Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 1 - Systems of Linear Equations - 1.2 Gaussian Elimination and Gauss-Jordan Elimination - 1.2 Exercises - Page 24: 68


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Work Step by Step

The difference between the reduced row-echelon form and reduced echelon form is that; by using the reduced row-echelon form one can get the solution of the corresponding system directly. But by using the reduced echelon form, one has to use the back substitution to get the solution of the corresponding system. For example, the following augmented matrix in the reduced row-echelon form $$\left[ \begin{array} {cc} 1&0&1\\0&1&-2 \end{array} \right].$$ The corresponding system is \begin{align*} x&=1\\ y&=-2.\\ \end{align*} Which gives the solution $x=1, y=-2$. But, the following augmented matrix in the row-echelon form $$\left[ \begin{array} {cc} 1&2&1\\0&3&3 \end{array} \right].$$ The corresponding system is \begin{align*} x+2y&=1\\ 3y&=3.\\ \end{align*} Using the back substitution we obtain the solution $x=-1, y=1$.
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