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: 1

Answer

$(2,-1)$

Work Step by Step

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