# The given matrix is the augmented matrix for a system of linear equations. Give the vector form for the general solution. displaystyle{left[begin{matrix}{1}&{0}&-{1}&-{2}{0}&{1}&{2}&{3}end{matrix}right]}

The given matrix is the augmented matrix for a system of linear equations. Give the vector form for the general solution.
$\left[\begin{array}{cccc}1& 0& -1& -2\\ 0& 1& 2& 3\end{array}\right]$
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Theodore Schwartz
The given matrix is
$B=\left[\begin{array}{cccc}1& 0& -1& -2\\ 0& 1& 2& 3\end{array}\right]$
Reduce the given augmented matrix B in to system of linear equation
Ax=b
The matrix form of the first equation is
$\left[\begin{array}{ccc}1& 0& -1\\ 0& 1& 2\end{array}\right]\cdot \left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right]=\left[\begin{array}{c}-2\\ 3\end{array}\right]$
where
$A=\left[\begin{array}{ccc}1& 0& -1\\ 0& 1& 2\end{array}\right],x=\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right],b=\left[\begin{array}{c}-2\\ 3\end{array}\right]$
We use another equation to find a. general solution
${x}_{1}-{x}_{3}=-2\left(3\right)$
${x}_{2}+2{x}_{3}=3\left(4\right)$
${x}_{3}={x}_{3}\left(5\right)$
Then
$\left(3\right)⇒{x}_{1}=-2+{x}_{3}$
$\left(4\right)⇒{x}_{2}=3-2{x}_{3}$
$\left(5\right)⇒{x}_{3}={x}_{3}$
In vectors form, the general solution, we obtain
$x=\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right]=\left[\begin{array}{c}-2+{x}_{3}\\ 3-2{x}_{3}\\ {x}_{3}\end{array}\right]=\left[\begin{array}{c}-2\\ 3\\ 0\end{array}\right]+{x}_{3}\left[\begin{array}{c}1\\ -2\\ 1\end{array}\right]$
answer $x=\left[\begin{array}{c}-2\\ 3\\ 0\end{array}\right]+{x}_{3}\left[\begin{array}{c}1\\ -2\\ 1\end{array}\right]$