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: 10

Answer

$(1,2,4)$

Work Step by Step

To solve the system $\begin{cases}5x \hspace{15mm}=5 \\ 2x+y \hspace{5mm}=4 \\ 3x+y-5z=-15 \\ \end{cases},$ 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 $$\left[ \begin{array}{ccc|c} 5 & 0 & 0 & 5 \\ 2 & 1 & 0 & 4\\ 3 & 1 & -5 & -15\\ \end{array} \right].$$ First, we multiply Row_1 by $\dfrac{1}{5}$ to obtain the equivalent matrix $$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 1 \\ 2 & 1 & 0 & 4\\ 3 & 1 & -5 & -15\\ \end{array} \right].$$ Next, we replace Row_2 with Row_2-2Row_1 to obtain the equivalent matrix $$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2\\ 3 & 1 & -5 & -15\\ \end{array} \right].$$ Thirdly, we replace Row_3 with Row_3-3Row_1 to obtain the equivalent matrix $$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2\\ 0 & 1 & -5 & -18\\ \end{array} \right].$$ Fourthly, we replace Row_3 with Row_3-Row_2 to obtain the equivalent matrix $$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2\\ 0 & 0 & -5 & -20\\ \end{array} \right].$$ Fifthly, we multiply Row_3 by $-\dfrac{1}{5}$ to obtain the equivalent matrix $$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2\\ 0 & 0 & 1 & 4\\ \end{array} \right].$$ Now, we form the system of equations corresponding to this matrix: $$\begin{cases}x \hspace{15mm} =1 \\ \hspace{5mm} y \hspace{9mm}=2 \\ \hspace{10mm} z=4 \\ \end{cases}.$$ We see $x=1$, $y=2$, and $z=4$. Thus we have exactly one solution, which is $(1,2,4).$
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