# Write a Vector as a Linear Combination of Three Vectors

## Problem 653

Write the vector $\begin{bmatrix} 1 \\ 3 \\ -1 \end{bmatrix}$ as a linear combination of the vectors
$\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} , \, \begin{bmatrix} 2 \\ -2 \\ 1 \end{bmatrix} , \, \begin{bmatrix} 2 \\ 0 \\ 4 \end{bmatrix}.$

## Solution.

We want to find real numbers $x_1 , x_2 , x_3$ which solve the equation
$\begin{bmatrix} 1 \\ 3 \\ -1 \end{bmatrix} = x_1 \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + x_2 \begin{bmatrix} 2 \\ -2 \\ 1 \end{bmatrix} + x_3 \begin{bmatrix} 2 \\ 0 \\ 4 \end{bmatrix} = \begin{bmatrix} x_1 + 2 x_2 + 2 x_3 \\ – 2 x_2 \\ x_2 + 4 x_3 \end{bmatrix}.$

Each row in this matrix now gives an equality, and so the three rows define a system of linear equations. To find a solution, we reduce its augmented matrix:
\begin{align*}
\left[\begin{array}{rrr|r} 1 & 2 & 2 & 1 \\ 0 & -2 & 0 & 3 \\ 0 & 1 & 4 & -1 \end{array} \right] \xrightarrow[R_2 + 2 R_3]{R_1 – 2 R_3} \left[\begin{array}{rrr|r} 1 & 0 & -6 & 3 \\ 0 & 0 & 8 & 1 \\ 0 & 1 & 4 & -1 \end{array} \right] \xrightarrow{ R_2 \leftrightarrow R_3 } \left[\begin{array}{rrr|r} 1 & 0 & -6 & 3 \\ 0 & 1 & 4 & -1 \\ 0 & 0 & 8 & 1 \end{array} \right] \\[6pt] \xrightarrow{ \frac{1}{8} R_3} \left[\begin{array}{rrr|r} 1 & 0 & -6 & 3 \\[3pt] 0 & 1 & 4 & -1 \\[3pt] 0 & 0 & 1 & \frac{1}{8} \end{array} \right] \xrightarrow[R_2 – 4 R_3 ]{ R_1 + 6 R_3} \left[\begin{array}{rrr|r} 1 & 0 & 0 & \frac{15}{4} \\[3pt] 0 & 1 & 0 & \frac{-3}{2} \\[3pt] 0 & 0 & 1 & \frac{1}{8} \end{array} \right].
\end{align*}

Now we can read off the solution $x_1 = \frac{15}{4}$, $x_2 = \frac{-3}{2}$, and $x_3 = \frac{1}{8}$.