## Problem Statement

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

$x=\{13 ; 2 ; 7\}$

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

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

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

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

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

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

Add the third row to the first row:

$$
\left\{\begin{array}{l}
10 \beta=10 \\
\alpha-\beta=2 \\
3 \beta-\gamma=7
\end{array}\right.
$$

$$
\begin{aligned}
& \left\{\begin{array}{l}
\beta=1 \\
\alpha-\beta=2 \\
3 \beta-\gamma=7
\end{array}\right. \\
& \left\{\begin{array}{l}
\beta=1 \\
\alpha-1=2 \\
3 \cdot 1-\gamma=7
\end{array}\right. \\
& \left\{\begin{array}{l}
\beta=1 \\
\alpha=3 \\
\gamma=-4
\end{array}\right.
\end{aligned}
$$

The desired decomposition:

$x=3 p+q-4 r$

## Problem Kuznetsov Analytic Geometry 2-6