GATE 2025 EC Q.11: solution and answer

GATE 2025 · ECQ.11 · 1 mark · MCQ

Consider the matrix $A$ below:

A = 20003600 47 α 0 58 β γ

For which of the following combinations of $α, β$ , and $γ$ , is the rank of $A$ at least three?

(i) $α = 0$ and $β = γ \neq = 0$ .

(ii) $α = β = γ = 0$ .

(iii) $β = γ = 0$ and $α \neq = 0$ .

(iv) $α = β = γ \neq = 0$ .

(A)
Only (i), (iii), and (iv)
(B)
Only (iv)
(C)
Only (ii)
(D)
Only (i) and (iii)

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Answer and solution

ANSWER

The rank of $A$ is at least three for combinations (i), (iii) and (iv): option (A), "Only (i), (iii), and (iv)".

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

Rows 1 and 2 are always independent, because the leading $2 \times 2$ minor is $2036 = 12 \neq = 0$ . So $ρ (A) \geq 2$ for every choice, and $ρ (A) \geq 3$ exactly when some $3 \times 3$ minor is non-zero.
$A$ is upper triangular, so $det A = 2 \cdot 6 \cdot α \cdot γ = 12 α γ$ .
(i) $α = 0, β = γ \neq = 0$ : rows 3 and 4 are both $[000 β]$ . The minor from rows 1, 2, 3 and columns 1, 2, 4 is $200360 58 β = 12 β \neq = 0$ , so $ρ (A) = 3$ . It is not 4, because $det A = 12 α γ = 0$ . The rank is at least 3.
(ii) $α = β = γ = 0$ : rows 3 and 4 are zero rows, so only the two independent rows 1 and 2 are left and $ρ (A) = 2$ . This case fails.
(iii) $β = γ = 0, α \neq = 0$ : row 3 is $[00 α 0]$ and row 4 is zero. The leading $3 \times 3$ minor is $2 \cdot 6 \cdot α = 12 α \neq = 0$ , so $ρ (A) = 3$ . The rank is at least 3.
(iv) $α = β = γ \neq = 0$ : $det A = 12 α^{2} \neq = 0$ , so $ρ (A) = 4$ . The rank is at least 3.
So (i), (iii) and (iv) give rank at least three, and only (ii) fails. That is option (A).

KEY IDEA

$ρ (A) \geq r$ exactly when some $r \times r$ minor is non-zero. Here the first two rows always give rank at least 2, so you only need to find one non-zero $3 \times 3$ minor for each case. For an upper triangular matrix, $det A$ is the product of the diagonal entries, $12 α γ$ . That settles rank 4 straight away, and a zero diagonal entry does not by itself push the rank below 3.

WATCH OUT FOR

Counting non-zero diagonal entries and taking that as the rank. In case (i) the diagonal is $2, 6, 0, γ$ , which looks like rank 3 for the wrong reason, and a student can easily talk themselves into " $α = 0$ breaks rank 3" and reject (i), which leads to option (D). Rank comes from independent rows or non-zero minors, not from the diagonal. In (i) the minor on columns 1, 2, 4 is $12 β \neq = 0$ , so the rank is 3. Rows 3 and 4 are identical, so it cannot be 4.

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GATE 2025 EC Q.11: solution and answer

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