Linear Algebra and Its Applications, 4th Edition

Published by Brooks Cole
ISBN 10: 0030105676
ISBN 13: 978-0-03010-567-8

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


l=5 Pivots: 2,-6

Work Step by Step

Multiply by l = $\frac{10}{2}$ = 5, and subtract to find 2x + 3y = 1 and -6y = 6. Pivots 2, -6. Our goal should be to form a triangular system of equations, such that in equation 1 all of our coefficients are non-zero, and in equation two our first coefficient is zero and second coefficient is non-zero. Start with your given system of equations: 2x+3y=1 10x+9y=11 Rewrite the equations as an augmented matrix: {2 3 | 1} {10 9 | 11} Multiply the first row vector of the matrix by 5 (the value of l), and subtract the first row vector from the second row vector to create a new value for the second row vector of your matrix. ($R_{2}$-$R_{1}$-->$R_{2}$) {10 9 | 11} - {10 15 | 5} = {0 -6 | 6} Our new matrix is now: {2 3 | 1} {0 -6 | 6} Which means our system of linear equations is: 2x+3y=1 (0x)-6y=6 This system of linear equations is now triangular and has pivot points. As a reminder, a pivot point is the first non-zero entry in a row when dealing with elimination. This would make our pivot coefficients 2 and -6.
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