## Problem Statement

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

$x=\{2 ; 7 ; 5\}$

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

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

$r=\{0 ; 3 ; 1\}$
## 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:

$$
\left\{\begin{array}{l}
\alpha+\beta=2 \\
-2 \beta+3 \gamma=7 \\
\alpha+\gamma=5
\end{array}\right.
$$

Add the first row multiplied by 2 to the second row:

$$
\left\{\begin{array}{l}
\alpha+\beta=2 \\
2 \alpha+3 \gamma=11 \\
\alpha+\gamma=5
\end{array}\right.
$$

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

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

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

The desired decomposition:

$x=4 p-2 q+r$

## Problem Kuznetsov Analytic Geometry 2-30