## Task Condition

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

$x=\{6 ; 5 ;-14\}$

$p=\{1 ; 1 ; 4\}$

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

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

$$
\left\{\begin{array}{l}
\alpha+2 \gamma=6 \\
\alpha-3 \beta+\gamma=5 \\
4 \alpha+2 \beta-\gamma=-14
\end{array}\right.
$$

Subtract the first row from the second:

$$
\left\{\begin{array}{l}
\alpha+2 \gamma=6 \\
-3 \beta-\gamma=-1 \\
4 \alpha+2 \beta-\gamma=-14
\end{array}\right.
$$

Add the first row multiplied by -4 to the third row:

$$
\left\{\begin{array}{l}
\alpha+2 \gamma=6 \\
-3 \beta-\gamma=-1 \\
2 \beta-9 \gamma=-38
\end{array}\right.
$$

Add the second row multiplied by -9 to the third row:

$$
\begin{aligned}
& \left\{\begin{array}{l}
\alpha+2 \gamma=6 \\
-3 \beta-\gamma=-1 \\
29 \beta=-29
\end{array}\right. \\
& \left\{\begin{array}{l}
\alpha+2 \gamma=6 \\
-3 \beta-\gamma=-1 \\
\beta=-1
\end{array}\right. \\
& \left\{\begin{array}{l}
\alpha+2 \gamma=6 \\
-3 \cdot(-1)-\gamma=-1 \\
\beta=-1
\end{array}\right. \\
& \left\{\begin{array}{l}
\alpha+2 \gamma=6 \\
\gamma=4 \\
\beta=-1
\end{array}\right. \\
& \left\{\begin{array}{l}
\alpha+2 \cdot 4=6 \\
\gamma=4 \\
\beta=-1
\end{array}\right. \\
& \left\{\begin{array}{l}
\alpha=-2 \\
\gamma=4 \\
\beta=-1
\end{array}\right.
\end{aligned}
$$

The desired decomposition:

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

## Problem Kuznetsov Analytic Geometry 2-11