Matrix Operations, Determinants and Trace Practice Questions for GATE CS

Solve 94+ Matrix Operations, Determinants and Trace practice 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 (10 Problems)

Question 1 · Engineering Mathematics MCQ

According to the Cayley-Hamilton theorem, a $2 \times 2$ matrix $A$ satisfies its characteristic equation $A^{2} - tr (A) A + det (A) I = 0$ . If $tr (A) = - 2$ and $det (A) = 5$ , which of the following is the correct characteristic equation for $A$ ?

A.
$A^{2} + 2 A + 5 I = 0$
B.
$A^{2} - 2 A + 5 I = 0$
C.
$A^{2} + 2 A - 5 I = 0$
D.
$A^{2} - 2 A - 5 I = 0$

Correct Answer:

Step-by-Step Solution

Key idea: Substitute trace and determinant into the characteristic equation.

Step 1: The characteristic equation is $A^{2} - tr (A) A + det (A) I = 0$ .

Step 2: Substitute $tr (A) = - 2$ and $det (A) = 5$ .

Step 3: $A^{2} - (- 2) A + 5 I = 0 ⟹ A^{2} + 2 A + 5 I = 0$ .

Answer: $A^{2} + 2 A + 5 I = 0$ .

Question 2 · Engineering Mathematics MCQ

Let $A$ be a $2 \times 2$ diagonalizable matrix such that $A^{2} = 5 I$ . What is the maximum possible value of $tr (A)$ ?

A.
$25$
B.
$5$
C.
0
D.
5

Correct Answer:

Step-by-Step Solution

Key idea: Relate matrix powers to eigenvalues.

Step 1: Since $A^{2} = 5 I$ , the eigenvalues $λ$ of $A$ must satisfy $λ^{2} = 5$ .

Step 2: Thus, the eigenvalues can be $5$ or $- 5$ .

Step 3: The trace of $A$ is the sum of its eigenvalues. The possible values for $tr (A)$ are $5 + 5 = 25$ , $5 - 5 = 0$ , or $- 5 - 5 = - 25$ .

Step 4: The maximum possible value is $25$ .

Answer: $25$ .

Question 3 · Engineering Mathematics MCQ

Assertion (A): If $A = (12 2 - 1)$ and $A^{2} = 5 I$ , then the trace of $A^{6}$ is 250.

Reason (R): The trace of $c I$ for a $2 \times 2$ matrix is $2 c$ .

A.
Both A and R are true and R is the correct explanation of A
B.
Both A and R are true but R is NOT the correct explanation of A
C.
A is true but R is false
D.
A is false but R is true

Correct Answer:

Step-by-Step Solution

Key idea: Evaluate the assertion and reason independently, then check their link.

Step 1: Evaluate Assertion (A). $A^{2} = 5 I ⟹ A^{6} = (A^{2})^{3} = (5 I)^{3} = 125 I$ . The trace of $125 I$ for a $2 \times 2$ matrix is $125 + 125 = 250$ . So, A is true.

Step 2: Evaluate Reason (R). The trace of $c I$ for an $n \times n$ matrix is $n c$ . For $n = 2$ , it is $2 c$ . So, R is true.

Step 3: Check the link. The calculation in A directly relies on the property stated in R to find the final trace value. Thus, R is the correct explanation of A.

Answer: Both A and R are true and R is the correct explanation of A.

Question 4 · Engineering Mathematics MCQ

Match the matrix equation in List I with its correct consequence in List II.

List I

P. $X^{3} = - I$

Q. $Y^{2} = Y^{- 1}$

R. $Z^{4} = 0$

List II

$Z^{12} = 0$
$X^{11} = - X^{2}$
$Y^{14} = Y^{2}$
$X^{11} = X^{2}$

A.
P-4, Q-3, R-1
B.
P-2, Q-4, R-1
C.
P-2, Q-3, R-1
D.
P-4, Q-4, R-1

Correct Answer:

Step-by-Step Solution

Key idea: This is a matrix polynomial reduction question. We use the given equations to find the periodicity or nilpotency of each matrix, then reduce the high exponents.

