## Problem Statement

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

$x=\{-13 ; 2 ; 18\}$

$p=\{1 ; 1 ; 4\}$

$q=\{-3 ; 0 ; 2\}$

$r=\{1 ; 2 ;-1\}$
## 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:

$$
\left\{\begin{array}{l}
\alpha-3 \beta+\gamma=-13 \\
\alpha+2 \gamma=2 \\
4 \alpha+2 \beta-\gamma=18
\end{array}\right.
$$

Add the first row multiplied by -2 to the second row:

$$
\left\{\begin{array}{l}
\alpha-3 \beta+\gamma=-13 \\
-\alpha+6 \beta=28 \\
4 \alpha+2 \beta-\gamma=18
\end{array}\right.
$$

Add the first row to the third row:

$$
\left\{\begin{array}{l}
\alpha-3 \beta+\gamma=-13 \\
-\alpha+6 \beta=28 \\
5 \alpha-\beta=5
\end{array}\right.
$$

Add the third row multiplied by 6 to the second row:

$$
\begin{aligned}
& \left\{\begin{array}{l}
\alpha-3 \beta+\gamma=-13 \\
29 \alpha=58 \\
5 \alpha-\beta=5
\end{array}\right. \\
& \left\{\begin{array}{l}
\alpha-3 \beta+\gamma=-13 \\
\alpha=2 \\
5 \alpha-\beta=5
\end{array}\right. \\
& \left\{\begin{array}{l}
2-3 \beta+\gamma=-13 \\
\alpha=2 \\
5 \cdot 2-\beta=5
\end{array}\right. \\
& \left\{\begin{array}{l}
-3 \beta+\gamma=-15 \\
\alpha=2 \\
\beta=5
\end{array}\right. \\
& \left\{\begin{array}{l}
-3 \cdot 5+\gamma=-15 \\
\alpha=2 \\
\beta=5
\end{array}\right. \\
& \left\{\begin{array}{l}
\gamma=0 \\
\alpha=2 \\
\beta=5
\end{array}\right.
\end{aligned}
$$

The desired decomposition:

$x=2 p+5 q$

## Problem Kuznetsov Analytic Geometry $2-27$