GATE 2023 IN Q.11: solution and answer

GATE 2023 · INQ.11 · 1 mark · MCQ

Choose solution set $S$ corresponding to the systems of two equations

x - 2 y + z x - z = 0 = 0

Note: $ℜ$ denotes the set of real numbers

(A)
$S = ⎩ ⎨ ⎧ α 111 α \in ℜ ⎭ ⎬ ⎫$
(B)
$S = ⎩ ⎨ ⎧ α 111 + β 101 α, β \in ℜ ⎭ ⎬ ⎫$
(C)
$S = ⎩ ⎨ ⎧ α 111 + β 212 α, β \in ℜ ⎭ ⎬ ⎫$
(D)
$S = ⎩ ⎨ ⎧ α 101 α \in ℜ ⎭ ⎬ ⎫$

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Answer and solution

ANSWER

The solution set is all scalar multiples of $(1, 1, 1)$ , so the answer is (A): $S = ⎩ ⎨ ⎧ α 111 : α \in ℜ ⎭ ⎬ ⎫$ .

Official answer: ASource: Official answer key, GATE 2023 IN (IIT Kanpur)

Write the system as $A x = 0$ with $A = [11 - 2 0 1 - 1]$ and $x = x y z$ . The two rows are not multiples of each other, so $rank (A) = 2$ .
By the rank–nullity theorem, the solution space has dimension $n - rank (A) = 3 - 2 = 1$ . So $S$ is a line through the origin, spanned by one vector. Options (B) and (C) each use two linearly independent vectors, so each describes a plane, which is 2-dimensional. Both are ruled out.
Now solve directly. The second equation gives $x = z$ .
Put $z = x$ into the first equation: $x - 2 y + x = 0 \Rightarrow 2 x = 2 y \Rightarrow y = x$ .
Let $x = α$ . Then $(x, y, z) = (α, α, α) = α 111$ for any $α \in ℜ$ . Check: $1 - 2 (1) + 1 = 0$ and $1 - 1 = 0$ .
Check (D): $(1, 0, 1)$ gives $1 - 0 + 1 = 2 \neq = 0$ in the first equation, so it is not a solution. The same vector lies in the set in (B) (take $α = 0, β = 1$ ). In (C), $(2, 1, 2)$ gives $2 - 2 + 2 = 2 \neq = 0$ . This confirms that (B), (C) and (D) are all wrong.
So $S = ⎩ ⎨ ⎧ α 111 : α \in ℜ ⎭ ⎬ ⎫$ , which is option (A).

KEY IDEA

For a homogeneous system $A x = 0$ in $n$ unknowns, the solution set is a subspace of dimension $n - rank (A)$ . Here that is $3 - 2 = 1$ , so $S$ is a line spanned by one vector. Two independent equations in three unknowns leave one free parameter. Any proposed spanning vector must satisfy every equation.

WATCH OUT FOR

Choosing (D) because $(1, 0, 1)$ satisfies the second equation $x - z = 0$ , without checking the first equation, where it gives $2 \neq = 0$ . A second mistake is choosing (B) or (C) because they contain the correct vector $(1, 1, 1)$ . Adding a second independent vector turns the set into a plane, and that plane includes non-solutions such as $(1, 0, 1)$ or $(2, 1, 2)$ .

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GATE 2023 IN Q.11: solution and answer

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