## Problem Statement

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

$x=\{-9 ; 5 ; 5\}$

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

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

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

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

Add the third row to the first:

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

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

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

Add the first row multiplied by 6 to the second row:

$$
\begin{aligned}
& \left\{\begin{array}{l}
5 \alpha-\beta=-4 \\
29 \alpha=-29 \\
\alpha-3 \beta+\gamma=5
\end{array}\right. \\
& \left\{\begin{array}{l}
5 \alpha-\beta=-4 \\
\alpha=-1 \\
\alpha-3 \beta+\gamma=5
\end{array}\right. \\
& \left\{\begin{array}{l}
5 \cdot(-1)-\beta=-4 \\
\alpha=-1 \\
-1-3 \beta+\gamma=5
\end{array}\right. \\
& \left\{\begin{array}{l}
\beta=-1 \\
\alpha=-1 \\
-1-3 \beta+\gamma=5
\end{array}\right. \\
& \left\{\begin{array}{l}
\beta=-1 \\
\alpha=-1 \\
-1-3 \cdot(-1)+\gamma=5
\end{array}\right. \\
& \left\{\begin{array}{l}
\beta=-1 \\
\alpha=-1 \\
\gamma=3
\end{array}\right.
\end{aligned}
$$

The desired decomposition:

$x=-p-q+3 r$

## Problem Kuznetsov Analytic Geometry $2-4$