Elementary Linear Algebra 7th Edition

Published by Cengage Learning

Chapter 1 - Systems of Linear Equations - 1.2 Gaussian Elimination and Gauss-Jordan Elimination - 1.2 Exercises: 14

Answer

There are infinitely many solutions to this system and they are $$x_1=1-2t\qquad x_2 = t\qquad x_3=-1$$

Work Step by Step

This matrix is associated with the following system: \begin{align*} x_1+2x_2+x_3=&0\\ x_3=&-1\\ \end{align*} There is no third equation since the third row only gives the identity $0=0$ This is a system of two equations with three variables so it will have infinitely many solutions. To find them we have to set one of them to be equal to parameter $t$ that can take any real value. Note that we cannot set $x_3 = t$ because we have a precise condition $x_3=-1$, so we will put $x_2 = t.$ Now we get \begin{align*} x_1+2x_2+x_3=&0\\ x_2\qquad=&t\\ x_3=&-1\\ \end{align*} Putting all of this into the firs equation we have $$x_1+2t-1=0$$ and this gives $$x_1=1-2t.$$

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