GATE 2026 EC Q.11: solution and answer

GATE 2026 · ECQ.11 · 1 mark · MCQ

Consider the differential equation $\dot{w} = A w$ , with $w (t = 0) = [11]$ .

If $w (t) = e^{t} u_{x} + e^{- 2 t} u_{y}$ be the solution to the equation where $u_{x}$ and $u_{y}$ are unit vectors along the positive $x$ and $y$ axes respectively, then which of the following options is the correct matrix representing $A$ ?

(A)
$[10 0 - 2]$
(B)
$[- 1 0 02]$
(C)
$[01 - 2 0]$
(D)
$[0 - 1 20]$

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Answer and solution

ANSWER

The answer is (A), $A = [10 0 - 2]$ . The solution $e^{t} u_{x} + e^{- 2 t} u_{y}$ means $u_{x}$ is an eigenvector with eigenvalue $1$ and $u_{y}$ is an eigenvector with eigenvalue $- 2$ , so $A$ is diagonal with entries $1$ and $- 2$ .

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

Write the solution in components: $w (t) = [e^{t} e^{- 2 t}]$ . At $t = 0$ this is $[11]$ , so it satisfies the initial condition.
Differentiate each component: $\dot{w} (t) = [e^{t} - 2 e^{- 2 t}]$ .
Split the solution into its modes: $w (t) = e^{t} [10] + e^{- 2 t} [01]$ . Substituting into $\dot{w} = A w$ gives $e^{t} A [10] + e^{- 2 t} A [01] = e^{t} [10] - 2 e^{- 2 t} [01]$ for every $t$ .
The functions $e^{t}$ and $e^{- 2 t}$ are linearly independent, so their coefficients must match separately: $A [10] = 1 \cdot [10]$ and $A [01] = - 2 [01]$ . So $u_{x}$ and $u_{y}$ are eigenvectors of $A$ with eigenvalues $1$ and $- 2$ . (A mode $e^{λ t} v$ solves the system only when $A v = λ v$ .)
The columns of $A$ are $A u_{x}$ and $A u_{y}$ , so $A = [10 0 - 2] .$
Check: $[10 0 - 2] [e^{t} e^{- 2 t}] = [e^{t} - 2 e^{- 2 t}] = \dot{w}$ . The other options fail. (B) gives $[- e^{t} 2 e^{- 2 t}]$ , which has the wrong signs. (C) gives $[- 2 e^{- 2 t} e^{t}]$ and (D) gives $[2 e^{- 2 t} - e^{t}]$ , and both swap the two components. Also, (C) and (D) both have trace $0$ and determinant $2$ , so their eigenvalues are $\pm j 2$ . That would give oscillatory solutions, not real exponentials. So the answer is (A).

KEY IDEA

For $\dot{w} = A w$ , each term $e^{λ t} v$ of the solution is a mode, and a mode only works when $A v = λ v$ . Here the modes lie along $u_{x}$ (with $λ = 1$ ) and $u_{y}$ (with $λ = - 2$ ). The eigenvectors are the standard basis vectors, so $A$ is simply $diag (1, - 2)$ : its eigenvalues sit on the diagonal in the same order as the axes.

WATCH OUT FOR

One mistake is picking (B) by reading the exponent $e^{- 2 t}$ as a pole at $+ 2$ , which flips the signs. The eigenvalue is the coefficient of $t$ in the exponent itself, so it is $- 2$ and not $+ 2$ . Another is choosing (C) or (D) because they contain the right numbers $1$ and $- 2$ in off-diagonal places. But those matrices mix $x$ and $y$ , and their eigenvalues $\pm j 2$ would give oscillation, not $e^{t}$ and $e^{- 2 t}$ .

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GATE 2026 EC Q.11: solution and answer

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