Solve Row-Echelon Form: 3x -y + 2z = -4

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The discussion revolves around solving a system of equations using Row-Echelon Form, with the original equations provided. The user expresses familiarity with elimination and substitution methods but seeks guidance on applying Row-Echelon techniques. An augmented matrix is established, and the process of transforming it into Row-Echelon Form is outlined, including necessary operations like row interchanges and scalar multiplication. However, a critical error is pointed out regarding the proposed solutions, indicating they do not satisfy all equations in the system. The conversation emphasizes the importance of accuracy in both the method and the solutions derived.
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Homework Statement


NOTE: This is not a homework question, rather it's a question I made up to practice using Row-Echelon. I know how to solve it easily using elimination or substitution, and by using Cramer's Rule (but my instructor suggests I rather not since it's too much work :P) The solutions are x = -1, y= 5, and z= 2

3x -y + 2z = -4
-4x +y -5z = -1
x +4y -z = 9

Can anyone give me an example of how to solve it using Row-Echelon Form?

Homework Equations





The Attempt at a Solution


I'm sorry! I'd try it myself, but I don't understand it at all, and I can't seem to study it without a textbook. Again, I know how to solve it easily by using Elimination and Substitution.
 
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Well, first we'll need to put your system of equations into an augmented matrix, like so:

<br /> \left[ <br /> \begin{array}{ccc|c} <br /> 3&amp;-1&amp;2&amp;-4\\ <br /> -4&amp;1&amp;-5&amp;-1\\ <br /> 1&amp;4&amp;-1&amp;9 <br /> \end{array} <br /> \right] <br />

Before we go any father, do you see where I got that?
 
By setting up the x's on the first column, the y's on the second column, the z's on the third column, and the ='s (which is how my instructor names them) on the fourth column?
 
Yep, pretty much. Now, you can work with this matrix until it is in row echelon form using certain rules, which I believe are as follows:

You can add any scalar multiple of a line to another line.
You can interchange two lines.
You can multiply any line by a nonzero scalar.

Now, first thing you'll want to do is get the top left entry to be 1, and then use that to zero both of the other left entries. What's the best way you can see to do that?
 
By multiplying the first row by -4, and the second one by -3?
 
Actually, if I were you, my first step would be to interchange rows 1 and 3, to get that 1 in the top left. Then you just need to add multiples of the first row to the second and third rows to zero out everything else in the left column.
 
I'm sorry. I'll try analyzing it within a few hours, because I'm late for a meeting. I do apologize. :(
 
Oh, no worries. It's past midnight for me here, so I'll probably go to bed anyway. Cheers!
 
Bogrune said:

Homework Statement


NOTE: This is not a homework question, rather it's a question I made up to practice using Row-Echelon. I know how to solve it easily using elimination or substitution, and by using Cramer's Rule (but my instructor suggests I rather not since it's too much work :P) The solutions are x = -1, y= 5, and z= 2

3x -y + 2z = -4
-4x +y -5z = -1
x +4y -z = 9

Can anyone give me an example of how to solve it using Row-Echelon Form?

Homework Equations





The Attempt at a Solution


I'm sorry! I'd try it myself, but I don't understand it at all, and I can't seem to study it without a textbook. Again, I know how to solve it easily by using Elimination and Substitution.

When you are doing elimination or substitution, you are essentially getting the row-echelon form, although without explicitly using matrices: you are doing so-called Gaussian elimination.

Your instructor is right when he/she says not to use Cramer's rule. This is rarely used to solve real-world sized problems unless they have special structure that makes determinants easy to compute. It does, however, have its theoretical uses.

RGV
 
  • #10
You are going to get some really messy numbers because x = -1, y= 5, and z= 2 are not solutions to those equations. In particular (-1)+ 4(5)- (2)= 17, not 9,.
 
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