## Problem Statement

Write the decomposition of vector $x$ in terms of vectors $p, q, r$:

$x=\{3 ; 1 ; 8\}$

$p=\{0 ; 1 ; 3\}$

$q=\{1 ; 2 ;-1\}$

$r=\{2 ; 0 ;-1\}$
## Solution

The desired decomposition of vector $x$ is:

$x=\alpha \cdot p+\beta \cdot q+\gamma \cdot r$

Or in the form of a system:

$$
\left\{\begin{array}{l}
\alpha \cdot p_{1}+\beta \cdot q_{1}+\gamma \cdot r_{1}=x_{1} \\
\alpha \cdot p_{2}+\beta \cdot q_{2}+\gamma \cdot r_{2}=x_{2} \\
\alpha \cdot p_{3}+\beta \cdot q_{3}+\gamma \cdot r_{3}=x_{3}
\end{array}\right.
$$

We obtain:

$$
\left\{\begin{array}{l}
\beta+2 \gamma=3 \\
\alpha+2 \beta=1 \\
3 \alpha-\beta-\gamma=8
\end{array}\right.
$$

Add the third row multiplied by 2 to the first row:

$$
\left\{\begin{array}{l}
6 \alpha-\beta=19 \\
\alpha+2 \beta=1 \\
3 \alpha-\beta-\gamma=8
\end{array}\right.
$$

Add the first row multiplied by 2 to the second row:

$$
\left\{\begin{array}{l}
6 \alpha-\beta=19 \\
13 \alpha=39 \\
3 \alpha-\beta-\gamma=8
\end{array}\right.
$$

$$
\begin{aligned}
& \left\{\begin{array}{l}
6 \alpha-\beta=19 \\
\alpha=3 \\
3 \alpha-\beta-\gamma=8
\end{array}\right. \\
& \left\{\begin{array}{l}
6 \cdot 3-\beta=19 \\
\alpha=3 \\
3 \cdot 3-\beta-\gamma=8
\end{array}\right. \\
& \left\{\begin{array}{l}
\beta=-1 \\
\alpha=3 \\
-\beta-\gamma=-1
\end{array}\right. \\
& \left\{\begin{array}{l}
\beta=-1 \\
\alpha=3 \\
-(-1)-\gamma=-1
\end{array}\right. \\
& \left\{\begin{array}{l}
\beta=-1 \\
\alpha=3 \\
\gamma=2
\end{array}\right.
\end{aligned}
$$

The desired decomposition:

$x=3 p-q+2 r$

## Problem Kuznetsov Analytic Geometry 2-17