GATE 2024 IN Q.19: solution and answer

GATE 2024 · INQ.19 · 1 mark · MCQ

A matrix $M$ is constructed by stacking three column vectors $v_{1}, v_{2}, v_{3}$ as
$M = [v_{1} v_{2} v_{3}] .$
Choose the set of vectors from the following options such that $r ank (M) = 3$ .

(A)
$v_{1} = 101, v_{2} = 0 - 1 0, v_{3} = 1 - 1 1$
(B)
$v_{1} = 111, v_{2} = - 1 01, v_{3} = 000$
(C)
$v_{1} = 101, v_{2} = - 1 01, v_{3} = 1 - 1 1$
(D)
$v_{1} = 111, v_{2} = - 1 1 - 1, v_{3} = 0 - 1 0$

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Answer and solution

ANSWER

(C). Its columns give $det M = 2 \neq = 0$ , so $r ank (M) = 3$ . In (A) and (D) one column is a combination of the other two ( $v_{3} = v_{1} + v_{2}$ in (A), $v_{1} + v_{2} = - 2 v_{3}$ in (D)), and (B) contains the zero vector, so each of those has rank 2.

Official answer: CSource: Official answer key, GATE 2024 IN (IISc Bengaluru)

A $3 \times 3$ matrix has $r ank (M) = 3$ exactly when its three columns are linearly independent, which is the same as $det M \neq = 0$ . So we look for the option in which no column is a combination of the others.
Option (A): $v_{1} + v_{2} = 1 - 1 1 = v_{3}$ . The columns are dependent. Since $v_{1}$ and $v_{2}$ are not multiples of each other, the rank is $2$ .
Option (B): $v_{3}$ is the zero vector, and any set containing the zero vector is linearly dependent. The rank is therefore at most $2$ (it is exactly $2$ , because $v_{1}$ and $v_{2}$ are not parallel).
Option (D): $v_{1} + v_{2} = 020 = - 2 v_{3}$ . The columns are dependent. Since $v_{1}$ and $v_{2}$ are not parallel, the rank is $2$ .
Option (C): $M = 101 - 1 01 1 - 1 1$ . Expand the determinant along row 2. Its only nonzero entry is $m_{23} = - 1$ , with cofactor sign $(- 1)^{2 + 3} = - 1$ , and deleting row 2 and column 3 leaves the minor $[11 - 1 1]$ : $det M = (- 1) \cdot (- 1)^{5} \cdot det [11 - 1 1] = (1) (1 \cdot 1 - (- 1) \cdot 1) = 2 \neq = 0.$
Because $det M = 2 \neq = 0$ , the three columns in (C) are linearly independent and $r ank (M) = 3$ . The answer is (C).

KEY IDEA

A square $n \times n$ matrix has full rank $n$ exactly when its columns are linearly independent, i.e. when its determinant is nonzero. To rule out an option quickly, look for an obvious dependence: a zero column, or one column equal to a sum or multiple of the others. Then confirm the remaining option with a determinant, expanding along the row or column with the most zeros.

WATCH OUT FOR

Seeing that no two columns are parallel and concluding the rank is 3. Pairwise non-parallel columns are not enough: in (A) and (D) no column is a multiple of another, yet $v_{3}$ is a combination of $v_{1}$ and $v_{2}$ , so the rank is only 2. A second slip is dropping the cofactor sign $(- 1)^{i + j}$ in a row expansion. Here it does not change whether $det M$ is zero, but it does give the wrong determinant.

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GATE 2024 IN Q.19: solution and answer

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