Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 3 - Section 3.4 - Matrix Solutions to Linear Systems - Exercise Set - Page 229: 39

Answer

$\Rightarrow \left\{\begin{matrix} w -x +y +z =\;3 \\ \; \; \;\;\;x -2 y - z=\;0 \\ \;\; \;\;\; \;\;\;\;\;\; y + 6z =17\\ \;\; \;\;\; \;\;\;\;\;\;\;\;\;\;\;\; z=\; 3 \end{matrix}\right.$ $\Rightarrow \{(2,1,-1,3)\}$.

Work Step by Step

The given augmented matrix is $\Rightarrow \left[\begin{array}{cccc|c} 1 & -1 & 1&1& 3\\ 0&1&-2&-1& 0 \\ 0 & 0& 1& 6&17\\ 0&0&0&1&3 \end{array}\right]$ The above matrix is the reduced row-echelon form. Convert the matrix into a system of equations as shown below. $\Rightarrow \left\{\begin{matrix} 1w -1x +1y +1z =\;3 \\ 0w +1x -2 y -1 z=\;0 \\ 0w +0 x +1 y + 6z =17\\ 0 w +0 x +0 y +1 z=\; 3 \end{matrix}\right.$ or we can write. $\Rightarrow \left\{\begin{matrix} w -x +y +z =\;3 \\ \; \; \;\;\;x -2 y - z=\;0 \\ \;\; \;\;\; \;\;\;\;\;\; y + 6z =17\\ \;\; \;\;\; \;\;\;\;\;\;\;\;\;\;\;\; z=\; 3 \end{matrix}\right.$ Substitute back the value of $z$ into the third equation. $\Rightarrow y+6(3)=17$ Simplify. $\Rightarrow y+18=17$ Solve for $y$. $\Rightarrow y=-1$. Substitute back the value of $z$ and $y$ into the second equation. $\Rightarrow x -2 (-1) - (3)=0$ Simplify. $\Rightarrow x +2 - 3=0$ Solve for $x$. $\Rightarrow x=1$. Substitute back the value of $z,y$ and $x$ into the first equation. $\Rightarrow w -(1) +(-1) +(3) =3$ Simplify. $\Rightarrow w -1 -1 +3 =3$ Solve for $x$. $\Rightarrow w=2$. The solution set is $\{(w,x,y,z)\}=\{(2,1,-1,3)\}$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.