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Answer:
x = 9/25, y = 7/25, z = 4/25
Explanation:
step 1: solve for y in first equation to use substitution
y = -2x + 1
step 2: insert into second equation to eliminate y variable and work with two variables
3(-2x+1) + z = 1 (now distribute)
-6x + 3 + z = 1 (bring over the -3)
-6x + z = -2
step 3: now left with -6x + z = -2 and x + 4z = 1, so eliminate z
-6x + z = -2 (multiply by 4)
4(-6x + z = -2)
-24x + 4z = -8
-
x + 4z = 1 (brought down third equation)
-25x = -9
x = 9/25
then just plug in x into the third equation to solve for z, followed by plugging it into the first equation to solve for y. you should end with y = 7/25 and z = 4/25
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.