GATE 2026 IN Q.14: solution and answer

GATE 2026 · INQ.14 · 1 mark · MCQ

Consider a matrix

A = 1 - 1 0 011202

Let $b = 130$ and $x = x_{1} x_{2} x_{3}$

The number of solutions to the linear system of equations $A x = b$ is ______.

(A)
0
(B)
1
(C)
6
(D)
infinitely many

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Answer and solution

ANSWER

The system is inconsistent: $ρ (A) = 2$ but $ρ ([A ∣ b]) = 3$ , so $A x = b$ has no solution. Answer: (A) 0.

Official answer: ASource: Official answer key, GATE 2026 IN (IIT Guwahati)

Check whether $A$ is singular by expanding $det A$ along the first row: $det A = 1 (1 \cdot 2 - 0 \cdot 1) - 0 + 2 ((- 1) \cdot 1 - 1 \cdot 0) = 2 - 2 = 0$ So $A$ is singular and a unique solution is ruled out. The system has either no solution or infinitely many, and the rank test decides which.
Find the dependence among the rows: $R_{1} + R_{2} = [1, 0, 2] + [- 1, 1, 0] = [0, 1, 2] = R_{3}$ . Rows 1 and 2 are not multiples of each other, so $ρ (A) = 2$ .
Form the augmented matrix and apply the same row operation, $R_{3} \to R_{3} - R_{1} - R_{2}$ : $[A ∣ b] = 1 - 1 0 011202130 \to 1 - 1 0 010200 13 - 4$
The last row reads $0 = - 4$ , which is impossible. Put another way, the system is consistent only if $b$ obeys the same dependence as the rows, $b_{3} = b_{1} + b_{2} = 1 + 3 = 4$ , but here $b_{3} = 0$ . The three rows of the reduced augmented matrix are independent, so $ρ ([A ∣ b]) = 3$ . (Direct check: the first equation gives $x_{1} = 1 - 2 x_{3}$ , and the second then gives $x_{2} = 3 + x_{1} = 4 - 2 x_{3}$ . So $x_{2} + 2 x_{3} = 4$ for every candidate, while the third equation demands $x_{2} + 2 x_{3} = 0$ .)
Rank (Rouché–Capelli) test: $ρ (A) = 2 \neq = ρ ([A ∣ b]) = 3$ , so the system is inconsistent. The number of solutions is $0$ , which is option (A).

KEY IDEA

When a square $A$ is singular ( $det A = 0$ ), $A x = b$ has either no solution or infinitely many. It has infinitely many exactly when $b$ satisfies the same linear dependence as the rows of $A$ , i.e. when $ρ (A) = ρ ([A ∣ b])$ . Here $R_{3} = R_{1} + R_{2}$ but $b_{3} \neq = b_{1} + b_{2}$ , so there are no solutions.

WATCH OUT FOR

Seeing $det A = 0$ and choosing "infinitely many" (D) straight away. A zero determinant only rules out a unique solution. You still have to check $b$ against the row dependence, and here $b_{3} = 0 \neq = b_{1} + b_{2} = 4$ . A sign slip in the cofactor $(- 1 \cdot 1 - 1 \cdot 0) = - 1$ can also make $det A$ look nonzero, which leads to the wrong answer (B).

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GATE 2026 IN Q.14: solution and answer

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