## Problem Statement

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

$x=\{8 ; 1 ; 12\}$

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

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

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

$$
\left\{\begin{array}{l}
\alpha+3 \beta-\gamma=8 \\
2 \alpha+\gamma=1 \\
-\alpha+2 \beta+\gamma=12
\end{array}\right.
$$

Add the first row to the third:

$$
\begin{aligned}
& \left\{\begin{array}{l}
\alpha+3 \beta-\gamma=8 \\
2 \alpha+\gamma=1 \\
5 \beta=20
\end{array}\right. \\
& \left\{\begin{array}{l}
\alpha+3 \beta-\gamma=8 \\
2 \alpha+\gamma=1 \\
\beta=4
\end{array}\right. \\
& \left\{\begin{array}{l}
\alpha+3 \cdot 4-\gamma=8 \\
2 \alpha+\gamma=1 \\
\beta=4
\end{array}\right.
\end{aligned}
$$

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

Add the first row to the second:

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

The desired decomposition:

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

## Problem Kuznetsov Analytic Geometry 2-18