## Problem Statement

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

$x=\{0 ; -8 ; 9\}$

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

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

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

$$
\left\{\begin{array}{l}
3 \beta+4 \gamma=0 \\
-2 \alpha+\beta=-8 \\
\alpha-\beta+\gamma=9
\end{array}\right.
$$

Add the third row multiplied by 2 to the second row:

$$
\left\{\begin{array}{l}
3 \beta+4 \gamma=0 \\
-\beta+2 \gamma=10 \\
\alpha-\beta+\gamma=9
\end{array}\right.
$$

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

$$
\left\{\begin{array}{l}
10 \gamma=30 \\
-\beta+2 \gamma=10 \\
\alpha-\beta+\gamma=9
\end{array}\right.
$$

$$
\begin{aligned}
& \left\{\begin{array}{l}
\gamma=3 \\
-\beta+2 \gamma=10 \\
\alpha-\beta+\gamma=9
\end{array}\right. \\
& \left\{\begin{array}{l}
\gamma=3 \\
-\beta+2 \cdot 3=10 \\
\alpha-\beta+3=9
\end{array}\right. \\
& \left\{\begin{array}{l}
\gamma=3 \\
\beta=-4 \\
\alpha-\beta=6
\end{array}\right. \\
& \left\{\begin{array}{l}
\gamma=3 \\
\beta=-4 \\
\alpha-(-4)=6
\end{array}\right. \\
& \left\{\begin{array}{l}
\gamma=3 \\
\beta=-4 \\
\alpha=2
\end{array}\right.
\end{aligned}
$$

The desired decomposition:

$x=2 p-4 q+3 r$

## Problem Kuznetsov Analytic Geometry $2-28$