GO Classes CS/DA 2025 | Weekly Quiz 3 | Fundamental Course and Linear Algebra | Question: 11

Answer 1

$\text{X}=\left[\begin{array}{lll}0 & 0 & 0\end{array}\right]^{\top}$ is a trivial solution, but there might exist other solution to the equation, $\text{X}=\left[\begin{array}{lll}-7 & 3 & 1\end{array}\right]^{\top}$ is one such solution which in turn shows that the column vectors in $\text{A}$ are linearly dependent. Hence, B and C are incorrect.

Comments

See Buddy, The def$^{^{n}}$ of Linearly Independent is as follows ---

A set of vectors $\left \{ v_{1} , v_{2} , v_{3} ... , v_{n} \right \}$ is linearly independent If the ONLY  linear combination of the vectors,  i.e. ( c$_{1}$v$_{1}$ + c$_{2}$v$_{2}$ + c$_{3}$v$_{3}$ + … + c$_{n}$v$_{n}$ ) that equals 0 is the trivial linear combination i.e. ( c$_{1}$ = c$_{2}$ = … = c$_{n}$ = 0 ).

Note, $\vec{X} = \vec{0} $ is always “a” sol$^{^{n}}$ (Trivial Sol$^{^{n}}$)  to $ A\vec{X} = \vec{0}$    

But Kuldeep, is it the only one?

If YES, then only $\left \{ v_{1} , v_{2} , ... , v_{n} \right \}$ is linearly independent…

 

And Thus Option:B is not “Sufficient” to conclude The Columns of A are linearly independent.

 

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If we reframe option B like
Showing that $\text{X}=\left[\begin{array}{lll}0 & 0 & 0\end{array}\right]^{\top}$ is the only solution to $\text{AX}=\overrightarrow{0}$ is sufficient to conclude that the columns of $\text{A}$ are Linearly Independent.

Then it would be correct.

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@Sachin Mittal 1 Sir her answer should be A option as

det(A)=0 then A as infinitely many solutions.

det(a)=0

X!=0

Answer 2

• A. X = [-7 3 1]$^{T}$ is not a solution of AX = 0. AX = (-7)*[2 -5 2]$^{T}$ + 3*[3 -1 -4]$^{T}$ + 1*[-5 32 -26]$^{T}$ = [-10 64 -52]$^{T}$ != [0 0 0]$^{T}$
• B. X = [0 0 0]$^{T}$ is not sufficient to conclude that the columns of A are LI. The sufficient condition is x = [0 0 0]$^{T}$ is the only solution.
• C. .
• D. X = [0 0 0]$^{T}$ is a trivial solution. 

But let’s look option C in terms of propositional logic. 

Statement :- Showing that X=[-7 3 1]$^{T}$ is a solution to AX=0 is sufficient to conclude that the columns of A are Linearly Independent. 

OR  If X=[-7 3 1]$^{T}$ is a solution to AX=0, then the columns of A are Linearly Independent.

As we know X=[-7 3 1]$^{T}$ is not a solution to AX=0, so the condition/hypothesis is false. Therefore we can say that the statement is true.

So, according to me answer should be both C & D. 

Comments

If p is false, then the truth value of p-->q is true.

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Truth Table of p → q
p | q | p → q

T | T | T

T | F | F

F | T | T

F | F | T

So yes, failing p does not imply q has failed nor does it imply p → q has failed.

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From Kenneth Rosen.

Answer 3

option d is only correct 

Answer 4

If X=[-7 3 1]T then D is the correct option but if we reform X=[-7 3 -1]T then X=[-7 3 -1] is a solution of AX=0 it is sufficient to say that {v1,v2,v3} are linearly dependent.where v1=[2 -5 2],v2=[3 -1 4],v3=[-5 32 -26] and c1v1+c2v2+c3v3 =0 here if we not find at least one ci that not equals to zero then we can say that {v1,v2,v3} is linearly independent.so from this A and D are correct answer.