Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 032198238X
ISBN 13: 978-0-32198-238-4

Chapter 2 - Matrix Algebra - 2.5 Exercises - Page 131: 2

Answer

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Work Step by Step

\[ L=\left[\begin{array}{ccc} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 2 & 0 & 1 \end{array}\right] \] Lower Triangular matrix \[ U=\left[\begin{array}{ccc} 4 & 3 & -5 \\ 0 & -2 & 2 \\ 0 & 0 & 2 \end{array}\right] \] Upper Triangular Matrix \[ \mathbf{b}=\left[\begin{array}{c} 2 \\ -4 \\ 6 \end{array}\right] \] Given b for $\mathbf{A x}=\mathbf{b}$ \[ L \quad \mathbf{b}]=\left[\begin{array}{cccc} 1 & 0 & 0 & 2 \\ -1 & 1 & 0 & -4 \\ 2 & 0 & 1 & 6 \end{array}\right]=\left[\begin{array}{cccc} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & 2 \end{array}\right] \] \[ \begin{array}{c} y=\left[\begin{array}{c} 2 \\ -2 \\ 2 \end{array}\right] \\ {\left[\begin{array}{cc} U \quad \mathrm{y} \end{array}\right]=\left[\begin{array}{cccc} 4 & 3 & -5 & 2 \\ 0 & -2 & 2 & -2 \\ 0 & 0 & 2 & 2 \end{array}\right]=\left[\begin{array}{cccc} 4 & 3 & -5 & 2 \\ 0 & -2 & 2 & -2 \\ 0 & 0 & 1 & 1 \end{array}\right]} \end{array} \] Next solve $U x=y,$ using back-substitution \[ \left[\begin{array}{ll} U & \mathbf{y} \end{array}\right]=\left[\begin{array}{cccc} 4 & 3 & 0 & 7 \\ 0 & -2 & 0 & -2 \\ 0 & 0 & 1 & 1 \end{array}\right]=\left[\begin{array}{cccc} 4 & 3 & 0 & 7 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 1 \end{array}\right]=\left[\begin{array}{cccc} 1 & 0 & 0 & 1 / 4 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 1 \end{array}\right] \] Reducing to echelon form \[ \mathbf{x}=\left[\begin{array}{c} 1 / 4 \\ 2 \\ 1 \end{array}\right] \]
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