GATE 2026 IN Q.44: solution and answer

GATE 2026 · INQ.44 · 2 marks · MCQ

A linear time invariant (LTI) system is described by:

[\overset{x}{˙}_{1} \overset{x}{˙}_{2}] = [0 - 10 1 - 7] [x_{1} x_{2}] + [11] u

y = [11] [x_{1} x_{2}]

The poles of the system are ____.

(A)
$- 2, - 5$
(B)
$2, 5$
(C)
$- 2, 5$
(D)
$2, - 5$

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Answer and solution

ANSWER

The poles of the system are $- 2$ and $- 5$ , which is option (A).

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

The poles are the roots of the characteristic equation $∣ s I - A ∣ = 0$ , which are the eigenvalues of $A$ . This holds as long as no pole cancels against a zero in $G (s) = C (s I - A)^{- 1} B$ , and step 5 checks that.
$s I - A = [s 10 - 1 s + 7]$
$∣ s I - A ∣ = s (s + 7) + 10 = s^{2} + 7 s + 10 = (s + 2) (s + 5)$
The eigenvalues of $A$ are $s = - 2$ and $s = - 5$ . Check: trace $A = - 7 = (- 2) + (- 5)$ and $det A = 10 = (- 2) (- 5)$ .
Check for cancellation using $G (s) = C (s I - A)^{- 1} B$ : $(s I - A)^{- 1} B = \frac{1}{( s + 2 ) ( s + 5 )} [s + 7 - 10 1 s] [11] = \frac{1}{( s + 2 ) ( s + 5 )} [s + 8 s - 10]$
$G (s) = [11] \frac{1}{( s + 2 ) ( s + 5 )} [s + 8 s - 10] = \frac{2 ( s - 1 )}{( s + 2 ) ( s + 5 )}$ The only zero is at $s = 1$ , so it cancels neither pole.
So the poles are $- 2$ and $- 5$ : option (A).

KEY IDEA

For a state-space model the poles are the roots of $∣ s I - A ∣ = 0$ , the eigenvalues of $A$ . Check them quickly with trace $A$ = sum of the eigenvalues and $det A$ = their product. A pole drops out of the transfer function only if a zero of $C adj (s I - A) B$ cancels it. Here the only zero is at $s = 1$ , so both eigenvalues are poles.

WATCH OUT FOR

Getting the sign of the $- 10$ entry wrong when forming $s I - A$ . Writing $s (s + 7) - 10$ gives $s^{2} + 7 s - 10$ , whose roots are not integers and match no option. Another mistake is to take the roots of $s^{2} - 7 s + 10$ (that is, $det (A - s I)$ with the signs slipped) and pick $2, 5$ , option (B). Two eigenvalues whose sum is the negative trace $- 7$ and whose product is $+ 10$ must both be negative.

From the GATE 2026 IN paper. Every question of this paper · Download the full paper · All 16 State space questions, with notes

GATE 2026 IN Q.44: solution and answer

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