## Problem Statement

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

$x=\{-1 ; 7 ; 0\}$

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

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

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
\beta+2 \gamma=-1 \\
3 \alpha-\beta-\gamma=7 \\
\alpha+2 \beta=0
\end{array}\right. \\
& \left\{\begin{array}{l}
\beta+2 \gamma=-1 \\
3 \alpha-\beta-\gamma=7 \\
\alpha=-2 \beta
\end{array}\right. \\
& \left\{\begin{array}{l}
\beta+2 \gamma=-1 \\
3 \cdot(-2 \beta)-\beta-\gamma=7 \\
\alpha=-2 \beta
\end{array}\right. \\
& \left\{\begin{array}{l}
\beta+2 \gamma=-1 \\
-7 \beta-\gamma=7 \\
\alpha=-2 \beta
\end{array}\right.
\end{aligned}
$$

$$
\begin{aligned}
& \left\{\begin{array}{l}
\beta+2 \gamma=-1 \\
13 \gamma=0 \\
\alpha=-2 \beta
\end{array}\right. \\
& \left\{\begin{array}{l}
\beta+2 \gamma=-1 \\
\gamma=0 \\
\alpha=-2 \beta
\end{array}\right. \\
& \left\{\begin{array}{l}
\beta+2 \cdot 0=-1 \\
\gamma=0 \\
\alpha=-2 \beta
\end{array}\right. \\
& \left\{\begin{array}{l}
\beta=-1 \\
\gamma=0 \\
\alpha=-2 \beta
\end{array}\right. \\
& \begin{cases}\beta=-1 \\
\gamma=0 \\
\alpha=2\end{cases}
\end{aligned}
$$

The desired decomposition:

$x=2 p-q$

## Problem Kuznetsov Analytic Geometry $2-25$