To find the scalars [tex]\( c_1 \)[/tex], [tex]\( c_2 \)[/tex], and [tex]\( c_3 \)[/tex] such that the linear combination of the given vectors [tex]\( \vec{u} = (1, 0, 1) \)[/tex], [tex]\( \vec{v} = (3, 2, 0) \)[/tex], and [tex]\( \vec{w} = (0, 1, 1) \)[/tex] equals the vector [tex]\( (-1, 1, 5) \)[/tex], we need to solve the following vector equation:
[tex]\[ c_1 \vec{u} + c_2 \vec{v} + c_3 \vec{w} = (-1, 1, 5) \][/tex]
First, we write this equation in component form:
[tex]\[
c_1 (1, 0, 1) + c_2 (3, 2, 0) + c_3 (0, 1, 1) = (-1, 1, 5)
\][/tex]
Next, we distribute the scalars [tex]\( c_1 \)[/tex], [tex]\( c_2 \)[/tex], and [tex]\( c_3 \)[/tex] across the components of the vectors:
[tex]\[
( c_1 \cdot 1 + c_2 \cdot 3 + c_3 \cdot 0, c_1 \cdot 0 + c_2 \cdot 2 + c_3 \cdot 1, c_1 \cdot 1 + c_2 \cdot 0 + c_3 \cdot 1 ) = (-1, 1, 5)
\][/tex]
This gives us a system of linear equations:
1. [tex]\( c_1 + 3 c_2 = -1 \)[/tex]
2. [tex]\( 2 c_2 + c_3 = 1 \)[/tex]
3. [tex]\( c_1 + c_3 = 5 \)[/tex]
Now, we solve this system step by step.
First, let's isolate [tex]\( c_1 \)[/tex] and [tex]\( c_3 \)[/tex] from the equations involving them.
From equation (3): [tex]\( c_1 + c_3 = 5 \)[/tex], we can express [tex]\( c_3 \)[/tex] in terms of [tex]\( c_1 \)[/tex]:
[tex]\[ c_3 = 5 - c_1 \][/tex]
Now substitute [tex]\( c_3 = 5 - c_1 \)[/tex] into equation (2) to express everything in terms of [tex]\( c_2 \)[/tex] and [tex]\( c_1 \)[/tex]:
[tex]\[ 2 c_2 + (5 - c_1) = 1 \][/tex]
[tex]\[ 2 c_2 + 5 - c_1 = 1 \][/tex]
[tex]\[ 2 c_2 - c_1 = 1 - 5 \][/tex]
[tex]\[ 2 c_2 - c_1 = -4 \][/tex]
[tex]\[ c_1 = 2 c_2 + 4 \][/tex]
Next, substitute [tex]\( c_1 = 2 c_2 + 4 \)[/tex] into equation (1) to solve for [tex]\( c_2 \)[/tex]:
[tex]\[ (2 c_2 + 4) + 3 c_2 = -1 \][/tex]
[tex]\[ 2 c_2 + 4 + 3 c_2 = -1 \][/tex]
[tex]\[ 5 c_2 + 4 = -1 \][/tex]
[tex]\[ 5 c_2 = -1 - 4 \][/tex]
[tex]\[ 5 c_2 = -5 \][/tex]
[tex]\[ c_2 = -1 \][/tex]
Now that we have [tex]\( c_2 = -1 \)[/tex], we can find [tex]\( c_1 \)[/tex] using [tex]\( c_1 = 2 c_2 + 4 \)[/tex]:
[tex]\[ c_1 = 2 (-1) + 4 \][/tex]
[tex]\[ c_1 = -2 + 4 \][/tex]
[tex]\[ c_1 = 2 \][/tex]
Finally, we find [tex]\( c_3 \)[/tex] using [tex]\( c_3 = 5 - c_1 \)[/tex]:
[tex]\[ c_3 = 5 - 2 \][/tex]
[tex]\[ c_3 = 3 \][/tex]
So, the scalars are:
[tex]\[ c_1 = 2 \][/tex]
[tex]\[ c_2 = -1 \][/tex]
[tex]\[ c_3 = 3 \][/tex]
Therefore, the solution to the equation [tex]\( c_1 \vec{u} + c_2 \vec{v} + c_3 \vec{w} = (-1, 1, 5) \)[/tex] is:
[tex]\[ c_1 = 2 \][/tex]
[tex]\[ c_2 = -1 \][/tex]
[tex]\[ c_3 = 3 \][/tex]
