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: 6


b=4; g=32; Two solutions are (8, 0) and (0, 4)

Work Step by Step

As a reminder, a system of equations that is singular means that it is impossible to eliminate without creating a pivot equal to 0. To do this, we first must determine l, or the number we are multiplying our first equation by to be able to eliminate the first term. Take our system: 2x+by=16 4x+8y=g If we were to eliminate 4x, we would have to multiply our first equation by 2. Therefore, to find the value to make it impossible to eliminate 4x without simultaneously eliminating 8y, we must divide 8y by 2, which leaves us with 4y. And if by=4y, then b=4. Now that we have a singular system of equations, there are two possibilities; the system will either have zero solutions or infinitely many solutions. Since we need two infinitely many solutions, notice that we can divide 2 from the second equation and make it similar to the first equation: 2x+4y=16 4x+8y=g 2x+4y=16 2x+4y=$\frac{g}{2}$ Now that we have similar equations, we can use substitution to solve for g. Substitute $\frac{g}{2}$ into the first equation $\frac{g}{2}$=16 g=32 Now we must find two of the infinite solutions. Choose any value for x or, and solve for the other variable's value algebraically. I chose y=0 for my first solution and x=0 for my second solution, because zero is an easy number to work with. 2x+0=16 2x=16 x=8 (8, 0) 0+4y=16 4y=16 y=4 (0, 4)
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