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Construct three different augmented matrices for linear systems whose solution set is $x_{1}=-2, x_{2}=1, x_{3}=0$.

Perform different elementary row operations to obtain augmented matrices.

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 1

Systems of Linear Equations

Introduction to Matrices

McMaster University

Harvey Mudd College

Baylor University

Lectures

01:32

In mathematics, the absolu…

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04:30

Construct $3 \times 2$ mat…

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Construct the augmented ma…

00:35

00:58

Write the augmented matrix…

01:08

The Augmented Matrix Write…

03:54

Use matrices to solve each…

01:22

02:36

Apply matrix algebra to so…

00:55

As systems of linear equat…

04:44

Solve each system using ma…

So this problem we're asked to construct three matrices Who solution sets is X one is equal to negative two X two is equal to one, and x three is equal to zero. Let's begin with the simple matrix whose solution sets is obviously gonna be negative 210 So for that, we're gonna have our first row cell for X one. Just gonna be negative too. Then our second row is gonna solve Rex too. And that's one. And then our last row is gonna solve for X three, which is zero. So here we have our variables going in a diagonal, and here we have their answers in the last column. Now, the simple principle of this question is that row operations do not change the solution set of a system. So she gets to other major cities from this matrix that we have right here, we can simply perform any to grow operations that we want on this matrix. So, for a second matrix, let's choose to add two times Rah Juan to wrote to That's going to give us a new road, too, of two one zero negative three. So this is our second matrix that's gonna have a solution. Now let's perform any cooperation that we want on this matrix right here. Let's choose to add Row two to Row three. That's gonna give us our original first heroes and then a new row three of two one one negative three. So here we have three matrices that have this'll solution sets, and we get that by performing rob orations on our original first matrix.

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