Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 10 - Section 10.3 - Matrices and Systems of Linear Equations - 10.3 Exercises - Page 711: 63

Answer

$(\frac{7}{4}-\frac{7}{4}t, -\frac{7}{4}+\frac{3}{4}t, \frac{9}{4}+\frac{3}{4}t, t)$

Work Step by Step

Step 1. Establish the augmented matrix of the system and use the Gauss Eliminations method: $\begin{vmatrix} 1 & 0 & 1 & 1 & 4 \\0 & 1 & -1 & 0 & -4\\1 & -2 & 3 & 1 & 12 \\2 & 0 & -2 & 5 & -1 \end{vmatrix} \begin{array}(\\. \\R_1-R_3\to R_3\\2R_1-R_4\to R_4\\ \end{array}$ Step 2. Do the operations given on the right side of the matrix. $\begin{vmatrix} 1 & 0 & 1 & 1 & 4 \\0 & 1 & -1 & 0 & -4\\0 & 2 & -2 & 0 & -8 \\0 & 0 & 4 & -3 & 9 \end{vmatrix} \begin{array}(\\. \\2R_2-R_3\to R_3\\.\\ \end{array}$ Step 3. Do the operations given on the right side of the matrix. $\begin{vmatrix} 1 & 0 & 1 & 1 & 4 \\0 & 1 & -1 & 0 & -4\\0 & 0 & 0 & 0 & 0 \\0 & 0 & 4 & -3 & 9 \end{vmatrix} \begin{array}(\\. \\.\\.\\ \end{array}$ Step 4. Let $w=t$, row 4 gives $z=\frac{3}{4}t+\frac{9}{4}$. Back substitute to row 2, we have $y-z=-4$ and $y=\frac{3}{4}t-\frac{7}{4}$. And row 1 gives $x+z+t=4$ and $x=-\frac{7}{4}t+\frac{7}{4}$ Step 5. We write the final solutions as $(\frac{7}{4}-\frac{7}{4}t, -\frac{7}{4}+\frac{3}{4}t, \frac{9}{4}+\frac{3}{4}t, t)$
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