Answer
$x_{1}=30,\ x_{2}=20,\ x_{3}=-10$
is also a solution to the given
system of equations.
Work Step by Step
Let $A$ be the coefficient matrix of the given system of equations.
$\mathrm{x}=\left[\begin{array}{l}
3\\
2\\
-1
\end{array}\right]$ is a solution to $Ax=0$.
This means that $\mathrm{x} \in$ Nul A.
Nul A is a subspace of $\mathbb{R}^{3}$, so
(a) it contains 0,
(b) it is closed under addition, and
most importantly,
(c) it is closed under scalar multiplication.
$\left[\begin{array}{l}
30\\
20\\
-10
\end{array}\right]=10x\in$ Nul A, meaning A$\left[\begin{array}{l}
30\\
20\\
-10
\end{array}\right]$=0
So
$x_{1}=30,\ x_{2}=20,\ x_{3}=-10$
is also a solution to the given
system of equations.
