## Task Condition

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

$x=\{15 ;-20 ;-1\}$

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

$q=\{0 ; 1 ;-1\}$

$r=\{5 ;-3 ; 2\}$
## 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:

$$
\left\{\begin{array}{l}
5 \gamma=15 \\
2 \alpha+\beta-3 \gamma=-20 \\
\alpha-\beta+2 \gamma=-1
\end{array}\right.
$$

Add the third row to the second:

$$
\begin{aligned}
& \left\{\begin{array}{l}
\gamma=3 \\
3 \alpha-\gamma=-21 \\
\alpha-\beta+2 \gamma=-1
\end{array}\right. \\
& \Rightarrow\left\{\begin{array}{l}
\gamma=3 \\
\alpha=-6 \\
\alpha-\beta=-7
\end{array}\right. \\
& \Rightarrow\left\{\begin{array}{l}
\gamma=3 \\
\alpha=-6 \\
\beta=1
\end{array}\right.
\end{aligned}
$$

The desired decomposition:

$x=-6 p+q+3 r$

## Problem Kuznetsov Analytic Geometry 2-31