Official PYQ · 2017Set 1 · Q3Linear Algebra: Matrices, Determinants, System of Linear EquationsMCQ1 mark

GATE CS 2017 Set 1 Q3 — Linear Algebra: Matrices, Determinants, System of Linear Equations

Let $c_{1}.....c_{n}$ be scalars, not all zero, such that $\sum_{i=1}^{n}c_{i}a_{i}$ = 0 where $a_{i}$ are column vectors in $R^{n}$ .

Consider the set of linear equations

$Ax = b$

where $A=\left [ a_{1}.....a_{n} \right ]$ and $b=\sum_{i=1}^{n}a_{i}$ . The set of equations has

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Answer: C

We are given that $c_1, \dots, c_n$ are scalars, not all zero, such that:

\sum_{i=1}^{n} c_i a_i = 0

This implies that the column vectors $a_1, \dots, a_n$ of the matrix $A = [a_1 \dots a_n]$ are linearly dependent. Consequently, the null space of $A$ contains a non-zero vector $c = [c_1, c_2, \dots, c_n]^T \neq \mathbf{0}$ such that:

Ac = \mathbf{0}

We are also given the system of linear equations $Ax = b$ , where:

b = \sum_{i=1}^{n} a_i

This can be written in matrix form as:

A \begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{bmatrix} = b \implies A J_n = b

where $J_n = [1, 1, \dots, 1]^T$ is the $n$ -dimensional vector of all $1$ s. Thus, $x = J_n$ is a particular solution to the system.

Since the system is consistent (having at least one solution $J_n$ ) and the columns of $A$ are linearly dependent ( $Ac = \mathbf{0}$ for $c \neq \mathbf{0}$ ), we can construct infinitely many solutions of the form:

x = J_n + k c \quad \text{for any scalar } k \in \mathbb{R}

Checking this:

A(J_n + k c) = A J_n + k(Ac) = b + k(\mathbf{0}) = b

Since $c \neq \mathbf{0}$ , each distinct value of $k$ yields a unique solution. Therefore, the system has infinitely many solutions.

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Published 10 Oct 2026

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