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In Exercises $1-4,$ determine if the system has a nontrivial solution. Try to use as few row operations as possible. $\begin{aligned}-5 x_{1}+7 x_{2}+9 x_{3} &=0 \\ x_{1}-2 x_{2}+6 x_{3} &=0 \end{aligned}$

The variable $x_{3}$ is free; the system has nontrivial solutions.

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 5

Solution Sets of Linear Systems

Introduction to Matrices

McMaster University

Harvey Mudd College

University of Michigan - Ann Arbor

Idaho State University

Lectures

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All right, so in this problem were given to eke wishes. So that's it. They are negative. Five x one seven x two plus nine x three He was sterile and the second equation is x one minus two, x two class six X x three and it's one that is zero again. We need to apply the gushing eliminations here. So we're set down the, uh, corporations for each question that is Echo 579 and one elected to six. We first interchange interchanged the first rule in the second row. So it will be one elective to six connected 579 And we at the throwing a rock at the second row by five times the first roll. So it will be Excuse me. So that will be one active to six on the first roll and, uh, naked fire plus five times one that is zero and seven plus neck of two times five. That is 10. Minus that it's seven minus 10. So it's a negative. Three or end. Ah, nine plus five times six. That is 30 30 plus night. So that's Ah, 39 and we can simplify the second roll so that is one nectar to six and zero one and 13 connective 30. All right, so right now we have x one connective two, x two plus six Next three, which is zero and x two minus 13 x three We just do so you can see that X two is dependent on X three and X one is dependent on X to an x three. So that means X one is also dependent on X three. That means X three. It's a free bearable. And since we have a free bearable here, so that means there are there. There exists a nontrivial solution to this equation, so we're done.

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