## Problem Statement

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

$x=\{-1 ; 7 ;-4\}$

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

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

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

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

Add the third row to the first:

$$
\begin{aligned}
& \left\{\begin{array}{l}
5 \beta=-5 \\
2 \alpha+\gamma=7 \\
\alpha+3 \beta-\gamma=-4
\end{array}\right. \\
& \left\{\begin{array}{l}
\beta=-1 \\
2 \alpha+\gamma=7 \\
\alpha+3 \beta-\gamma=-4
\end{array}\right. \\
& \left\{\begin{array}{l}
\beta=-1 \\
2 \alpha+\gamma=7 \\
\alpha+3 \cdot(-1)-\gamma=-4
\end{array}\right.
\end{aligned}
$$

$$
\left\{\begin{array}{l}
\beta=-1 \\
2 \alpha+\gamma=7 \\
\alpha-\gamma=-1
\end{array}\right.
$$

Add the third row to the second:

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

The desired decomposition:

$x=2 p-q+3 r$

Problem Kuznetsov Analytic Geometry 2-10