## Linear Algebra and Its Applications, 4th Edition

Published by Brooks Cole

# Chapter 1 - Section 1.3 - An Example of Gaussian Elimination - Problem Set - Page 15: 3

#### Answer

-$\frac{1}{2}$; the original solution is (5, 1); the new solution is (-5, -1)

#### Work Step by Step

Rewrite the system of equations as an augmented matrix: {2 -4 | 6 -1 5 | 0} Multiply the first row vector by -$\frac{1}{2}$, and use $R_{2}$-$R_{1}$=$R_{2'}$ to convert the system to a triangular matrix {-1 5 | 0} - {-1 2 | -3} = {0 3 | 3} Our new augmented matrix is now: {2 -4 | 6 0 3 | 3} Now that our augmented matrix is triangular, we can convert it back to a system of equations, and solve using back-substitution: 2x-4y=6 3y=3 y=1 Substituting y=1 into equation 1: 2x-4(1)=6 2x-4=6 2x=10 x=5 The solution is (5, 1) If the system of equations were changed to: 2x-4y=-6 -x+5y=0 we can solve again using Gaussian elimination. The values of our matrix will be the same, except for value (1,3), which instead of 6 will be -6, and value (2,3), which instead of 3 will be -3. Our new triangular system of equations will be: 2x-4y=-6 3y=-3 and we can once again solve using back-substitution. 3y=-3 y=-1 Substituting y=-1 into equation 1: 2x-4(-1)=-6 2x+4=-6 2x=-10 x=-5 The new solution is (-5, -1)

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