3.1 Substitution and Gaussian Elimination
To solve a system of linear equations by substitution, we start by making one variable the subject of the formula in one of the equations. Then we substitute for this variable in one of the remaining equations. We keep substituting until the fewest number of variables possible remain. The solution can then be obtained.
Example 3.1.1. Use the substitution method to solve the system of linear equations. \begin {align*} -x+y+z &=1\\x-y+z &=2\\x+y-z &=3\\ \end {align*}
Solution. From the first equation, \(x=y+z-1\) substitute for \(x\) in the second equation. \(y+z-1-y+z=2\) or \(2z=3\); Thus \(z=3/2\). From the third equation, we obtain \(x=-y+9/2\) and from \(x=y+z-1\), we have \(x=y+1/2\). Thus \(x=5/2\) and \(y=2\).
The next example illustrates the Gaussian elimination method for solving systems of linear equations.
Example 3.1.2. Use the method of Gaussian elimination to solve the system of linear equations. \begin {align*} x-2y+z &-1\\3x+y-4z&=3\\-2x+4y+3z&=-2\\ \end {align*}
Solution. Multiply the first equation by 2 and add to the third equation \(5z=-4\) of \(z=-4/5\). Therefore the system reduces to \begin {align*} z&=-4/5\\ 3x+y-4z&=3\\-2x+4y+3z &=-2 \end {align*}
Multiply the second equation by 2 and the third equation by 3 then add \begin {align*} z&=-4/5\\14y+z&=0\\ -2x+4y+3z&=2\\\\ \implies \hspace {0.4cm}(x,y,z)&=\Big (\frac {7}{2},\frac {3}{70},\frac {-4}{5}\Big )\\ \end {align*}
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