Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 0-32198-238-X
ISBN 13: 978-0-32198-238-4

Chapter 1 - Linear Equations in Linear Algebra - 1.2 Exercises: 10


$\begin{cases}x_1=-4+2x_2 \\ x_2 \text{ is free}\\ x_3=-7 \end{cases}$

Work Step by Step

To find the general solution, we use row operations on the augmented matrix $\left[ \begin{array}{cccc} 1 & -2 & -1 & 3\\ 3 & -6 & -2 & 2 \end{array} \right]$ to obtain an equivalent matrix in reduced row echelon form. Then we solve the corresponding system of equations. First, we want to eliminate the 3 in the second row since it is below the leading term of the first row. To do this, we use row-replacement. So we replace row 2 with -3*(row 1)+(row 2), which gives $$\left[ \begin{array}{cccc} 1 & -2 & -1 & 3\\ 0 & 0 & 1 & -7 \end{array} \right].$$ Next, we need to eliminate the -1 above the leading one of the second row. We will again use row-replacement. We replace row 1 with (row 2) + (row 1) to get $$\left[ \begin{array}{cccc} 1 & -2 & 0 & -4\\ 0 & 0 & 1 & -7 \end{array} \right].$$ Now we solve the corresponding system of equations, which is $$x_1-2x_2=-4\\ x_3=-7.$$ From the reduced row echelon form of the matrix, we see that $x_1$ and $x_3$ are the leading variables (since they correspond to the pivot positions of the matrix) and $x_2$ is free. So we want to solve for $x_1$ in terms of $x_2$. Thus our system becomes $$x_1=-4+2x_2\\ x_3=-7, $$ and, hence our general solution is $$\begin{cases}x_1=-4+2x_2 \\ x_2 \text{ is free}\\ x_3=-7 \end{cases}.$$
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