GATE 2025 EE Q.14: solution and answer

GATE 2025 · EEQ.14 · 1 mark · MCQ

Let $P = 2 - 1 0 100001$ and let $I$ be the identity matrix. Then $P^{2}$ is equal to

(A)
$2 P - I$
(B)
$P$
(C)
$I$
(D)
$P + I$

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Answer and solution

ANSWER

$P^{2} = 2 P - I$ , which is option (A).

Official answer: ASource: Official answer key, GATE 2025 EE (IIT Roorkee)

Characteristic polynomial. $P$ is block-diagonal. Its upper $2 \times 2$ block $[2 - 1 10]$ has trace $2$ and determinant $2 \cdot 0 - (1) (- 1) = 1$ , so that block contributes $λ^{2} - 2 λ + 1 = (λ - 1)^{2}$ . The $1 \times 1$ block $[1]$ contributes $(1 - λ)$ . So $det (P - λ I) = (λ - 1)^{2} (1 - λ) = - (λ - 1)^{3}$ , and the only eigenvalue is $λ = 1$ (repeated three times).
By the Cayley-Hamilton theorem, $(P - I)^{3} = 0$ . That is a cubic, so check whether a lower power already vanishes.
$P - I = 1 - 1 0 1 - 1 0 000$ . Squaring it: the third row and third column are zero, so only the top-left $2 \times 2$ entries need checking. $(1, 1)$ : $1 \cdot 1 + 1 \cdot (- 1) = 0$ . $(1, 2)$ : $1 \cdot 1 + 1 \cdot (- 1) = 0$ . $(2, 1)$ : $(- 1) \cdot 1 + (- 1) (- 1) = 0$ . $(2, 2)$ : $(- 1) \cdot 1 + (- 1) (- 1) = 0$ . So $(P - I)^{2} = 0$ . Since $P - I \neq = 0$ , the minimal polynomial is $(λ - 1)^{2}$ .
Expand, remembering that $P$ and $I$ commute: $(P - I)^{2} = P^{2} - 2 P + I = 0 \Rightarrow P^{2} = 2 P - I$ .
Direct check: $P^{2} = 2 - 1 0 100001 2 - 1 0 100001 = 3 - 2 0 2 - 1 0 001$ and $2 P - I = 4 - 1 - 2 0 2 0 - 1 0 00 2 - 1 = 3 - 2 0 2 - 1 0 001 .$ The two agree.
Rule out the other options: $P^{2} \neq = P$ (the $(1, 1)$ entry of $P^{2}$ is $3$ , not $2$ ); $P^{2} \neq = I$ (its off-diagonal entries $2$ and $- 2$ are nonzero); and $P + I = 3 - 1 0 110002$ has $(2, 2)$ entry $1$ , while $P^{2}$ has $- 1$ there. So the answer is (A).

KEY IDEA

Every eigenvalue of $P$ is $1$ , and $(P - I)^{2} = 0$ , so the minimal polynomial is $(λ - 1)^{2}$ . Expanding $(P - I)^{2} = 0$ gives $P^{2} = 2 P - I$ directly. For a $3 \times 3$ matrix like this one, simply multiplying $P$ by itself and comparing with each option is just as quick, and it is a good check on the theory.

WATCH OUT FOR

Going only as far as Cayley-Hamilton, which gives the cubic $(P - I)^{3} = 0$ , and then trying to force a quadratic answer out of it without first checking that $(P - I)^{2}$ is already zero. The other slip is multiplying carelessly: a single sign error in the $(2, 1)$ or $(2, 2)$ entry of $P^{2}$ (true values $- 2$ and $- 1$ ) makes $P + I$ or $P$ look like a match.

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