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: 34

Answer

$T_{1}=20$ $T_{2}=27.5$ $T_{3}=30$ $T_{4}=22.5$

Work Step by Step

$4T_{1}-T_{2}-T_{4}=30$ $4T_{2}-T_{1}-T_{3}=60$ $4T_{3}-T_{2}-T_{4}=70$ $4T_{4}-T_{1}-T_{3}=40$ Rewrite system as an augmented matrix: \begin{bmatrix} 4&-1&0&-1&30\\ -1&4&-1&0&60\\ 0&-1&4&-1&70\\ -1&0&-1&4&40\\ \end{bmatrix} Swap R1 and R4 \begin{bmatrix} -1&0&-1&4&40\\ -1&4&-1&0&60\\ 0&-1&4&-1&70\\ 4&-1&0&-1&30\\ \end{bmatrix} R2=R2-R1 R4=R4+4R1 \begin{bmatrix} -1&0&-1&4&40\\ 0&4&0&-4&20\\ 0&-1&4&-1&70\\ 0&-1&-4&15&190\\ \end{bmatrix} R2=$\frac{1}{4}$R2 \begin{bmatrix} -1&0&-1&4&40\\ 0&1&0&-1&5\\ 0&-1&4&-1&70\\ 0&-1&-4&15&190\\ \end{bmatrix} R3=R3+R2 R4=R4+R2 \begin{bmatrix} -1&0&-1&4&40\\ 0&1&0&-1&5\\ 0&0&4&-2&75\\ 0&0&-4&14&195\\ \end{bmatrix} R4=R4+R3 \begin{bmatrix} -1&0&-1&4&40\\ 0&1&0&-1&5\\ 0&0&4&-2&75\\ 0&0&0&12&270\\ \end{bmatrix} R4=$\frac{1}{12}$R4 \begin{bmatrix} -1&0&-1&4&40\\ 0&1&0&-1&5\\ 0&0&4&-2&75\\ 0&0&0&1&22.5\\ \end{bmatrix} R3=R3+2R4 R2=R2+R4 R1=R1-4R4 \begin{bmatrix} -1&0&-1&0&-50\\ 0&1&0&0&27.5\\ 0&0&4&0&120\\ 0&0&0&1&22.5\\ \end{bmatrix} R3=$\frac{1}{4}$R3 R1=R1+R3 \begin{bmatrix} -1&0&0&0&-20\\ 0&1&0&0&27.5\\ 0&0&1&0&30\\ 0&0&0&1&22.5\\ \end{bmatrix} R1=-R1 \begin{bmatrix} 1&0&0&0&20\\ 0&1&0&0&27.5\\ 0&0&1&0&30\\ 0&0&0&1&22.5\\ \end{bmatrix} Solution: $T_{1}=20$ $T_{2}=27.5$ $T_{3}=30$ $T_{4}=22.5$
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