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Unit 14.3 Gaussian Elimination

Gaussian Elimination

Gaussian Elimination is another way to solve systems of equations. The goal of it is to get the matrix into Reduced Row Echelon Form (RREF)

[latex]\left[\begin{matrix}1&0\\0&1\\\end{matrix}\left|\begin{matrix}a\\b\\\end{matrix}\right.\right][/latex]

Where

1x + 0y = a

0x + 1y = b

We can accomplish this through a series of elimination steps

Gaussian Elimination

Given:

[latex]5x-20y=35[/latex]

[latex]-6x+28y=-62[/latex]

Solve the systems of equations using Gaussian Elimination

First build the matrix

[latex]\left[\begin{matrix}5&-20\\-6&28\\\end{matrix}\left|\begin{matrix}35\\-62\\\end{matrix}\right.\right][/latex] = [latex]\begin{matrix}R1C1&R1C2\\R2C1&R2C2\\\end{matrix}\left|\begin{matrix}a\\b\\\end{matrix}\right.[/latex]

Get R1C1 to equal 1 by multiplication. We are dealing with the first row ONLY. Since R1C1 = 5, we multiply by [latex]\frac{1}{5}[/latex]

[latex]\left[\begin{matrix}(5\ \cdot\ \frac{1}{5}\ )&(-20\ \cdot\ \frac{1}{5}\ )\\-6&28\\\end{matrix}\left|\begin{matrix}(35\ \cdot\frac{1}{5}\ )\\-62\\\end{matrix}\right.\right]=\left[\begin{matrix}1&-4\\-6&28\\\end{matrix}\left|\begin{matrix}7\\-62\\\end{matrix}\right.\right][/latex]

Next we get R2C1 to equal 0 by multiplication and addition. What value would we need to multiply R1C1 by to make R2C1 = 0 when added?

m1 * R1C1 + R2C1 = 0

m1 * 1 + -6 = 0

m1 = 6

Multiply the values in row 1 by 6 and add it to row 2

[latex]\left[\begin{matrix}1&-4\\\left(6\cdot1\right)+-6&\left(6\cdot-4\right)+28\\\end{matrix}\left|\begin{matrix}7\\\left(6\cdot7\right)+\ -62\\\end{matrix}\right.\right]=\left[\begin{matrix}1&-4\\0&4\\\end{matrix}\left|\begin{matrix}7\\-20\\\end{matrix}\right.\right][/latex]

Now we look at R2C2. We need that to equal 1. Looking at the second row only, what would you need to multiply 4 by so that it is equal to 1? We multiply by [latex]\frac{1}{4}[/latex]

[latex]\left[\begin{matrix}1&-4\\(0\cdot\frac{1}{4})&(4\cdot\frac{1}{4})\\\end{matrix}\left|\begin{matrix}7\\(-20\cdot\frac{1}{4})\\\end{matrix}\right.\right]=\left[\begin{matrix}1&-4\\0&1\\\end{matrix}\left|\begin{matrix}7\\-5\\\end{matrix}\right.\right][/latex]

Lastly, we get R1C2 to equal 0 by multiplication and addition. What value would we need to multiply R2C2 by to make R1C2 = 0 when added?

m2 * R2C2 + R1C2 = 0

m2 * 1 + -4 = 0

m2 = 4

Multiply the values in row 2 by 4 and add it to row 1

[latex]\left[\begin{matrix}\left(0\cdot4\right)+1&\left(1\cdot4\right)+-4\\0&1\\\end{matrix}\left|\begin{matrix}\left(-5\cdot4\right)+7\\-5\\\end{matrix}\right.\right]=\left[\begin{matrix}1&0\\0&1\\\end{matrix}\left|\begin{matrix}-13\\-5\\\end{matrix}\right.\right][/latex]

Now that we have it in RREF, we can see that x = -13 and y = -5

 

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