## Problem Statement

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

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

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

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

$r=\{0 ; 4 ; 1\}$
## 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:

$$
\left\{\begin{array}{l}
-2 \alpha+\beta=-5 \\
3 \beta+4 \gamma=-5 \\
\alpha-\beta+\gamma=5
\end{array}\right.
$$

Add the first row multiplied by -3 to the second row:

$$
\left\{\begin{array}{l}
-2 \alpha+\beta=-5 \\
6 \alpha+4 \gamma=10 \\
\alpha-\beta+\gamma=5
\end{array}\right.
$$

Add the first row to the third row:

$$
\left\{\begin{array}{l}
-2 \alpha+\beta=-5 \\
6 \alpha+4 \gamma=10 \\
-\alpha+\gamma=0
\end{array}\right.
$$

$$
\begin{aligned}
& \left\{\begin{array}{l}
-2 \alpha+\beta=-5 \\
6 \alpha+4 \gamma=10 \\
\gamma=\alpha
\end{array}\right. \\
& \left\{\begin{array}{l}
-2 \alpha+\beta=-5 \\
6 \alpha+4 \alpha=10 \\
\gamma=\alpha
\end{array}\right. \\
& \left\{\begin{array}{l}
-2 \alpha+\beta=-5 \\
\alpha=1 \\
\gamma=\alpha
\end{array}\right. \\
& \left\{\begin{array}{l}
-2 \cdot 1+\beta=-5 \\
\alpha=1 \\
\gamma=1
\end{array}\right. \\
& \left\{\begin{array}{l}
\beta=-3 \\
\alpha=1 \\
\gamma=1
\end{array}\right.
\end{aligned}
$$

The desired decomposition:

$x=p-3 q+r$

## Problem Kuznetsov Analytic Geometry 2-5