## Linear Algebra and Its Applications, 4th Edition

Published by Brooks Cole

# Chapter 1 - Section 1.3 - An Example of Gaussian Elimination - Problem Set: 10

#### Answer

Pivots: 2, 1, 8; Solution: (2, 1, 1)

#### Work Step by Step

Start by converting your system of equations to an augmented matrix: [2 3 1 | 8 4 7 5 | 20 0 -2 2 | 0] Using Gaussian-elimination, create a triangular matrix: $R_{2}$-(2$\times$$R_{1})=R_{2'} [4 7 5 | 20]-[4 6 2 | 16] = [0 1 3 | 4] Our new augmentred matrix is: [2 3 1 | 8 0 1 3 | 4 0 -2 2 | 0] R_{3}-((-2)\times$$R_{2}$)=$R_{3'}$ [0 -2 2 | 0]-[0 -2 -6 | -8]=[0 0 8 | 8] We now have a triangular matrix, [2 3 1 | 8 0 1 3 | 4 0 0 8 | 8 We can convert this matrix back to a system of equations: 2x+3y+z=8 y+3z=4 8z=8 Now that we have a triangular system of equations, we can identify our pivot points (the first non-zero coefficient in the individual equations) as 2, 1, and 8. We can now also solve this system using back-substitution: 8z=8 z=1 y+3(1)=4 y+3=4 y=1 2x+3(1)+(1)=8 2x+3+1=8 2x+4=8 2x=4 x=2 Our solution is: (2, 1, 1)

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