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Solve the systems in Exercises $11-14$$$\begin{aligned} x_{1}-3 x_{2} &=5 \\-x_{1}+x_{2}+5 x_{3} &=2 \\ x_{2}+x_{3} &=0 \end{aligned}$$

$x 1=2$$x 2=-1$$x 3=1$

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 1

Systems of Linear Equations

Introduction to Matrices

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

Idaho State University

Lectures

01:32

In mathematics, the absolu…

01:11

07:13

Solve the systems in Exerc…

05:22

02:57

03:43

Determine the solution set…

01:48

Solve and check each equat…

07:56

09:30

Solve each system in Exerc…

01:06

Find all solutions to the …

04:09

Use $A^{-1}$ to find the s…

04:21

Solve the system of linear…

Okay, so this problem were given this system and were asked to solve it, we could do us using elementary row operations. So is the first step. Let's look at X one. We can see that if we add row, Want to row to weaken, Get a new road to with this X one right here. Eliminated. So again, that's row one Plus wrote to And that gives us a new row two of negative X two plus five X three equals seven and we still have our original first row and second row right here. Now we can look at, uh, ex too, and we can see that if we add two times Row three Ciro, too. We can get rid off this negative x two right here, and that gives us a new second row of seven. X three is equal to seven now. Using this, we can solve for X three by dividing by seven on both sides. And that gives us X three is equal to one. Now that we've self Rex three, we can plug that into Road three right here and gets X two is equal to negative one since negative one plus run is equal to zero, and now that would have X two. We can plug that into our first equation right here, and that's improvised to X. One plus three is equal to five, which gives us an answer of X one is equal to two. So the answer to the system would be too common. Negative one comma one.

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