Express the vector (4,5,5) as a linear combination of the vectors (1,2,3), (-1,1,4) and (3,3,2).

To express the vector \((4, 5, 5)\) as a linear combination of the given vectors \((1, 2, 3)\), \((-1, 1, 4)\), and \((3, 3, 2)\), we solve for scalars \(a\), \(b\), and \(c\) such that:

\[
(4, 5, 5) = a(1, 2, 3) + b(-1, 1, 4) + c(3, 3, 2)
\]

This expands to the system of equations:
\[
4 = a(1) + b(-1) + c(3) \tag{1}
\]
\[
5 = a(2) + b(1) + c(3) \tag{2}
\]
\[
5 = a(3) + b(4) + c(2) \tag{3}
\]

### Step 1: Write as a matrix equation
We can express the system as:
\[
\begin{bmatrix}
1 & -1 & 3 \\
2 &  1 & 3 \\
3 &  4 & 2
\end{bmatrix}
\begin{bmatrix}
a \\
b \\
c
\end{bmatrix}
=
\begin{bmatrix}
4 \\
5 \\
5
\end{bmatrix}
\]

### Step 2: Solve the system using Gaussian elimination
#### Augmented matrix:
\[
\begin{bmatrix}
1 & -1 & 3 & | & 4 \\
2 &  1 & 3 & | & 5 \\
3 &  4 & 2 & | & 5
\end{bmatrix}
\]

#### Row operations:
1. \(R_2 \to R_2 - 2R_1\), \(R_3 \to R_3 - 3R_1\):
\[
\begin{bmatrix}
1 & -1 &  3 & | &  4 \\
0 &  3 & -3 & | & -3 \\
0 &  7 & -7 & | & -7
\end{bmatrix}
\]

2. \(R_3 \to R_3 - \frac{7}{3}R_2\):
\[
\begin{bmatrix}
1 & -1 &  3 & | &  4 \\
0 &  3 & -3 & | & -3 \\
0 &  0 &  0 & | &  0
\end{bmatrix}
\]

From the reduced matrix, we find:
\[
a - b + 3c = 4 \tag{4}
\]
\[
3b - 3c = -3 \tag{5}
\]

#### Solve for \(b\) and \(c\) from Equation (5):
\[
b - c = -1 \quad \implies \quad b = c - 1
\]

#### Substitute \(b = c - 1\) into Equation (4):
\[
a - (c - 1) + 3c = 4
\]
\[
a - c + 1 + 3c = 4
\]
\[
a + 2c = 3 \quad \implies \quad a = 3 - 2c
\]

### Step 3: Final Expression
Substitute back:
- \(a = 3 - 2c\)
- \(b = c - 1\)

The vector \((4, 5, 5)\) can be expressed as:
\[
(4, 5, 5) = (3 - 2c)(1, 2, 3) + (c - 1)(-1, 1, 4) + c(3, 3, 2)
\]

This provides a parametric representation where \(c\) can take any value.

Express the vector (4,5,5) as a linear combination of the vectors (1,2,3), (-1,1,4), and (3,3,2).

Learn how to express the vector (4,5,5) as a linear combination of (1,2,3), (-1,1,4), and (3,3,2). This detailed solution covers Gaussian elimination, matrix representation, and parametric equations. Perfect for students studying introductory linear algebra.

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