise both sides to the 4th power:

$(A^{2})^{4} = (5 I)^{4}$

$A^{8} = 5^{4} I^{4} = 625 I$ .

Step 3: Write out the final matrix.

$A^{8} = (62500625)$ .

Answer: Option C.

Question 4 · Engineering Mathematics · 2024_Set2 MSQ

Let $A$ be an $n \times n$ matrix over the set of all real numbers $R$ . Let $B$ be a matrix obtained from $A$ by swapping two rows. Which of the following statements is/are TRUE?

A.
The determinant of $B$ is the negative of the determinant of $A$
B.
If $A$ is invertible, then $B$ is also invertible
C.
If $A$ is symmetric, then $B$ is also symmetric
D.
If the trace of $A$ is zero, then the trace of $B$ is also zero

Correct Answer:

["A","B"]

Step-by-Step Solution

Key idea: This is a row operations question, recognizable because it asks about the effect of swapping two rows on various matrix properties (determinant, invertibility, symmetry, trace). The trigger is "B is obtained from A by swapping two rows."

Step 1: Analyze the effect on the determinant. Swapping any two rows of a matrix multiplies its determinant by $- 1$ . Thus, $det (B) = - det (A)$ . Statement A is TRUE.

Step 2: Analyze invertibility. A matrix is invertible if and only if its determinant is non-zero. Since $det (B) = - det (A)$ , if $det (A) \neq = 0$ , then $det (B) \neq = 0$ . Thus, if $A$ is invertible, $B$ is also invertible. Statement B is TRUE.

Step 3: Analyze symmetry. A matrix is symmetric if $A = A^{T}$ . Swapping rows $i$ and $j$ changes the diagonal elements $a_{ii}$ and $a_{j j}$ to $a_{j i}$ and $a_{ij}$ respectively. In general, this destroys symmetry. Statement C is FALSE.

Step 4: Analyze the trace. The trace is the sum of the main diagonal elements. As shown in Step 3, swapping rows replaces diagonal elements with off-diagonal elements, changing the trace in general. Statement D is FALSE.

Answer: Statements A and B are TRUE.

Question 5 · Engineering Mathematics · 2023 MCQ

Let

A = 1432214332144321

and

B = 3412412312342341 .

Let

det (A)

and

det (B)

denote the determinants of the matrices

A

and

B

, respectively.

Which one of the options given below is TRUE?

A.
$det (A) = det (B)$
B.
$det (B) = - det (A)$
C.
$det (A) = 0$
D.
$det (A B) = det (A) + det (B)$

Correct Answer:

Step-by-Step Solution

Key idea: This is a determinant permutation question, recognizable because it compares the determinants of two matrices that are row-permutations of each other. The trigger is the cyclic shifting of rows between $A$ and $B$ .

Step 1: Write down the rows of $A$ and $B$ .

$A$ : R1=(1,2,3,4), R2=(4,1,2,3), R3=(3,4,1,2), R4=(2,3,4,1).

$B$ : R1=(3,4,1,2), R2=(4,1,2,3), R3=(1,2,3,4), R4=(2,3,4,1).

Step 2: Compare the rows to find the permutation.

Notice that Row 1 of $B$ is exactly Row 3 of $A$ .

Row 2 of $B$ is exactly Row 2 of $A$ .

Row 3 of $B$ is exactly Row 1 of $A$ .

Row 4 of $B$ is exactly Row 4 of $A$ .

Step 3: Determine the effect on the determinant.

Matrix $B$ is obtained from $A$ by swapping Row 1 and Row 3. This is a single elementary row swap.

Step 4: A single row swap multiplies the determinant by $- 1$ .

Therefore, $det (B) = - det (A)$ .

Answer: Option B.

Question 6 · Engineering Mathematics · 2022 MCQ

Consider the following two statements with respect to the matrices

A_{m \times n}

B_{n \times m}

C_{n \times n}

and

D_{n \times n}

.

Statement 1:

tr (A B) = tr (B A)

Statement 2:

tr (C D) = tr (D C)

where

tr ()

represents the trace of a matrix. Which one of the following holds?

A.
Statement 1 is correct and Statement 2 is wrong.
B.
Statement 1 is wrong and Statement 2 is correct.
C.
Both Statement 1 and Statement 2 are correct.
D.
Both Statement 1 and Statement 2 are wrong.

Correct Answer:

Step-by-Step Solution

Key idea: This is a trace identity question, recognizable because it tests the cyclic property of the trace operator for matrix products. The trigger is the presence of $tr (A B)$ and $tr (B A)$ for matrices of compatible dimensions.

Step 1: Analyze Statement 1. Let $A$ be an $m \times n$ matrix and $B$ be an $n \times m$ matrix. The product $A B$ is an $m \times m$ matrix, and $B A$ is an $n \times n$ matrix. Both traces are well-defined.

Step 2: By definition, $tr (A B) = \sum_{i = 1}^{m} (A B)_{ii} = \sum_{i = 1}^{m} \sum_{j = 1}^{n} A_{ij} B_{j i}$ .

Step 3: Similarly, $tr (B A) = \sum_{j = 1}^{n} (B A)_{j j} = \sum_{j = 1}^{n} \sum_{i = 1}^{m} B_{j i} A_{ij}$ .

Step 4: Since scalar multiplication is commutative ( $A_{ij} B_{j i} = B_{j i} A_{ij}$ ) and finite sums can be swapped, $tr (A B) = tr (B A)$ . Statement 1 is correct.

Step 5: Analyze Statement 2. $C$ and $D$ are both $n \times n$ matrices. This is a special case of Statement 1 where $m = n$ . Thus, $tr (C D) = tr (D C)$ is also correct.

Step 6: Both statements are universally true for matrices of compatible dimensions.

Answer: Both Statement 1 and Statement 2 are correct. (Option C)

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 Trace PYQs for GATE CS

Chapter Roadmap: Matrix Operations, Determinants and Trace

Matrix Operations, Determinants and Trace

Matrix Powers and Polynomial Identities

Matrix Powers and Polynomial Identities

What you'll learn

Matrix Operations, Determinants and Trace: Solved Questions with Step-by-Step Explanations (6 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

More previous year questions (pyqs) in this unit