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. displaystyle{left[begin{matrix}{1}&{0}&-{1}&-{2}{0}&{1}&{2}&{3}end{matrix}right]}

Question
Forms of linear equations
asked 2020-11-08
The given matrix is the augmented matrix for a system of linear equations. Give the vector form for the general solution.
\(\displaystyle{\left[\begin{matrix}{1}&{0}&-{1}&-{2}\\{0}&{1}&{2}&{3}\end{matrix}\right]}\)

Answers (1)

2020-11-09
The given matrix is
\(\displaystyle{B}={\left[\begin{matrix}{1}&{0}&-{1}&-{2}\\{0}&{1}&{2}&{3}\end{matrix}\right]}\)
Reduce the given augmented matrix B in to system of linear equation
Ax=b
The matrix form of the first equation is
\(\displaystyle{\left[\begin{matrix}{1}&{0}&-{1}\\{0}&{1}&{2}\end{matrix}\right]}\cdot{\left[\begin{matrix}{x}_{{1}}\\{x}_{{2}}\\{x}_{{3}}\end{matrix}\right]}={\left[\begin{matrix}-{2}\\{3}\end{matrix}\right]}\)
where
\(\displaystyle{A}={\left[\begin{matrix}{1}&{0}&-{1}\\{0}&{1}&{2}\end{matrix}\right]},{x}={\left[\begin{matrix}{x}_{{1}}\\{x}_{{2}}\\{x}_{{3}}\end{matrix}\right]},{b}={\left[\begin{matrix}-{2}\\{3}\end{matrix}\right]}\)
We use another equation to find a. general solution
\(x_1-x_3=-2 (3)\)
\(x_2+2x_3=3 (4)\)
\(x_3=x_3 (5)\)
Then
\(\displaystyle{\left({3}\right)}\Rightarrow{x}_{{1}}=-{2}+{x}_{{3}}\)
\(\displaystyle{\left({4}\right)}\Rightarrow{x}_{{2}}={3}-{2}{x}_{{3}}\)
\(\displaystyle{\left({5}\right)}\Rightarrow{x}_{{3}}={x}_{{3}}\)
In vectors form, the general solution, we obtain
\(\displaystyle{x}={\left[\begin{matrix}{x}_{{1}}\\{x}_{{2}}\\{x}_{{3}}\end{matrix}\right]}={\left[\begin{matrix}-{2}+{x}_{{3}}\\{3}-{2}{x}_{{3}}\\{x}_{{3}}\end{matrix}\right]}={\left[\begin{matrix}-{2}\\{3}\\{0}\end{matrix}\right]}+{x}_{{3}}{\left[\begin{matrix}{1}\\-{2}\\{1}\end{matrix}\right]}\)
answer \(\displaystyle{x}={\left[\begin{matrix}-{2}\\{3}\\{0}\end{matrix}\right]}+{x}_{{3}}{\left[\begin{matrix}{1}\\-{2}\\{1}\end{matrix}\right]}\)
0

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