GATE DA Linear Algebra Previous Year Questions

GATE DA 2024 | Question 47

MSQ · 2 marks

Select all choices that are subspaces of \(\mathbb{R}^{3}\).

Note: \(\mathbb{R}\) denotes the set of real numbers.

\(\left\{\mathbf{x} = \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} \in \mathbb{R}^{3} : \mathbf{x} = \alpha \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + \beta \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \alpha, \beta \in \mathbb{R}\right\}\) \(\left\{\mathbf{x} = \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} \in \mathbb{R}^{3} : \mathbf{x} = \alpha^{2} \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} + \beta^{2} \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, \alpha, \beta \in \mathbb{R}\right\}\) \(\left\{\mathbf{x} = \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} \in \mathbb{R}^{3} : 5x_{1} + 2x_{3} = 0, 4x_{1} - 2x_{2} + 3x_{3} = 0\right\}\) \(\left\{\mathbf{x} = \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} \in \mathbb{R}^{3} : 5x_{1} + 2x_{3} + 4 = 0\right\}\)

Explanation

Answer: (A), (C)

A subset of \(\mathbb{R}^{3}\) is a subspace when it contains \(\mathbf{0}\) and is closed under addition and scalar multiplication.

(A) is the span of \((1, 1, 0)\) and \((1, 0, 0)\), the plane \(x_{3} = 0\): a subspace.
(B) The coefficients \(\alpha^{2}\) and \(\beta^{2}\) are never negative, so the set holds only the non-negative combinations of the two vectors (a cone, not a plane). It contains \((1, 2, 0)\) but not \((-1, -2, 0)\), which would need \(\alpha^{2} = -1\).
(C) is the solution set of a homogeneous linear system, the null space of a \(2 \times 3\) matrix: a subspace.
(D) The zero vector gives \(5(0) + 2(0) + 4 = 4 \ne 0\), so \(\mathbf{0}\) is missing; this is a shifted plane.

Answer: A, C

GATE DA 2024 | Question 48

MSQ · 2 marks

Which of the following statements is/are TRUE?

Note: \(\mathbb{R}\) denotes the set of real numbers.

There exist \(\boldsymbol{M} \in \mathbb{R}^{3 \times 3}\), \(\boldsymbol{p} \in \mathbb{R}^{3}\), and \(\boldsymbol{q} \in \mathbb{R}^{3}\) such that \(\boldsymbol{M}\mathbf{x} = \boldsymbol{p}\) has a unique solution and \(\boldsymbol{M}\mathbf{x} = \boldsymbol{q}\) has infinite solutions. There exist \(\boldsymbol{M} \in \mathbb{R}^{3 \times 3}\), \(\boldsymbol{p} \in \mathbb{R}^{3}\), and \(\boldsymbol{q} \in \mathbb{R}^{3}\) such that \(\boldsymbol{M}\mathbf{x} = \boldsymbol{p}\) has no solutions and \(\boldsymbol{M}\mathbf{x} = \boldsymbol{q}\) has infinite solutions. There exist \(\boldsymbol{M} \in \mathbb{R}^{2 \times 3}\), \(\boldsymbol{p} \in \mathbb{R}^{2}\), and \(\boldsymbol{q} \in \mathbb{R}^{2}\) such that \(\boldsymbol{M}\mathbf{x} = \boldsymbol{p}\) has a unique solution and \(\boldsymbol{M}\mathbf{x} = \boldsymbol{q}\) has infinite solutions. There exist \(\boldsymbol{M} \in \mathbb{R}^{3 \times 2}\), \(\boldsymbol{p} \in \mathbb{R}^{3}\), and \(\boldsymbol{q} \in \mathbb{R}^{3}\) such that \(\boldsymbol{M}\mathbf{x} = \boldsymbol{p}\) has a unique solution and \(\boldsymbol{M}\mathbf{x} = \boldsymbol{q}\) has no solutions.

Explanation

Answer: (B), (D)

A consistent system \(\boldsymbol{M}\mathbf{x} = \boldsymbol{b}\) has a unique solution exactly when the null space of \(\boldsymbol{M}\) is \(\{\mathbf{0}\}\); otherwise every consistent right-hand side has infinitely many solutions.

