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: 3

Answer

$(-4,2)$

Work Step by Step

To solve the system $\begin{cases}x+3y=2 \\ x+2y=0 \\ \end{cases},$ we perform elementary row operations on the corresponding augmented matrix to obtain an equivalent matrix with $1s$ along the main diagonal. The corresponding augmented matrix is $$\left[ \begin{array}{cc|c} 1 & 3 & 2 \\ 1 & 2 & 0\\ \end{array} \right].$$ Next, we replace Row_2 with Row_2-Row_1 to obtain the equivalent matrix $$\left[ \begin{array}{cc|c} 1 & 3 & 2 \\ 0 & -1 & -2\\ \end{array} \right].$$ Next, we multiply Row_2 by $-1$ to obtain the equaivalent matrix $$\left[ \begin{array}{cc|c} 1 & 3 & 2 \\ 0 & 1 & 2\\ \end{array} \right].$$ Now, we form the system of equations corresponding to this matrix $$\begin{cases}x+3y=2 \\ y=2 \\ \end{cases}.$$ We see $y=2$, and to find our value for $x$, we plug $y=2$ into the first equation and solve it for $x$. Thus $$x+3(2)=2 \\ x+6=2\\ x=-4.$$ Thus we have exactly one solution, which is $(-4,2).$
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