## Problem Statement

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

$x=\{11 ;-1 ; 4\}$

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

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

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

$$
\left\{\begin{array}{l}
\alpha+3 \beta-\gamma=11 \\
-\alpha+2 \beta+\gamma=-1 \\
2 \alpha+\gamma=4
\end{array}\right.
$$

Add the first row to the second:

$$
\begin{aligned}
& \left\{\begin{array}{l}
\alpha+3 \beta-\gamma=11 \\
5 \beta=10 \\
2 \alpha+\gamma=4
\end{array}\right. \\
& \left\{\begin{array}{l}
\alpha+3 \beta-\gamma=11 \\
\beta=2 \\
2 \alpha+\gamma=4
\end{array}\right. \\
& \left\{\begin{array}{l}
\alpha+3 \cdot 2-\gamma=11 \\
\beta=2 \\
2 \alpha+\gamma=4
\end{array}\right.
\end{aligned}
$$

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

Add the first row to the third:

$$
\begin{aligned}
& \left\{\begin{array}{l}
\alpha-\gamma=5 \\
\beta=2 \\
3 \alpha=9
\end{array}\right. \\
& \left\{\begin{array}{l}
\alpha-\gamma=5 \\
\beta=2 \\
\alpha=3
\end{array}\right. \\
& \left\{\begin{array}{l}
3-\gamma=5 \\
\beta=2 \\
\alpha=3
\end{array}\right. \\
& \left\{\begin{array}{l}
\gamma=-2 \\
\beta=2 \\
\alpha=3
\end{array}\right.
\end{aligned}
$$

The desired decomposition:

$x=3 p+2 q-2 r$

## Problem Kuznetsov Analytic Geometry $2-26$