Step 1: Analyze P. $X^{3} = - I$ .

To find $X^{11}$ , divide 11 by 3: $11 = 3 (3) + 2$ .

So, $X^{11} = (X^{3})^{3} \cdot X^{2} = (- I)^{3} \cdot X^{2} = - I \cdot X^{2} = - X^{2}$ .

This matches item 2. (P $\to$ 2)

Step 2: Analyze Q. $Y^{2} = Y^{- 1}$ . Multiply both sides by $Y$ to get $Y^{3} = I$ .

To find $Y^{14}$ , divide 14 by 3: $14 = 3 (4) + 2$ .

So, $Y^{14} = (Y^{3})^{4} \cdot Y^{2} = I^{4} \cdot Y^{2} = Y^{2}$ .

This matches item 3. (Q $\to$ 3)

Step 3: Analyze R. $Z^{4} = 0$ . This means $Z$ is nilpotent of index 4.

Any power $k \geq 4$ can be written as $Z^{k} = Z^{4} \cdot Z^{k - 4} = 0 \cdot Z^{k - 4} = 0$ .

So, $Z^{12} = 0$ . This matches item 1. (R $\to$ 1)

The correct matching is P-2, Q-3, R-1.

Answer: C

Question 5 · Engineering Mathematics MCQ

A student applies a sequence of elementary row operations to a $3 \times 3$ matrix $A$ . The operations are:

Swap $R_{1}$ and $R_{2}$
$R_{3} \to R_{3} + 2 R_{1}$
$R_{2} \to 2 R_{2}$
$R_{1} \to R_{1} - R_{3}$

If the determinant of the resulting matrix is $8$ , what is the determinant of $A$ ?

A.
-1
B.
-4
C.
4
D.
1

Correct Answer:

Step-by-Step Solution

Key idea: We track the cumulative multiplier applied to the determinant by each elementary row operation.

Step 1: Swap $R_{1}$ and $R_{2} ⟹$ multiplier is $- 1$ .

Step 2: $R_{3} \to R_{3} + 2 R_{1} ⟹$ adding a multiple of a row does not change the determinant. Multiplier is $1$ .

Step 3: $R_{2} \to 2 R_{2} ⟹$ scaling a single row by $2$ multiplies the determinant by $2$ . Multiplier is $2$ .

Step 4: $R_{1} \to R_{1} - R_{3} ⟹$ adding a multiple of a row. Multiplier is $1$ .

Step 5: Calculate the total multiplier:

Total = $(- 1) \times 1 \times 2 \times 1 = - 2$ .

Step 6: Relate to the final determinant:

$det (final) = Total \times det (A)$

$8 = - 2 \times det (A) ⟹ det (A) = - 4$ .

Answer: B

Question 6 · Engineering Mathematics MCQ

Let $A$ be a $3 \times 3$ matrix with $det (A) = 5$ . If matrix $B$ is obtained by adding $3$ times the first row of $A$ to the second row, what is the value of $det (B)$ ?

A.
15
B.
5
C.
45
D.
10

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct formula application question recognisable by the description of a single row operation.

Step 1: Identify the operation. The problem states that $B$ is formed by adding a multiple of one row to another row ( $R_{2} \to R_{2} + 3 R_{1}$ ).

Step 2: Recall the property of determinants under elementary row operations. Adding a scalar multiple of one row to a different row leaves the determinant completely unchanged.

Step 3: Apply the property. Since the determinant is unchanged, $det (B) = det (A) = 5$ .

Answer: B

Question 7 · Engineering Mathematics MCQ

Let $A$ be a $3 \times 3$ matrix with $det (A) = 6$ . Matrix $B$ is obtained by first swapping row $1$ and row $2$ of $A$ , and then multiplying the new row $3$ by $- 2$ . What is the value of $det (B)$ ?

A.
12
B.
-12
C.
-6
D.
6

Correct Answer:

Step-by-Step Solution

Key idea: This is a casework question where multiple row operations are applied sequentially. We track the cumulative multiplier.

Step 1: Initial determinant is $det (A) = 6$ .

Step 2: First operation is swapping row $1$ and row $2$ . A row swap multiplies the determinant by $- 1$ .

