Elementary Linear Algebra 7th Edition

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

Chapter 2 - Matrices - 2.1 Operations with Matrices - 2.1 Exercises - Page 49: 53

Answer

$$A=\left[\begin{array}{cc}{-5} &{2} \\{3}& {-1} \end{array}\right].$$

Work Step by Step

Assume that the matrix $A$ is given by $$A=\left[\begin{array}{cc}{x_1} &{x_2} \\{x_3}& {x_4} \end{array}\right],$$ then we have the system $$ \begin{aligned} x_{1}+2x_{3}&=1 \\ x_{2} +2x_4 &=0\\ 3x_1+5x_{3} &=0\\ 3x_{2} +5x_4 &=1 \end{aligned}. $$ The augmented matrix is given by $$ \left[ \begin {array}{ccccc} 1&0&2&0&1\\ 0&1&0&2&0 \\ 3&0&5&0&0\\ 0&3&0&5&1 \end {array} \right] . $$ Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows $$\left[ \begin {array}{ccccc} 1&0&0&0&-5\\ 0&1&0&0&2 \\ 0&0&1&0&3\\ 0&0&0&1&-1 \end {array} \right] .$$ From which the solution is $$x_1=-5, \quad x_2=2, \quad x_3=3, \quad x_4=-1 .$$ Hence, we have $$A=\left[\begin{array}{cc}{-5} &{2} \\{3}& {-1} \end{array}\right].$$
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