## Task Condition

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

$x=\{6 ; 12 ;-1\}$

$p=\{1 ; 3 ; 0\}$

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

$r=\{0 ;-1 ; 2\}$
## 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:

$$
\left\{\begin{array}{l}
\alpha+2 \beta=6 \\
3 \alpha-\beta-\gamma=12 \\
\beta+2 \gamma=-1
\end{array}\right.
$$

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

$$
\left\{\begin{array}{l}
\alpha+2 \beta=6 \\
-7 \beta-\gamma=-6 \\
\beta+2 \gamma=-1
\end{array}\right.
$$

Add the third row multiplied by 7 to the second row:

$$
\left\{\begin{array}{l}
\alpha+2 \beta=6 \\
13 \gamma=-13 \\
\beta+2 \gamma=-1
\end{array}\right.
$$

$$
\begin{aligned}
& \left\{\begin{array}{l}
\alpha+2 \beta=6 \\
\gamma=-1 \\
\beta+2 \gamma=-1
\end{array}\right. \\
& \left\{\begin{array}{l}
\alpha+2 \beta=6 \\
\gamma=-1 \\
\beta+2 \cdot(-1)=-1
\end{array}\right. \\
& \left\{\begin{array}{l}
\alpha+2 \beta=6 \\
\gamma=-1 \\
\beta=1
\end{array}\right. \\
& \left\{\begin{array}{l}
\alpha+2 \cdot 1=6 \\
\gamma=-1 \\
\beta=1
\end{array}\right. \\
& \left\{\begin{array}{l}
\alpha=4 \\
\gamma=-1 \\
\beta=1
\end{array}\right.
\end{aligned}
$$

The desired decomposition:

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

## Problem Kuznetsov Analytic Geometry $2-2$