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: 4

Answer

l=$\frac{c}{a}$; 0x + y(d-$\frac{bc}{a}$)=(g-$\frac{cf}{a}$); y=$\frac{ag-cf}{ad-bc}$

Work Step by Step

Start by creating a matrix from the given system of equations. Our goal is to create a triangular system of equations so we can solve for y. {a b | f c d | g} By multiplying the first row vector by $\frac{c}{a}$, and using the formula $R_{2}$-$R_{1}$=$R_{2'}$ {c d | g} - {c $\frac{bc}{a}$ | $\frac{cf}{a}$} = {0 (d-$\frac{bc}{a}$) | (g-$\frac{cf}{a}$)} Our new augmented matrix will be: {a b | f 0 (d-$\frac{bc}{a}$) | (g-$\frac{cf}{a}$)} Now that we have a triangular matrix, we can convert it back into a system of equations. Our new system of equations will be: ax+by=f y(d-$\frac{bc}{a}$)=(g-$\frac{cf}{a}$) We can now solve for y by dividing both sides of the second equation by (d-$\frac{bc}{a}$): y=(g-$\frac{cf}{a}$)$\div($d-$\frac{bc}{a}$) y=($\frac{ag-cf}{a}$)$\div$($\frac{ad-bc}{a}$) y=($\frac{ag-cf}{a}$)$\times$($\frac{a}{ad-bc}$) y=$\frac{ag-cf}{ad-bc}$
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