GATE 2026 EE Q.56: solution and answer

GATE 2026 · EEQ.56 · 2 marks · MSQ

Consider the system of linear equations: $A x = b$ , where $A$ is an $n \times n$ matrix, and $x$ and $b$ are $n$ -dimensional column vectors.

Suppose this system of equations has a unique solution. Which of the following statements is/are correct?

(A)
$A^{- 1}$ exists
(B)
The system of equations $A^{m} x = b$ also has a unique solution for $m = 1, 2, 3, \dots$
(C)
$rank (A) = rank (A^{m})$ , for $m = 1, 2, 3, \dots$
(D)
$rank (A) < rank ([A ∣ b])$ , where $[A ∣ b]$ denotes the augmented matrix.

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Answer and solution

ANSWER

The correct options are (A), (B) and (C). A unique solution means $A$ is non-singular, and so is every power of it, since $det (A^{m}) = (det A)^{m} \neq = 0$ . Option (D) is the condition for having no solution.

Official answer: A, B, CSource: Official answer key, GATE 2026 EE (IIT Guwahati)

Rouché–Capelli theorem. A system $A x = b$ with $n$ unknowns has a unique solution exactly when $rank (A) = rank ([A ∣ b]) = n$ . Here $A$ is $n \times n$ , so it has full rank $n$ , which means $det A \neq = 0$ .
(A) is true. A square matrix with $det A \neq = 0$ is non-singular, so $A^{- 1}$ exists. The solution is $x = A^{- 1} b$ .
(B) is true. The determinant is multiplicative, so $det (A^{m}) = (det A)^{m} \neq = 0$ . That makes $A^{m}$ invertible too, and $A^{m} x = b$ has exactly one solution, $x = (A^{- 1})^{m} b$ , for every $m = 1, 2, 3, \dots$ Uniqueness comes from $A^{m}$ being invertible, so it holds whatever $b$ is.
(C) is true. $A^{m}$ is an invertible $n \times n$ matrix, so it has full rank. Therefore $rank (A^{m}) = n = rank (A)$ for every $m$ .
(D) is false. $rank (A) < rank ([A ∣ b])$ is the condition for an inconsistent system, meaning one with no solution. This system has a solution, so $rank (A) = rank ([A ∣ b]) = n$ .
The correct options are (A), (B) and (C).

KEY IDEA

For a square system, a unique solution exists exactly when $A$ is non-singular: $rank (A) = rank ([A ∣ b]) = n$ , which is the same as $det A \neq = 0$ . Every power of $A$ is then non-singular too, because $det (A^{m}) = (det A)^{m} \neq = 0$ . So $A^{m}$ keeps full rank $n$ , and $A^{m} x = b$ also has exactly one solution.

WATCH OUT FOR

Picking (D) because "rank" and "augmented matrix" sound like the uniqueness test. $rank (A) < rank ([A ∣ b])$ actually means the system is inconsistent, with no solution at all. A second trap is rejecting (C) on the idea that powers of a matrix can lose rank. That only happens when $A$ is singular (a nilpotent matrix, for example), and a unique solution rules that out.

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GATE 2026 EE Q.56: solution and answer

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