GATE CSE 2026 | Set 1 | Question: 10

Question

Let $n>1$. Consider an $n \times n$ matrix $M$ with its elements from $\mathbb{R}$. Let the vector $(0,1,0,0, \ldots, 0) \in \mathbb{R}^{n}$ be in the null space of $M$.

Which of the following options is/are always correct?

• A. Determinant of $M$ is $1$
• B. Determinant of $M$ is $0$
• C. Rank of $M$ is $1$
• D. There are at least two non-zero vectors in the null space of $M$

Comments

https://youtu.be/IUAIytbb2mw?t=15183

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I was confused in option D during examination as too many MSQ was there....and I maked not correct(I think, I was thinking about Linearly Independent vectors)...🙂

Answer 1

$$
\begin{array}{c}
\text{Given that vector } (0,1,0,0,.....0) \in \mathbb{R}^n  \text{ is in null space of M} \\
\downarrow \\
\text{there is one non-trivial solution to Ax=0}  \\
\downarrow \\
\text{Columns of M are linearly dependent} \\ \downarrow \\ \text{determinant of M is 0} \\ \downarrow \\ \text{A is incorrect, B is correct}
\end{array}
$$

---

$$
\begin{array}{c}
\text{Given that vector } (0,1,0,0,.....0) \in \mathbb{R}^n \text{ is in null space of M} \\
\downarrow \\
\text{Rank(M) < n, but we cannot tell exact rank of matrix in this case}  \\
\downarrow \\
\text{C is incorrect}
\end{array}
$$

---

$$
\begin{array}{c}
\text{Given that vector } (0,1,0,0,.....0) \in \mathbb{R}^n \text{ is in null space of M} \\
\downarrow \\
\text{For any value of k} \ne 0, \text{the vector } (0,k,0,0,.....0) \in \mathbb{R}^n \text{ is also in null space of M}  \\
\downarrow \\
\text{There are infinitely many non-zero vectors in the null space of M, but} \\ \text{only one linearly independent vector in the null space of M } \\

\downarrow \\

\text{D is correct}
\end{array}
$$

 

Comments

For option D

\begin{array}{c}
\text{Given that the vector } (0, 1, 0, 0, \dots, 0) \in \mathbb{R}^n \text{ is in the null space of } M \\
\downarrow \\
\text{For any value of } k \neq 0, \text{ the vector } (0, k, 0, 0, \dots, 0) \text{ is also in the null space of } M \\
\downarrow \\
\text{There are infinitely many non-zero vectors in the null space of } M, \text{ but} \\
\text{only one linearly independent vector in the null space of } M \\
\downarrow \\
\text{D is correct}
\end{array}

Rendered Output:

$$\begin{array}{c} \text{Given that the vector } (0, 1, 0, 0, \dots, 0) \in \mathbb{R}^n \text{ is in the null space of } M \\ \downarrow \\ \text{For any value of } k \neq 0, \text{ the vector } (0, k, 0, 0, \dots, 0) \text{ is also in the null space of } M \\ \downarrow \\ \text{There are infinitely many non-zero vectors in the null space of } M, \text{ but} \\ \text{only one linearly independent vector in the null space of } M \\ \downarrow \\ \text{D is correct} \end{array}$$

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Edited my answer, It should render properly now.

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@sri321 why the statement atleast 2 cant we work with atleast 1 it would also be true right??
 

Answer 2

.

Answer 3

.

 

\[
\mathbf{e}_1 = (1,0,0,\ldots,0)^T
\]

\[
\mathbf{e}_2 = (0,1,0,\ldots,0)^T \quad \text{(This is the one from our problem)}
\]

\[
\mathbf{e}_3 = (0,0,1,\ldots,0)^T
\]

\[
M\mathbf{e}_1 = \text{Extracts the 1st column of } M
\]

\[
M\mathbf{e}_2 = \text{Extracts the 2nd column of } M
\]

\[
M\mathbf{e}_3 = \text{Extracts the 3rd column of } M
\]

According to our problem,

\[
M\mathbf{e}_2 = \mathbf{0}
\]

The 2nd column of matrix $M$ is made entirely of zeros.

 

Let's say: $M$ is just a $3\times3$ matrix.
Multiply it by your vector $\mathbf{e}_2$:

\[
\begin{bmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{bmatrix}
\begin{bmatrix}
0 \\ 1 \\ 0
\end{bmatrix}
=
\begin{bmatrix}
b \\ e \\ h
\end{bmatrix}
\]

Will give second column of matrix $M = \begin{bmatrix} b \\ e \\ h \end{bmatrix}$.

According to our problem, this vector is in the null space of $M$.

\[
\Rightarrow
\begin{bmatrix}
b \\ e \\ h
\end{bmatrix}
=
\begin{bmatrix}
0 \\ 0 \\ 0
\end{bmatrix}
\]

Therefore, if $M\mathbf{e}_2 = \mathbf{0}$, it means the entire second column of $M$ is just zeros.

 

A matrix with an entire column of zeros always has a determinant of $0$.

Option B = Correct

1 zero column means the rank is at most $= n - 1$.

Option C = False

 

Null Space is a continuous vector space. If there is 1 non-zero vector, then there can be infinite non-zero vectors.

Option D = Correct

B + D = CORRECT

 

Answer 4

(0,1,0,0,0...0) is in nullspace of A mean (0,1,0,0,0...0) is a solution to Ax=0. It is a non-trivial solution.
So, columns of A are linearly dependent. So, there are infinitely many non-trivial solutions of A (it implies that there are at least two non-zero vectors in the null space of A). Determinant of A will also be zero.

B,D correct, A incorrect. We can't infer option C. So, C is not necessarily true.