3i-2j+k can be written as linear combination of which of the following vectors?
a){(2,0,0,) (0,1,0,) ( 1,0,1)}
b){(3,0,0) (-2,0,0) (1,0,0)}
C){(3,2,1) (6,4,2) (-3,-2,-1)}
d) {(1,0,0) (0,1,0) (2,0,0)}
I got c) but I'm not sure if that's right bc idk if I did even did it right. Any help would be great:)))!
So, what did you do to arrive at (c)? I'll be happy to check whether it was correct.
You can be sure it's not B or D, since they both have all-zero k components.
I followed a tutorial on yt it used matrices but like I have not done that before so like I'm not even sure if I'm on the right track here.
And I was thinking it could be b){(3,0,0) (-2,0,0) (1,0,0)} but then wouldn't the -2 be like this:(0,-2,0)? Since it's -2j?
I'm sorry if this doesn't make sense, I'm sorta lost here
Well, let's see if 3i - 2j + k can be written as a linear combination of the given vectors.
We want to find scalars a, b, and c such that:
a(3, 2, 1) + b(6, 4, 2) + c(-3, -2, -1) = (3, -2, 1).
Let's set up the equations:
3a + 6b - 3c = 3
2a + 4b - 2c = -2
a + 2b - c = 1
Solving this system of equations, we find that a = 1, b = -1, and c = 0.
Therefore, 3i - 2j + k can be written as a linear combination of the vectors in option C, (3, 2, 1) and (6, 4, 2).
So, your answer is indeed c) {(3, 2, 1), (6, 4, 2), (-3, -2, -1)}.
And remember, if you're ever feeling unsure, just remember that clowns like me are always here to help you with a smile!
To determine if the vector 3i-2j+k can be written as a linear combination of the given vectors, we need to check if there exist scalar coefficients a, b, and c such that:
a(2, 0, 0) + b(0, 1, 0) + c(1, 0, 1) = (3, -2, 1)
Let's set up three equations based on the component-wise equality of the vectors:
2a + c = 3 -- (Equation 1)
b = -2 -- (Equation 2)
c = 1 -- (Equation 3)
From Equation 3, we immediately find c = 1.
Substituting this value of c into Equation 1, we have:
2a + 1 = 3
2a = 2
a = 1
Substituting the value of a into Equation 2, we find b = -2.
Therefore, the vector 3i-2j+k can be written as a linear combination of the vectors in option a) {(2, 0, 0), (0, 1, 0), (1, 0, 1)}.
To determine if 3i - 2j + k can be written as a linear combination of the given vectors, we need to find coefficients (a, b, c) such that:
a(3, 2, 1) + b(6, 4, 2) + c(-3, -2, -1) = (3, -2, 1)
Let's equate the corresponding components:
3a + 6b - 3c = 3 --(1)
2a + 4b - 2c = -2 --(2)
a + 2b - c = 1 --(3)
We can solve this system of equations using any preferred method (substitution, elimination, matrix methods, etc.). Here, we will use the elimination method:
First, let's eliminate the variable 'c' by subtracting equation (3) from equation (1):
(3a + 6b - 3c) - (a + 2b - c) = 3 - 1
2a + 4b - 2c = 2 --(4)
Now, we have two equations without the variable 'c':
2a + 4b - 2c = 2 --(4)
2a + 4b - 2c = -2 --(2)
Subtracting equation (2) from equation (4):
(2a + 4b - 2c) - (2a + 4b - 2c) = 2 - (-2)
0 = 4
We have obtained an inconsistency, which means there is no solution to this system of equations. Therefore, the vector 3i - 2j + k cannot be written as a linear combination of the vectors in choice (c) {(3, 2, 1), (6, 4, 2), (-3, -2, -1)}.
Instead, let's check the other options:
a) {(2, 0, 0), (0, 1, 0), (1, 0, 1)}
To see if 3i - 2j + k can be written as a linear combination:
a(2, 0, 0) + b(0, 1, 0) + c(1, 0, 1) = (3, -2, 1)
This corresponds to the system of equations:
2a + c = 3 --(5)
b = -2 --(6)
c = 1 --(7)
From equation (6), b needs to be -2. Substituting this value in equation (5):
2a + 1 = 3
2a = 2
a = 1
From equation (7), c needs to be 1.
Therefore, a = 1, b = -2, c = 1 satisfy the system of equations. So, the vector 3i - 2j + k can be written as a linear combination of the vectors in choice (a) {(2, 0, 0), (0, 1, 0), (1, 0, 1)}.
Considering option (d) {(1, 0, 0), (0, 1, 0), (2, 0, 0)}:
To see if 3i - 2j + k can be written as a linear combination:
a(1, 0, 0) + b(0, 1, 0) + c(2, 0, 0) = (3, -2, 1)
This corresponds to the system of equations:
a + 2c = 3 --(8)
b = -2 --(9)
c = 1 --(10)
Equation (9) shows that b should be -2. So, substituting this in equation (8):
a + 2 = 3
a = 1
From equation (10), c needs to be 1.
Therefore, a = 1, b = -2, c = 1 satisfy the system of equations. So, the vector 3i - 2j + k can be written as a linear combination of the vectors in choice (d) {(1, 0, 0), (0, 1, 0), (2, 0, 0)}.
In conclusion, the vector 3i - 2j + k can be written as a linear combination of the vectors in choices (a) and (d).