Either use an appropriate theorem to show that the given set, W, is a vector space, or find a specific example to the contrary. Wequals=StartSet [Start 4 By 1 Matrix 1st Row 1st Column p 2nd Row 1st Column q 3rd Row 1st Column r 4st Row 1st Column s EndMatrix ]: Start 2 By 1 Matrix 1st Row 1st Column negative p plus 3 q equals 5 s 2nd Row 1st Column p equals 2 s minus 3 r EndMatrix EndSet

W cannot form a vector space.

Step-by-step explanation:

Given set is,

\{[p,q,r,s] : -p+3q=5s,p=2s-3r\}" alt="W=\big{\Big[p,q,r,s\Big]^t : -p+3q=5s,p=2s-3r\big}" />\{[p,q,r,s] : -p+3q=5s,p=2s-3r\}" />

where is a column vector.

To show W is a vector space or not, taking two equations as,

Adding (1) and (2) we get,

Solving,

where k is constant.

Gives, q=15k, r=6k, s=9k. Substitute this in (2) we get p=0.

Since we cannot get p=q=r=s=0, given system is not linearly independent. Rather there is no zero vector [(p,q,r,s)=(0,0,0,0)] in the systen so there is no additive identity and W will not form a vector space.

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step-by-step explanation:

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step-by-step explanation: