Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 0-32198-238-X
ISBN 13: 978-0-32198-238-4

Chapter 1 - Linear Equations in Linear Algebra - 1.1 Exercises: 14

Answer

$x_{1} = 2$ $x_{2} = -1$ $x_{3} = 1$

Work Step by Step

The procedure is shown with the matrix notation for simpler understanding. $x_{1} - 3x_{2} = 5 $ $-x_{1} + x_{2} + 5x_{3} = 2$ $x_{2} + x_{3} = 0$ This can be depicted in the augmented matrix notation as follows : $\begin{bmatrix} 1 & -3 & 0 & 5\\ -1 & 1 & 5 & 2\\ 0 & 1 & 1 & 0 \end{bmatrix}$ To eliminate the $-1x_{1}$ term in the second equation, add $1$ times row 1 to row 2: $\begin{bmatrix} 1 & -3 & 0 & 5\\ 0 & -2 & 5 & 7\\ 0 & 1 & 1 & 0 \end{bmatrix}$ To get an $x_{2}$ term in the second row, interchange row 2 and row 3: $\begin{bmatrix} 1 & -3 & 0 & 5\\ 0 & 1 & 1 & 0\\ 0 & -2 & 5 & 7 \end{bmatrix}$ Next we use the $x_{2}$ term in the second equation to eliminate the $-2x_{2}$ term from the third equation. Add $2$ times row 2 to row 3: $\begin{bmatrix} 1 & -3 & 0 & 5\\ 0 & 1 & 1 & 0\\ 0 & 0 & 7 & 7 \end{bmatrix}$ Now the augmented matrix is in a triangular form. For interpreting it, we go back to the equation notation: $x_{1} - 3x_{2} = 5$ $x_{2} + x_{3} = 0$ $7x_{3} = 7$ From $7x_{3} = 7$, we get $x_{3} = 1$ Using this value in $x_{2} + x_{3} = 0$, we get: $x_{2} + 1 = 0$ or, $x_{2} = -1$ Using the value of $x_{2}$ in the equation $x_{1} - 3x_{2} = 5$, $x_{1} - 3(-1) = 5$ or, $x_{1} = 2$ Hence, $x_{1} = 2$ ; $x_{2} = -1$ ; $x_{3} = 1$ .
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