Question

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

Answer 1

Answer:

W cannot form a vector space.

Step-by-step explanation:

Given set is,

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

where $[p,q,r,s]^t$ is a column vector.

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

$-p+3q+(0\times r)-5s=0\hfill (1)$

$p+(0\times q)+3r-2s=0\hfill (2)$

Adding (1) and (2) we get,

$3q+(0\times r)-5s=0$

$(0\times q)+3r-2s=0$

Solving,

$\frac{q}{15}=\frac{r}{6}=\frac{s}{9}=k$

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.