## Task Condition

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

$x=\{8 ; 9 ; 4\}$

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

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

$r=\{1 ; 3 ; 0\}$
## 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:

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

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

$$
\left\{\begin{array}{l}
\alpha+\gamma=8 \\
-3 \alpha-2 \beta=-15 \\
\alpha+\beta=4
\end{array}\right.
$$

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

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
\alpha+\gamma=8 \\
\beta=-3 \\
\alpha+-3=4
\end{array}\right. \\
& \left\{\begin{array}{l}
\alpha+\gamma=8 \\
\beta=-3 \\
\alpha=7
\end{array}\right. \\
& \left\{\begin{array}{l}
7+\gamma=8 \\
\beta=-3 \\
\alpha=7
\end{array}\right. \\
& \left\{\begin{array}{l}
\gamma=1 \\
\beta=-3 \\
\alpha=7
\end{array}\right.
\end{aligned}
$$

The desired decomposition is:

$$
x=7 p-3 q+r
$$

## Problem Kuznetsov Analytic Geometry $2-22$