Question

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}&-{1}&{0}&-{2}&{0}&{0}{0}&{0}&{1}&{2}&{0}&{0}{0}&{0}&{0}&{0}&{1}&{0}end{matrix}right]}

Forms of linear equations
ANSWERED
asked 2020-11-01
The given matrix is the augmented matrix for a system of linear equations. Give the vector form for the general solution.
\(\displaystyle{\left[\begin{matrix}{1}&-{1}&{0}&-{2}&{0}&{0}\\{0}&{0}&{1}&{2}&{0}&{0}\\{0}&{0}&{0}&{0}&{1}&{0}\end{matrix}\right]}\)

Answers (1)

2020-11-02
The given matrix is
\(\displaystyle{B}={\left[\begin{matrix}{1}&-{1}&{0}&-{2}&{0}&{0}\\{0}&{0}&{1}&{2}&{0}&{0}\\{0}&{0}&{0}&{0}&{1}&{0}\end{matrix}\right]}\)
Reduce the given augmented matrix B in to system of linear equation
Ax=b
The matrix form of the first equation is
\(\displaystyle{\left[\begin{matrix}{1}&-{1}&{0}&-{2}&{0}\\{0}&{0}&{1}&{2}&{0}\\{0}&{0}&{0}&{0}&{1}\end{matrix}\right]}\cdot{\left[\begin{matrix}{x}_{{1}}\\{x}_{{2}}\\{x}_{{3}}\\{x}_{{4}}\\{x}_{{5}}\end{matrix}\right]}={\left[\begin{matrix}{0}\\{0}\\{0}\end{matrix}\right]}\)
where
\(\displaystyle{A}={\left[\begin{matrix}{1}&-{1}&{0}&-{2}&{0}\\{0}&{0}&{1}&{2}&{0}\\{0}&{0}&{0}&{0}&{1}\end{matrix}\right]},{x}={\left[\begin{matrix}{x}_{{1}}\\{x}_{{2}}\\{x}_{{3}}\\{x}_{{4}}\\{x}_{{5}}\end{matrix}\right]},{b}={\left[\begin{matrix}{0}\\{0}\\{0}\end{matrix}\right]}\)
We use another equation to find a. general solution
\(x_1-x_2-2x_4=0\)
\(x_3+2x_4=0\)
\(x_5=0\)
Then
\(\displaystyle{\left({3}\right)}\Rightarrow{x}_{{1}}={x}_{{2}}+{2}{x}_{{4}}\)
\(\displaystyle{\left({4}\right)}\Rightarrow{x}_{{3}}=-{2}{x}_{{4}}\)
\(\displaystyle{\left({5}\right)}\Rightarrow{x}_{{3}}={0}\)
In vectors form , the general solution , we obtain
\(\displaystyle{x}={\left[\begin{matrix}{x}_{{1}}\\{x}_{{2}}\\{x}_{{3}}\\{x}_{{4}}\\{x}_{{5}}\end{matrix}\right]}={\left[\begin{matrix}{x}_{{2}}+{2}{x}_{{4}}\\{x}_{{2}}\\{0}\\-{2}{x}_{{4}}\\{0}\end{matrix}\right]}={x}_{{2}}{\left[\begin{matrix}{1}\\{1}\\{0}\\{0}\\{0}\end{matrix}\right]}+{x}_{{4}}{\left[\begin{matrix}{2}\\{0}\\-{2}\\{1}\\{0}\end{matrix}\right]}\)
Hence,
\(\displaystyle{x}={x}_{{2}}{\left[\begin{matrix}{1}\\{1}\\{0}\\{0}\\{0}\end{matrix}\right]}+{x}_{{4}}{\left[\begin{matrix}{2}\\{0}\\-{2}\\{1}\\{0}\end{matrix}\right]}\)
Answer \(\displaystyle{x}={x}_{{2}}{\left[\begin{matrix}{1}\\{1}\\{0}\\{0}\\{0}\end{matrix}\right]}+{x}_{{4}}{\left[\begin{matrix}{2}\\{0}\\-{2}\\{1}\\{0}\end{matrix}\right]}\)
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