Rank, Invertibility and Linear Systems Previous Year Questions (PYQs) for GATE DA: 3+ Solved Questions with Step-by-Step Solutions

Solve 3+ Rank, Invertibility and Linear Systems previous year questions for GATE DA with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Rank, Invertibility and Linear Systems

Chapter Roadmap

Rank, Invertibility and Linear Systems

Rank, Nullity and Matrix Polynomials
Solution spaces, Rank-Nullity Theorem, and matrix powers.

Invertibility and Rank-One Updates
Outer products

x x^{T}

, invertibility shifts, and quick eigenvalue tricks.

Goal: Move from calculating determinants to understanding the structure of linear transformations.

Rank, Nullity and the Solution Space

For any $m \times n$ matrix $A$ , the fundamental accounting rule of linear algebra is:

Rank (A) + Nullity (A) = n

Rank: Dimension of the column space (independent columns).

Nullity: Dimension of the null space (free variables in

A x = 0

$n$ : Total number of columns (variables).

Exam Insight: If

Rank (A) < n

, the system

A x = 0

has non-trivial (infinite) solutions.

Rank, Invertibility and Linear Systems: Solved Questions with Step-by-Step Explanations (3 Problems)

Question 1 · Linear Algebra 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 2 · Linear Algebra MSQ

Let $A = I_{n} + x x^{T}$ , where $I_{n}$ is the $n \times n$ identity matrix and $x \in R^{n}$ , $x^{T} x = 1$ . Which of the following options is/are correct?

A.
Rank of $A$ is $n$
B.
$A$ is invertible
C.
$0$ is an eigenvalue of $A$
D.
$A^{- 1}$ has a negative eigenvalue

Correct Answer:

["A","B"]

Step-by-Step Solution

Key idea: The eigenvalues of a rank-one update $I + x x^{T}$ can be found directly from the eigenvalues of the rank-one matrix $x x^{T}$ .

Step 1: Analyze the rank-one matrix $M = x x^{T}$ . For any vector $v$ orthogonal to $x$ (i.e., $x^{T} v = 0$ ), we have $M v = x (x^{T} v) = x (0) = 0$ . Since the orthogonal complement of $x$ in $R^{n}$ has dimension $n - 1$ , $λ = 0$ is an eigenvalue of $M$ with multiplicity at least $n - 1$ .

Step 2: For the vector $x$ itself, $M x = (x x^{T}) x = x (x^{T} x)$ . We are given $x^{T} x = 1$ , so $M x = 1 \cdot x$ . Thus, $λ = 1$ is an eigenvalue of $M$ with multiplicity 1.

Step 3: The eigenvalues of $A = I_{n} + M$ are simply the eigenvalues of $M$ shifted by 1 (since $I_{n} v = 1 \cdot v$ for all $v$ ).

Step 4: Therefore, the eigenvalues of $A$ are $1 + 0 = 1$ (with multiplicity $n - 1$ ) and $1 + 1 = 2$ (with multiplicity 1).

Step 5: Evaluate the options:

Since all eigenvalues of $A$ are non-zero (they are 1 and 2), $A$ is invertible. (Option B is correct)
An invertible $n \times n$ matrix has full rank, so the rank of $A$ is $n$ . (Option A is correct)
$0$ is not an eigenvalue of $A$ . (Option C is incorrect)
The eigenvalues of $A^{- 1}$ are the reciprocals of the eigenvalues of $A$ , which are $1/1 = 1$ and $1/2$ . Neither is negative. (Option D is incorrect)

Answer: Rank of $A$ is $n$ , and $A$ is invertible.

Question 3 · Linear Algebra MCQ

The sum of the elements in each row of $A \in R^{n \times n}$ is $1$ . If $B = A^{3} - 2 A^{2} + A$ , which one of the following statements is correct (for $x \in R^{n}$ )?

A.
The equation $B x = 0$ has no solution
B.
The equation $B x = 0$ has exactly two solutions
C.
The equation $B x = 0$ has infinitely many solutions
D.
The equation $B x = 0$ has a unique solution

Correct Answer:

Step-by-Step Solution

Key idea: The row sum condition directly gives an eigenvalue and its eigenvector, which can be evaluated in the matrix polynomial.

Step 1: The statement "sum of the elements in each row of $A$ is $1$ " means that if we multiply $A$ by the all-ones column vector $1 = [1, 1, \dots, 1]^{T}$ , the result is $1$ . Thus, $A 1 = 1 \cdot 1$ .

Step 2: This implies $λ = 1$ is an eigenvalue of $A$ , and $1$ is a corresponding non-zero eigenvector.

Step 3: We are given $B = A^{3} - 2 A^{2} + A$ . We can factor this polynomial as $B = A (A^{2} - 2 A + I) = A (A - I)^{2}$ .

Step 4: Evaluate $B 1$ . Since $(A - I) 1 = A 1 - 1 = 1 - 1 = 0$ , we have $(A - I)^{2} 1 = (A - I) 0 = 0$ .

Step 5: Therefore, $B 1 = A (A - I)^{2} 1 = A 0 = 0$ .

Step 6: Since $1$ is a non-zero vector and $B 1 = 0$ , the homogeneous system $B x = 0$ has a non-trivial solution. Any homogeneous system with a non-trivial solution has infinitely many solutions.

Answer: The equation $B x = 0$ has infinitely many solutions.

Rank, Invertibility and Linear Systems Previous Year Questions (PYQs) for GATE DA: 3+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Rank, Invertibility and Linear Systems

Chapter Roadmap

Rank, Nullity and the Solution Space

Rank, Nullity and the Solution Space

Rank, Invertibility and Linear Systems: Solved Questions with Step-by-Step Explanations (3 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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