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:

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.