New determinant = $6 \times (- 1) = - 6$ .

Step 3: Second operation is multiplying row $3$ by $- 2$ . Scaling a single row by $k$ multiplies the determinant by $k$ .

New determinant = $- 6 \times (- 2) = 12$ .

Answer: A

Question 8 · Engineering Mathematics MCQ

Let $A$ be a $3 \times 3$ matrix with $det (A) = 2$ . What is the maximum possible value of $det (k A)$ if $k$ is an integer such that $- 2 \leq k \leq 2$ ?

A.
16
B.
8
C.
-16
D.
4

Correct Answer:

Step-by-Step Solution

Key idea: This is an observation question about the bounds of determinants under scalar multiplication.

Step 1: Recall the master scaling formula for determinants: $det (k A) = k^{n} det (A)$ , where $n$ is the order of the matrix.

Step 2: Identify the parameters. Here, $n = 3$ (since $A$ is $3 \times 3$ ) and $det (A) = 2$ . So, $det (k A) = k^{3} \times 2 = 2 k^{3}$ .

Step 3: Evaluate the expression for the integer values of $k$ in the range $[- 2, 2]$ :

$k = - 2 ⟹ 2 (- 2)^{3} = 2 (- 8) = - 16$
$k = - 1 ⟹ 2 (- 1)^{3} = - 2$
$k = 0 ⟹ 0$
$k = 1 ⟹ 2 (1)^{3} = 2$
$k = 2 ⟹ 2 (2)^{3} = 2 (8) = 16$

Step 4: The maximum value among these is $16$ .

Answer: A

Question 9 · Engineering Mathematics MCQ

If a matrix $A$ satisfies the polynomial equation $A^{2} = 4 A - 3 I$ , what is the expression for $A^{3}$ in terms of $A$ and $I$ ?

A.
$13 A - 12 I$
B.
$13 A + 12 I$
C.
$16 A - 12 I$
D.
$16 A + 12 I$

Correct Answer:

Step-by-Step Solution

Key idea: Use the given polynomial to reduce the power.

Step 1: Multiply the given equation by $A$ : $A^{3} = A (4 A - 3 I) = 4 A^{2} - 3 A$ .

Step 2: Substitute $A^{2} = 4 A - 3 I$ into the equation: $A^{3} = 4 (4 A - 3 I) - 3 A$ .

Step 3: Simplify: $A^{3} = 16 A - 12 I - 3 A = 13 A - 12 I$ .

Answer: $13 A - 12 I$ .

Question 10 · Engineering Mathematics MCQ

Let $A$ be a $3 \times 3$ matrix with $det (A) = 4$ . Let $B$ , $C$ , and $D$ be matrices obtained from $A$ by the following operations:

$B$ is obtained by swapping two rows of $A$ .
$C$ is obtained by multiplying the first row of $A$ by $- 2$ .
$D$ is obtained by adding twice the second row to the first row of $A$ .

Which of the following correctly ranks the determinants of these matrices in ascending order?

A.
$det (C) < det (B) < det (D)$
B.
$det (B) < det (C) < det (D)$
C.
$det (C) < det (D) < det (B)$
D.
$det (D) < det (B) < det (C)$

Correct Answer:

Step-by-Step Solution

Key idea: This is a language-to-math translation question. We convert each word description into its effect on the determinant.

Step 1: $det (A) = 4$ .

Step 2: $B$ is obtained by swapping two rows. A row swap multiplies the determinant by $- 1$ .

$det (B) = - 1 \times 4 = - 4$ .

Step 3: $C$ is obtained by multiplying a row by $- 2$ . Scaling a row by $k$ multiplies the determinant by $k$ .

$det (C) = - 2 \times 4 = - 8$ .

Step 4: $D$ is obtained by adding a multiple of one row to another. This operation leaves the determinant unchanged.

$det (D) = 4$ .

Step 5: Rank them in ascending order:

$- 8 < - 4 < 4 ⟹ det (C) < det (B) < det (D)$ .

Answer: A

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Matrix Operations, Determinants and Trace Practice Questions 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 (10 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

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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