Question

Construct three different augmented matrices for linear systems whose solution set is $x=1, y=0, z=-2$. There are many different correct answers.

   Construct three different augmented matrices for linear systems whose solution set is $x=1, y=0, z=-2$. There are many different correct answers.
 
Precalculus: A Right Triangle Approach
Precalculus: A Right Triangle Approach
Ratti, McWaters,… 5th Edition
Chapter 9, Problem 100 ↓
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Construct three different augmented matrices for linear systems whose solution set is $x=1, y=0, z=-2$. There are many different correct answers.
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Transcript

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00:01 So in this problem, we're asked to construct three matrices whose solution sets is x1 is equal to negative 2, x2 is equal to 1, and x3 is equal to 0.
00:11 Let's begin with a simple matrix whose solution sets is obviously going to be negative 2, 1, 0.
00:19 So for that, we're going to have our first row solve for x1, which is going to be negative 2.
00:25 Then our second row is going to solve for x2 and that's one and then our last row is going to solve for x3 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 to get two other matrices from this matrix that we have right here, we can simply perform any two row operations that we want on this matrix.
01:08 So for our second matrix, let's choose to add two times row one to row two.
01:19 That's going to give us a new row two of two, one, zero, nine...
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