## Task Condition

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

$x=\{5 ; 15 ; 0\}$

$p=\{1 ; 0 ; 5\}$

$q=\{-1 ; 3 ; 2\}$

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

$$
\left\{\begin{array}{l}
\alpha-\beta=5 \\
3 \beta-\gamma=15 \\
5 \alpha+2 \beta+\gamma=0
\end{array}\right.
$$

Add the second row to the third:

$$
\begin{aligned}
& \left\{\begin{array}{l}
\alpha-\beta=5 \\
3 \beta-\gamma=15 \\
5 \alpha+5 \beta=15
\end{array}\right. \\
& \left\{\begin{array}{l}
\alpha-\beta=5 \\
3 \beta-\gamma=15 \\
\alpha+\beta=3
\end{array}\right.
\end{aligned}
$$

Add the first row to the third:

$$
\begin{aligned}
& \left\{\begin{array}{l}
\alpha-\beta=5 \\
3 \beta-\gamma=15 \\
2 \alpha=8
\end{array}\right. \\
& \left\{\begin{array}{l}
\alpha-\beta=5 \\
3 \beta-\gamma=15 \\
\alpha=4
\end{array}\right. \\
& \left\{\begin{array}{l}
4-\beta=5 \\
3 \beta-\gamma=15 \\
\alpha=4
\end{array}\right. \\
& \left\{\begin{array}{l}
\beta=-1 \\
3 \beta-\gamma=15 \\
\alpha=4
\end{array}\right. \\
& \left\{\begin{array}{l}
\beta=-1 \\
3 \cdot(-1)-\gamma=15 \\
\alpha=4
\end{array}\right. \\
& \left\{\begin{array}{l}
\beta=-1 \\
\gamma=-18 \\
\alpha=4
\end{array}\right.
\end{aligned}
$$

The desired decomposition:

$x=4 p-q-18 r$

## Problem Kuznetsov Analytic Geometry 2-13