## Problem Statement

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

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

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

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

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

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

Add the second row multiplied by -3 to the first row:

$$
\left\{\begin{array}{l}
-7 \beta-\gamma=-6 \\
\alpha+2 \beta=3 \\
\beta+2 \gamma=-1
\end{array}\right.
$$

Add the third row multiplied by 7 to the first row:

$$
\left\{\begin{array}{l}
13 \gamma=-13 \\
\alpha+2 \beta=3 \\
\beta+2 \gamma=-1
\end{array}\right.
$$

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

The desired decomposition:

$x=p+q-r$

## Problem Kuznetsov Analytic Geometry $2-9$