Use Gaussian elimination to find the complete solution to the system of given equations, or show that none exists. {(5x+8y-6z=14),(3x+4y-2z=8),(x+2y-2z=3):}

Use Gaussian elimination to find the complete solution to the system of given equations, or show that none exists. {(5x+8y-6z=14),(3x+4y-2z=8),(x+2y-2z=3):}

Question
Equations
asked 2020-11-07
Use Gaussian elimination to find the complete solution to the system of given equations, or show that none exists.
\(\displaystyle{\left\lbrace\begin{array}{c} {5}{x}+{8}{y}-{6}{z}={14}\\{3}{x}+{4}{y}-{2}{z}={8}\\{x}+{2}{y}-{2}{z}={3}\end{array}\right.}\)

Answers (1)

2020-11-08
Step 1
Consider the following system of linear equations:
5x+8y-6z=14
3x+4y-2z=8
x+2y-2z=3
Convert above system of linear equations into augmented matrix form:
\(\displaystyle{\left[\begin{array}{ccc|c} {5}&{8}&-{6}&{14}\\{3}&{4}&-{2}&{8}\\{1}&{2}&-{2}&{3}\end{array}\right]}\)
Step 2
Transform the above matrix into reduced row echelon form:
\(\displaystyle{R}_{{2}}\rightarrow\frac{{5}}{{3}}{R}_{{2}}-{R}_{{1}}\)
\(\displaystyle{R}_{{3}}\rightarrow{5}{R}_{{3}}-{R}_{{1}}\)
\(\displaystyle{\left[\begin{array}{ccc|c} {5}&{8}&-{6}&{14}\\{0}&-\frac{{4}}{{3}}&\frac{{8}}{{3}}&-\frac{{2}}{{3}}\\{0}&{2}&-{4}&{1}\end{array}\right]}\)
\(\displaystyle{R}_{{3}}\rightarrow\frac{{2}}{{3}}{R}_{{3}}+{R}_{{2}}\)
\(\displaystyle{\left[\begin{array}{ccc|c} {5}&{8}&-{6}&{14}\\{0}&-\frac{{4}}{{3}}&\frac{{8}}{{3}}&-\frac{{2}}{{3}}\\{0}&{0}&{0}&{0}\end{array}\right]}\)
Step 3
Hence, the solution of a system of linear equations does not exist.
0

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