Solution. To check whether the vectors are linearly independent, we must answer the following question: if a linear combination of the vectors is the zero vector, is it necessarily true that all the coefficients are zeros?
Suppose that
x 1 ⃗v 1 + x 2 ⃗v 2 + x 3 ( ⃗v 1 + ⃗v 2 + ⃗v 3 ) = ⃗0
(a linear combination of the vectors is the zero vector). Is it necessarily true that x1 =x2 =x3 =0?
We have
x1⃗v1 + x2⃗v2 + x3(⃗v1 + ⃗v2 + ⃗v3) = x1⃗v1 + x2⃗v2 + x3⃗v1 + x3⃗v2 + x3⃗v3
=(x1 + x3)⃗v1 + (x2 + x3)⃗v2 + x3⃗v3 = ⃗0.
Since ⃗v1, ⃗v2, and ⃗v3 are linearly independent, we must have the coeffi-
cients of the linear combination equal to 0, that is, we must have
x1 + x3 = 0 x2 + x3 = 0 ,
x3 = 0
from which it follows that we must have x1 = x2 = x3 = 0. Hence the
vectors ⃗v1, ⃗v2, and ⃗v1 + ⃗v2 + ⃗v3 are linearly independent.
Answer. The vectors ⃗v1, ⃗v2, and ⃗v1 + ⃗v2 + ⃗v3 are linearly independent.
First you must divided my the dividend and then add the whole number by number on your sheet!
Answer: It can help pay for things that you could not get otherwise. Sometimes one makes enough money with 2 months of savings so a credit card covers the cost until the second month.
Explanation: Mark me brainly please
Answer:
Acceleration a =-1.75 m/s²
Explanation:
Given:
Initial speed u = 24 m/s
Final speed v = 10 m/s
Time taken t = 8 sec
Find:
Acceleration a
Computation:
a = (v-u)/t
a = (10-24)/8
a = -14 / 8
Acceleration a =-1.75 m/s²
Answer: It was okay, i took one of my standard assesments for writing i think i atleast passed.
Explanation:
