## Problem Statement

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

$x=\{-5 ; 9 ;-13\}$

$p=\{0 ; 1 ;-2\}$

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

$r=\{4 ; 1 ; 0\}$
## 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:

$$
\left\{\begin{array}{l}
3 \beta+4 \gamma=-5 \\
\alpha-\beta+\gamma=9 \\
-2 \alpha+\beta=-13
\end{array}\right.
$$

Add the second row multiplied by -4 to the first row:

$$
\left\{\begin{array}{l}
-4 \alpha+7 \beta=-41 \\
\alpha-\beta+\gamma=9 \\
-2 \alpha+\beta=-13
\end{array}\right.
$$

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

$$
\left\{\begin{array}{l}
5 \beta=-15 \\
\alpha-\beta+\gamma=9 \\
-2 \alpha+\beta=-13
\end{array}\right.
$$

$$
\begin{aligned}
& \left\{\begin{array}{l}
\beta=-3 \\
\alpha-\beta+\gamma=9 \\
-2 \alpha+\beta=-13
\end{array}\right. \\
& \left\{\begin{array}{l}
\beta=-3 \\
\alpha-(-3)+\gamma=9 \\
-2 \alpha+(-3)=-13
\end{array}\right. \\
& \left\{\begin{array}{l}
\beta=-3 \\
\alpha+\gamma=6 \\
-2 \alpha=-10
\end{array}\right. \\
& \left\{\begin{array}{l}
\beta=-3 \\
\alpha+\gamma=6 \\
\alpha=5
\end{array}\right. \\
& \left\{\begin{array}{l}
\beta=-3 \\
5+\gamma=6 \\
\alpha=5
\end{array}\right. \\
& \left\{\begin{array}{l}
\beta=-3 \\
\gamma=1 \\
\alpha=5
\end{array}\right.
\end{aligned}
$$

The desired decomposition:

$x=5 p-3 q+r$

## Problem Kuznetsov Analytic Geometry $2-20$