Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 4 - Section 4.4 - Solving Systems of Equations by Matrices - Exercise Set - Page 239: 11

Answer

$(1,-2,3)$

Work Step by Step

To solve the system $\begin{cases} \hspace{5mm} 2y-z=-7 \\ x+4y+z=-4 \\ 5x-y+2z=13 \\ \end{cases},$ we perform elementary row operations on the corresponding augmented matrix to obtain an equivalent matrix with $1s$ along the main diagonal (if possible). The corresponding augmented matrix is $$\left[ \begin{array}{ccc|c} 0 & 2 & -1 & -7 \\ 1 & 4 & 1 & -4\\ 5 & -1 & 2 & 13\\ \end{array} \right].$$ First, we switch Row_1 and Row_2 to obtain the equivalent matrix $$\left[ \begin{array}{ccc|c} 1 & 4 & 1 & -4 \\ 0 & 2 & -1 & -7\\ 5 & -1 & 2 & 13\\ \end{array} \right].$$ Next, we replace Row_3 with Row_3-5Row_1 to obtain the equaivalent matrix $$\left[ \begin{array}{ccc|c} 1 & 4 & 1 & -4 \\ 0 & 2 & -1 & -7\\ 0 & -21 & -3 & 33\\ \end{array} \right].$$ Thirdly, we multiply Row_2 by $\dfrac{1}{2}$ and multiply Row_3 by $-\dfrac{1}{3}$ to obtain the equaivalent matrix $$\left[ \begin{array}{ccc|c} 1 & 4 & 1 & -4 \\ 0 & 1 & -\frac{1}{2} & -\frac{7}{2}\\ 0 & 7 & 1 & 11\\ \end{array} \right].$$ Fourthly, we replace Row_3 with Row_3-7Row_2 to obtain the equaivalent matrix $$\left[ \begin{array}{ccc|c} 1 & 4 & 1 & -4 \\ 0 & 1 & -\frac{1}{2} & -\frac{7}{2}\\ 0 & 0 & \frac{9}{2} & \frac{27}{2}\\ \end{array} \right].$$ Fifthly, we multiply Row_3 by $\dfrac{2}{9}$ to obtain the equivalent matrix $$\left[ \begin{array}{ccc|c} 1 & 4 & 1 & -4 \\ 0 & 1 & -\frac{1}{2} & -\frac{7}{2}\\ 0 & 0 & 1 & 3\\ \end{array} \right].$$ Now, we form the system of equations corresponding to this matrix: $$\begin{cases} x+4y+z=-4 \\ \hspace{5mm} y-\dfrac{1}{2}z=-\dfrac{7}{2} \\ \hspace{15mm} z=3 \\ \end{cases}.$$ We see $z=3$. To find $y$, we plug $z=3$ into our second equation and solve for $y$: $$y-\dfrac{1}{2}(3)=-\dfrac{7}{2} \\ y-\dfrac{3}{2}=-\dfrac{7}{2} \\ y=-\dfrac{4}{2} \\ y=-2.$$ To find $x$, we plug $y=-2$ and $z=3$ into the first equation and solve for $x$: $$x+4(-2)+3=-4 \\ x-8+3=-4 \\ x-5=-4 \\ x=1.$$ Thus we have exactly one solution, which is $(1,-2,3).$
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.