GO Classes CS/DA 2025 | Weekly Quiz 4 | Linear Algebra | Question: 21

GO Classes asked Mar 29, 2023 • edited Mar 20 by shadymademe

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Consider a matrix $A$ of dimension $m \times n$ such that -

$A x=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$ has no solutions and $A x=\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]$ has exactly one solution

Which of the following CAN be true?

$\operatorname{Rank}(A)=2$
$m=3$
$n=1$
$\operatorname{Rank}(A)=1$

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GO Classes asked Mar 29, 2023 • edited Mar 20 by shadymademe

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Manish joshi

commented Jun 13, 2023

Since Ax in a 3x1 vector so m will be 3 so it is in R3 and if would have 3 LI vectors it could give any B but it is having no solution for some B so it dont have 3 LI vectors so rank coul be 2 or 1

Is is correct?

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ranjanKiran

commented Mar 24

From Ax = b, where b = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}

We know that there is exactly one solution. So from this we can infer that the columns of A are linearly independent.

Because with Linearly dependent vectors there are only two possibility of solutions for Ax = b.

No solution
Infinite Solution

m = 3, can be inferred from dimension of solution

if Rank(A) = 2, then it means that n is also equal to 2

If Rank(A) = 1, then it means that n is also equal to 1

If n = 1, It can be true, as then Rank(A) will be 1, as 1 Column vector is always independent.

As there are no constraints on n. These 3 options are true. Along with m = 3

Let me know if there are any corrections in explanation.

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option ; a , b, c, d are correct one .

Gutley>> answered Mar 30, 2023

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Swarnava Bose

commented Sep 15, 2023

but M = No. Of rows. In the previous image too,when n=1 , m was 3 only....

so what was the need for the 2nd image and taking n = 2 ( no. of columns as 2 ) ??

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minimalist

commented Nov 17, 2023

To check rank(A)=2 possible.

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Since the question is Which of them CAN be true, it means that we have to just find one example to show that the option is possible.

Option	No Solution	Unique Solution
A & B	$$\begin{bmatrix} a & b & 1 \\ 0 & c & 1 \\ 0 & 0 & 1 \end{bmatrix}$$	$$\begin{bmatrix} a & b & 0 \\ 0 & c & 1 \\ 0 & 0 & 0 \end{bmatrix}$$
C & D	$$\begin{bmatrix} a & 1 \\ 0 & 1 \\ 0 & 1 \end{bmatrix}$$	interchanging R1 and R2 $$\begin{bmatrix} a & 1 \\ 0 & 0 \\ 0 & 0 \end{bmatrix}$$

,where all the alphabets are non-zero numbers.

tejashmore25 answered Mar 23

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