## problem statement

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

$x=\{3 ;-3 ; 4\}$

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

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

$r=\{2 ;-1 ; 4\}$
## 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:

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

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

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

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

The desired decomposition:

$x=p-2 q+r$

## Problem Kuznetsov Analytical Geometry $2-8$