GO Classes CS/DA 2025 | Weekly Quiz 3 | Fundamental Course and Linear Algebra | Question: 12

Answer 1

Given: a set S of vectors in $R^{n}$ and a set X (X${\subset }$S) and Y (S${\subset}$Y).

• A. False. If the vectors of S are LD then it is not necessary that vectors of X are also LD they can be LI as well and the remaining vectors (S-X) can be dependent which will make the S LD.
• B. True. Since X ${\subset }$ S and S is LI then none of the vectors of S can be represented as a Linear Combination of other vectors of S, which include X as well.
• C. True. Since S ${\subset }$ Y and S is LD then Y is also LD because Y contains S i.e. at least 1 vector of Y (from set S) can be represented as a Linear Combination of another vector in Y. 
• D. False. If S is LI doesn’t mean Y will be LI as well because S ${\subset }$ Y and the remaining vectors from set Y (Y-S) can be LD as well.

Abbreviations: LD – Linearly Dependent, LI – Linearly Independent 

Answer 2

Given set of vectors S $\epsilon$ $R^{n}$ and other 2 set of vectors X, Y $\epsilon$ $R^{n}$ such that

$X\subset S\subset Y$

Option A – False

Counter Example: S = $\begin{Bmatrix} \begin{bmatrix} 0\\ 1 \end{bmatrix}, \begin{bmatrix} 1\\ 0 \end{bmatrix}, \begin{bmatrix} \alpha \\ \beta\end{bmatrix}&\end{Bmatrix}$ – Linearly Dependent

                                 X = $\begin{Bmatrix} \begin{bmatrix} 0\\ 1 \end{bmatrix}, \begin{bmatrix} 1\\ 0 \end{bmatrix}&\end{Bmatrix}$ – Linearly independent

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Option B – True

Let S = {$v_{1}$,$v_{2}$, $v_{3}$ }

       X = {$v_{1}$,$v_{2}$}

Given that vectors of set S are linearly independent i.e, $c_{1}v_{1} + c_{2}v_{2} + c_{3}v_{3}=0$, where

$c_{1}=c_{2}=c_{3}=0$

Now we need to show prove X = {$v_{1}$,$v_{2}$} is linearly independent

i.e,  $c_{1}v_{1}+c_{2}v_{2}=0$ has only trivial solution

Assume  $c_{1}v_{1}+c_{2}v_{2}=0$

Add 0 on both sides 

$c_{1}v_{1}+c_{2}v_{2}+0*v_{3}=0$

As it is given that vectors S = {$v_{1}$,$v_{2}$, $v_{3}$ } are linearly independent; $c_{1}=c_{2}=0$ is the

only possible solution.

(Subset of linearly independent set of vectors are always linearly independent)

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Option C – True

Superset of linearly dependent set of vectors are always linearly dependent.

This can be done by making newly added vector redundant

Example: S= $\begin{Bmatrix} \begin{bmatrix} 1\\ 2 \end{bmatrix}, \begin{bmatrix} 2\\ 4 \end{bmatrix}&\end{Bmatrix}$ – Linearly Dependent, 

X = $\begin{Bmatrix} \begin{bmatrix} 1\\ 2 \end{bmatrix}, \begin{bmatrix} 2\\ 4 \end{bmatrix}, \begin{bmatrix} 0 \\ 1\end{bmatrix}&\end{Bmatrix}$ – still linearly dependent

$\begin{bmatrix} 2\\ 4 \end{bmatrix}=2  *$ $\begin{bmatrix} 1\\ 2 \end{bmatrix}$$+ 0* \begin{bmatrix} 0\\ 1 \end{bmatrix}$

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Option D – False

Counter Example: 

S = $\begin{Bmatrix} \begin{bmatrix} 0\\ 1 \end{bmatrix}, \begin{bmatrix} 1\\ 0 \end{bmatrix}&\end{Bmatrix}$ – Linearly independent

Y = $\begin{Bmatrix} \begin{bmatrix} 0\\ 1 \end{bmatrix}, \begin{bmatrix} 1\\ 0 \end{bmatrix}, \begin{bmatrix} \alpha \\ \beta\end{bmatrix}&\end{Bmatrix}$ – Linearly dependent

Answer 3

option b and c seem correct