## Problem Statement

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

$x=\{11 ; 5 ;-3\}$

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

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

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
\alpha-\beta+2 \gamma=11 \\
5 \gamma=5 \\
2 \alpha+\beta-3 \gamma=-3
\end{array}\right. \\
& \left\{\begin{array}{l}
\alpha-\beta+2 \gamma=11 \\
\gamma=1 \\
2 \alpha+\beta-3 \gamma=-3
\end{array}\right. \\
& \left\{\begin{array}{l}
\alpha-\beta+2 \cdot 1=11 \\
\gamma=1 \\
2 \alpha+\beta-3 \cdot 1=-3
\end{array}\right. \\
& \left\{\begin{array}{l}
\alpha-\beta=9 \\
\gamma=1 \\
2 \alpha+\beta=0
\end{array}\right.
\end{aligned}
$$

Add the first row to the third:

$$
\begin{aligned}
& \left\{\begin{array}{l}
\alpha-\beta=9 \\
\gamma=1 \\
3 \alpha=9
\end{array}\right. \\
& \left\{\begin{array}{l}
\alpha-\beta=9 \\
\gamma=1 \\
\alpha=3
\end{array}\right. \\
& \left\{\begin{array}{l}
3-\beta=9 \\
\gamma=1 \\
\alpha=3
\end{array}\right. \\
& \left\{\begin{array}{l}
\beta=-6 \\
\gamma=1 \\
\alpha=3
\end{array}\right.
\end{aligned}
$$

The desired decomposition:

$x=3 p-6 q+r$

## Problem Kuznetsov Analytic Geometry 2-15