A vector w~ is called a linear combination of the vectors ~v1, ~v2, and ~v3 if w~ can be expressed as w~ = c1~v1 + c2~v2 + c3~v3, where c1, c2, and c3 are scalars. (a) Find scalars c1, c2, and c3 to express h−1, 1, 5i as a linear combination of ~v1 = h1, 0, 1i, ~v2 = h3, 2, 0i, and ~v3 = h0, 1, 1i.

We are tasked with finding scalars \( c_1 \), \( c_2 \), and \( c_3 \) such that the vector \( \mathbf{w} = \langle -1, 1, 5 \rangle \) can be written as a linear combination of the vectors \( \mathbf{v_1} = \langle 1, 0, 1 \rangle \), \( \mathbf{v_2} = \langle 3, 2, 0 \rangle \), and \( \mathbf{v_3} = \langle 0, 1, 1 \rangle \).

This means we want to solve the following equation for \( c_1 \), \( c_2 \), and \( c_3 \):

\[
\mathbf{w} = c_1 \mathbf{v_1} + c_2 \mathbf{v_2} + c_3 \mathbf{v_3}
\]

Substituting the vectors, we get:

\[
\langle -1, 1, 5 \rangle = c_1 \langle 1, 0, 1 \rangle + c_2 \langle 3, 2, 0 \rangle + c_3 \langle 0, 1, 1 \rangle
\]

This expands into the following system of equations for each component:

1. \( c_1(1) + c_2(3) + c_3(0) = -1 \) (from the \( x \)-component)
2. \( c_1(0) + c_2(2) + c_3(1) = 1 \) (from the \( y \)-component)
3. \( c_1(1) + c_2(0) + c_3(1) = 5 \) (from the \( z \)-component)

We now solve this system of equations step by step:

### Equation 1 (x-component):
\[
c_1 + 3c_2 = -1 \tag{1}
\]

### Equation 2 (y-component):
\[
2c_2 + c_3 = 1 \tag{2}
\]

### Equation 3 (z-component):
\[
c_1 + c_3 = 5 \tag{3}
\]

#### Step 1: Solve Equation (3) for \( c_1 \)
From Equation (3):
\[
c_1 = 5 - c_3 \tag{4}
\]

#### Step 2: Substitute \( c_1 \) from Equation (4) into Equation (1)
Substituting \( c_1 = 5 - c_3 \) into Equation (1):
\[
(5 - c_3) + 3c_2 = -1
\]
\[
5 + 3c_2 - c_3 = -1
\]
\[
3c_2 - c_3 = -6 \tag{5}
\]

#### Step 3: Solve the system of Equations (2) and (5)
We now have the system:
1. \( 2c_2 + c_3 = 1 \) (Equation 2)
2. \( 3c_2 - c_3 = -6 \) (Equation 5)

Add the two equations to eliminate \( c_3 \):

\[
(2c_2 + c_3) + (3c_2 - c_3) = 1 + (-6)
\]
\[
5c_2 = -5
\]
\[
c_2 = -1
\]

#### Step 4: Find \( c_3 \)
Substitute \( c_2 = -1 \) into Equation (2):
\[
2(-1) + c_3 = 1
\]
\[
-2 + c_3 = 1
\]
\[
c_3 = 3
\]

#### Step 5: Find \( c_1 \)
Substitute \( c_3 = 3 \) into Equation (4):
\[
c_1 = 5 - 3 = 2
\]

### Final Answer:
The scalars are:
\[
c_1 = 2, \quad c_2 = -1, \quad c_3 = 3
\]

Thus, the vector \( \mathbf{w} = \langle -1, 1, 5 \rangle \) can be written as:

\[
\mathbf{w} = 2 \mathbf{v_1} - \mathbf{v_2} + 3 \mathbf{v_3}
\]

Do you need any further details or clarifications?

Here are some related questions for practice:
1. How can you check if a set of vectors is linearly independent?
2. What is the geometric interpretation of a linear combination of vectors?
3. How do you express a vector as a linear combination in higher dimensions?
4. What conditions make a vector space span?
5. How would the solution change if one of the vectors was linearly dependent on another?

**Tip:** To solve systems of linear equations efficiently, matrix methods like Gaussian elimination or using matrix inverses can be very helpful.

Learn how to express the vector w = ⟨−1, 1, 5⟩ as a linear combination of vectors v1 = ⟨1, 0, 1⟩, v2 = ⟨3, 2, 0⟩, and v3 = ⟨0, 1, 1⟩. This problem involves solving a system of linear equations to find the scalars c1, c2, and c3. Suitable for students in advanced high school or undergraduate courses.

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