## Problem Statement

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

$x=\{8 ;-7 ;-13\}$

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

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

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

$$
\left\{\begin{array}{l}
3 \beta-\gamma=8 \\
\alpha-\beta=-7 \\
5 \alpha+2 \beta+\gamma=-13
\end{array}\right.
$$

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

$$
\left\{\begin{array}{l}
3 \beta-\gamma=8 \\
\alpha-\beta=-7 \\
7 \beta+\gamma=22
\end{array}\right.
$$

Add the first row to the third row:

$$
\left\{\begin{array}{l}
3 \beta-\gamma=8 \\
\alpha-\beta=-7 \\
10 \beta=30
\end{array}\right.
$$

$$
\begin{aligned}
& \left\{\begin{array}{l}
3 \beta-\gamma=8 \\
\alpha-\beta=-7 \\
\beta=3
\end{array}\right. \\
& \left\{\begin{array}{l}
3 \cdot 3-\gamma=8 \\
\alpha-3=-7 \\
\beta=3
\end{array}\right. \\
& \left\{\begin{array}{l}
\gamma=1 \\
\alpha=-4 \\
\beta=3
\end{array}\right.
\end{aligned}
$$

The desired decomposition:

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

## Problem Kuznetsov Analytic Geometry $2-29$