## Problem Statement

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

$x=\{6 ;-1 ; 7\}$

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

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

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

$$
\left\{\begin{array}{l}
\alpha-\beta+\gamma=6 \\
-2 \alpha+\beta=-1 \\
3 \beta+4 \gamma=7
\end{array}\right.
$$

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

$$
\left\{\begin{array}{l}
\alpha-\beta+\gamma=6 \\
-2 \alpha+\beta=-1 \\
-4 \alpha+7 \beta=-17
\end{array}\right.
$$

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

$$
\left\{\begin{array}{l}
\alpha-\beta+\gamma=6 \\
-2 \alpha+\beta=-1 \\
5 \beta=-15
\end{array}\right.
$$

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

The desired decomposition:

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

## Problem Kuznetsov Analytic Geometry 2-12