If u, v, and w are nonzero vectors in r 2 , is w a linear combination of u and v?
1 answer:
Not necessarily.
![\mathbf u](https://tex.z-dn.net/?f=%5Cmathbf%20u)
and
![\mathbf v](https://tex.z-dn.net/?f=%5Cmathbf%20v)
may be linearly dependent, so that their span forms a subspace of
![\mathbb R^2](https://tex.z-dn.net/?f=%5Cmathbb%20R%5E2)
that does not contain every vector in
![\mathbb R^2](https://tex.z-dn.net/?f=%5Cmathbb%20R%5E2)
.
For example, we could have
![\mathbf u=(0,1)](https://tex.z-dn.net/?f=%5Cmathbf%20u%3D%280%2C1%29)
and
![\mathbf v=(0,-1)](https://tex.z-dn.net/?f=%5Cmathbf%20v%3D%280%2C-1%29)
. Any vector
![\mathbf w](https://tex.z-dn.net/?f=%5Cmathbf%20w)
of the form
![(r,0)](https://tex.z-dn.net/?f=%28r%2C0%29)
, where
![r\neq0](https://tex.z-dn.net/?f=r%5Cneq0)
, is impossible to obtain as a linear combination of these
![\mathbf u](https://tex.z-dn.net/?f=%5Cmathbf%20u)
and
![\mathbf v](https://tex.z-dn.net/?f=%5Cmathbf%20v)
, since
![c_1\mathbf u+c_2\mathbf v=(0,c_1)+(0,-c_2)=(0,c_1-c_2)\neq(r,0)](https://tex.z-dn.net/?f=c_1%5Cmathbf%20u%2Bc_2%5Cmathbf%20v%3D%280%2Cc_1%29%2B%280%2C-c_2%29%3D%280%2Cc_1-c_2%29%5Cneq%28r%2C0%29)
unless
![r=0](https://tex.z-dn.net/?f=r%3D0)
and
![c_1=c_2](https://tex.z-dn.net/?f=c_1%3Dc_2)
.
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13
good job. really excelling :)
I think it’s infinitely many
Correct Answer: A. 2(n+8).
Answer:
91 hours
Step-by-step explanation:
Add 12 to itself 8 times and then add 7 once