(A) False. A unique solution forces a trivial null space, so the square \(\boldsymbol{M}\) is invertible and \(\boldsymbol{M}\mathbf{x} = \boldsymbol{q}\) also has exactly one solution.
(B) True. Take \(\boldsymbol{M} = \operatorname{diag}(1, 1, 0)\), \(\boldsymbol{p} = (0, 0, 1)^{\top}\) (no solution) and \(\boldsymbol{q} = \mathbf{0}\) (infinitely many).
(C) False. With 3 unknowns and rank at most 2, the null space is never trivial, so no solution is unique.
(D) True. Let \(\boldsymbol{M}\) have columns \((1, 0, 0)^{\top}\) and \((0, 1, 0)^{\top}\). Then \(\boldsymbol{p} = (1, 1, 0)^{\top}\) has the single solution \((1, 1)\), while \(\boldsymbol{q} = (0, 0, 1)^{\top}\) lies outside the column space.

Answer: B, D

GATE DA 2024 | Question 49

MSQ · 2 marks

Let \(\mathbb{R}\) be the set of real numbers, \(U\) be a subspace of \(\mathbb{R}^{3}\) and \(\boldsymbol{M} \in \mathbb{R}^{3 \times 3}\) be the matrix corresponding to the projection on to the subspace \(U\).

Which of the following statements is/are TRUE?

If \(U\) is a 1-dimensional subspace of \(\mathbb{R}^{3}\), then the null space of \(\boldsymbol{M}\) is a 1-dimensional subspace. If \(U\) is a 2-dimensional subspace of \(\mathbb{R}^{3}\), then the null space of \(\boldsymbol{M}\) is a 1-dimensional subspace. \(\boldsymbol{M}^{2} = \boldsymbol{M}\) \(\boldsymbol{M}^{3} = \boldsymbol{M}\)

Explanation

Answer: (B), (C), (D)

A projection onto \(U\) leaves every vector of \(U\) unchanged and sends the complementary directions to \(\boldsymbol{0}\). So its column space is \(U\), and by rank–nullity its null space has dimension \(3 - \dim U\).

A fails: a line \(U\) leaves a null space of dimension \(3 - 1 = 2\), a plane.
B holds: a plane \(U\) leaves a null space of dimension \(3 - 2 = 1\), a line.
C holds: \(\boldsymbol{M}\boldsymbol{x}\) already lies in \(U\), so projecting it again changes nothing, which gives \(\boldsymbol{M}^{2} = \boldsymbol{M}\).
D holds: \(\boldsymbol{M}^{3} = \boldsymbol{M}^{2}\boldsymbol{M} = \boldsymbol{M}\boldsymbol{M} = \boldsymbol{M}\).

Answer: B, C, D

GATE DA 2024 | Question 61

NAT · 2 marks

Let \(\boldsymbol{u} = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \\ 5 \end{bmatrix}\), and let \(\sigma_{1}, \sigma_{2}, \sigma_{3}, \sigma_{4}, \sigma_{5}\) be the singular values of the matrix \(\boldsymbol{M} = \boldsymbol{u}\boldsymbol{u}^{\boldsymbol{T}}\) (where \(\boldsymbol{u}^{\boldsymbol{T}}\) is the transpose of \(\boldsymbol{u}\)). The value of \(\sum_{i=1}^{5} \sigma_{i}\) is ______.

Your answer

Explanation

Answer: 55

\(\boldsymbol{M} = \boldsymbol{u}\boldsymbol{u}^{T}\) is a symmetric \(5 \times 5\) matrix of rank 1, since every column is a multiple of \(\boldsymbol{u}\). Moreover

\[ \boldsymbol{M}\boldsymbol{u} = \boldsymbol{u}(\boldsymbol{u}^{T}\boldsymbol{u}) = \|\boldsymbol{u}\|^{2}\,\boldsymbol{u}, \qquad \|\boldsymbol{u}\|^{2} = 1 + 4 + 9 + 16 + 25 = 55 \]

while every vector orthogonal to \(\boldsymbol{u}\) is sent to \(\boldsymbol{0}\). So the eigenvalues are 55 and 0 (four times). For a symmetric matrix the singular values are the absolute values of the eigenvalues, here 55, 0, 0, 0, 0.

Hence \(\sum_{i=1}^{5} \sigma_{i} = 55\), which is also the trace of \(\boldsymbol{M}\).

Answer: 55