Join the growing community of curious minds on IDNLearn.com. Discover reliable answers to your questions with our extensive database of expert knowledge.
Sagot :
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]
[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]
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. IDNLearn.com is committed to providing the best answers. Thank you for visiting, and see you next time for more solutions.
