GO Classes CS/DA 2025 | Weekly Quiz 4 | Linear Algebra | Question: 26

Answer 1

• A. Is not always true. We have have solution if $b$ is linear combination of columns of A.
   
• B. This is true, if there is always a solution it means we must have $m$ LI columns.
• C. If $m<n\;$, Let $m=2$ and $n=3$, there are 2 rows and three columns, which means we have three columns in $R^2$. 
  Statement P is saying that A has $m$ LI columns which means $2$ LI columns (remember total columns are 3).
  It is not always true that we must have 2 LI columns out of 3 columns, may be we just have 1 LI column. Hence it is False.
   
• D. (Let $m=2$ and $n=3$), This is true because if we have 2 LI columns in $R^2$ then we must have solution for every $b$.

Answer – B,D

Comments

Option B: R → P
Let R is true means columns of A fill the vector space of $R^{m}$ then it means that we have ‘m’ LI columns ( or  m pivots)  which means out of ‘n’ columns we have n-m LD columns
hence  filling the space we need ‘m’ LI columns not ‘n’
after Making R true P can be false: T → F == False

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In option D when (S and P) → R,

Whether the statement S is required? When we have (mxn) matrix and Rank(matrix) = m then it is obvious that m<=n. In that case only P → R is sufficient.

Answer 2

Option A: S → R  clearly false  let S is True then R has to be true means if we have m < n then then it does not means we have ‘m’ LI columns they might be LD for R may be false
hence  S→ R is false

Option B: R → P
Let R is true means columns of A fill the vector space of $R^{m}$ then it means that we have ‘m’ LI columns ( or  m pivots)  which means out of ‘n’ columns
hence  filling the space we need ‘m’ LI’ so P is true

: T → T == True

Option C :S → ~P False

S: m<n means we have more then $m$ vectors
~ P : A can’t have m LI column which is False  as out of $n$ columns we can have $m$ or$ < m$ L.I columns
$Hence, T → F ==False  $
Option D: (S and P) → R  True   if we have m < n and then we cant not have ‘n’ LI column we can have at most ‘m’ so Propositional Statement F → Anything is True

Comments

ArturoGangwar for option C , P can have m LI also so that stmt is false na 

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Yes False ,Actually when post my answer then
P: A has $n$ Linearly Independent
But it has been modified To $m$

Answer 3

B, D

Option A) False

when S is true i.e when m < n there is no guarantee that we will have at least m linearly independent vectors. There could be zero linearly independent vectors as well and b can be any vector in R^m. So, when S is true R may not be true. Therefore option A is false.

Option B)True

when there is a solution for every b in R^m, Then there must be m linearly independent vectors in A.

Option C) False

option C says “when S is true, P cannot be true” but when S is true i.e, when m < n, P can be true i.e, A can have m linearly independent columns like the example below. So option C is false. 

$\begin{bmatrix} 1 & 0 & 0 & 5 &6 \\ 0& 1& 0& 3& 7\\ 0& 0& 1& 2& 5 \end{bmatrix}$

Option D)True

when m<n and A and has m linearly independent vectors then there is a solution for every b in R^m.