## Problem Statement

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

$x=\{-15 ; 5 ; 6\}$

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

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

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

$$
\left\{\begin{array}{l}
3 \beta-\gamma=-15 \\
5 \alpha+2 \beta+\gamma=5 \\
\alpha-\beta=6
\end{array}\right.
$$

Add the third row multiplied by -5 to the second row:

$$
\left\{\begin{array}{l}
3 \beta-\gamma=-15 \\
7 \beta+\gamma=-25 \\
\alpha-\beta=6
\end{array}\right.
$$

Add the first row to the second row:

$$
\left\{\begin{array}{l}
3 \beta-\gamma=-15 \\
10 \beta=-40 \\
\alpha-\beta=6
\end{array}\right.
$$

$$
\begin{aligned}
& \left\{\begin{array}{l}
3 \beta-\gamma=-15 \\
\beta=-4 \\
\alpha-\beta=6
\end{array}\right. \\
& \left\{\begin{array}{l}
3 \cdot(-4)-\gamma=-15 \\
\beta=-4 \\
\alpha-(-4)=6
\end{array}\right. \\
& \left\{\begin{array}{l}
-\gamma=-3 \\
\beta=-4 \\
\alpha=2
\end{array}\right. \\
& \left\{\begin{array}{l}
\gamma=3 \\
\beta=-4 \\
\alpha=2
\end{array}\right.
\end{aligned}
$$

The desired decomposition:

$x=2 p-4 q+3 r$

## Problem Kuznetsov Analytic Geometry $2-21$