Write each vector as a linear combination of the vectors in

Write each vector as a linear combination of the vectors in S. (If it is not possible, leave all boxes in that row empty.) S ={S_1. S_2, S_3}, S_1 = (3, 0, 2), S_2 = (3, 3, 2), S_3 = (3, -7, 1)

Solution

a>
step 1:
We set up our augmented matrix and row reduce it

let x,y,z be the constants

Augmented matrix = [s1 s2 s3 | u]
                  
Augmented matrix = [3 3 3 |12]
                                [0 3 -7 |-4]
                                 [2 2   1 | 7]
performing row operations
R1 --> R1/3

Augmented matrix = [1 1   1 | 4]
                                 [0 3 -7 |-4]
                              [2 2   1 | 7]

R3 --> R3 - 2R1

Augmented matrix = [1 1   1 | 4]
                               [0 3 -7 |-4]
                                [0 0   -1|-1]

step 2:

We find if the matrix represents a consistent system of equations or not
Based on the reduced matrix, the underlying system is consistent. This happens as there
are no rows of all zeros in the coefficient part of the matrix and a single nonzero value in the
augment.

=> using back substitution to find x1,x2 and x3

=> -x3 = -1 , x3 = 1

3x2 - 7x3 = -4
3x2 - 7(1) = -4 , x2 = 1

x1+x2+x3 = 4
x1+1+1 =4 , x1 = 2

hence the linear combination is :

u(12, -4, 7) = x1(3, 0, 2) + x2(3, 3, 2) + x3(3, -7, 1)

u(12, -4, 7) = 2(3, 0, 2) + 1(3, 3, 2) + 1(3, -7, 1)

 Write each vector as a linear combination of the vectors in S. (If it is not possible, leave all boxes in that row empty.) S ={S_1. S_2, S_3}, S_1 = (3, 0, 2),

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