Solving a 3x3 System of Equations

[Amaz1ng]

Homework Statement

1 1 1 | 2
0 5 2 | 12
0 -4 8 | 0


Answer:

100 | -3
010 | 2
001 | 1

Homework Equations

The Attempt at a Solution

How did they get the answer? I went over how to do 2x2 w/ someone but 3x3 seems different.

Basically this is how I'm thinking to do it:

Find position 1 which is 1 so don't do anything.
Go to position 2 which is 0, so good.
Go to three 0, so good.
Go to position 4, which is 5. Multiply the entire row with that 5 by the reciprocol of 5 (1/5).
Go to position 5 which is -4, needs to be 0. Multiply the first row by 4 then add that result to the row with the 4 in it. So basically: 4(Row1) + Row2 -> New Row 2

Then pretty much repeat with 8, then 2, then 1, then 1...remembering that

1. To get a 1, multiply by the reciprocal.
2. To get a 0 in row 2 in a position that has a 3, for example: -3(row 1) + row 2 => row 2


I'm not getting the right answer though. D:

 

[DivisionByZro]

I would begin by adding 5*R₃ to 4*R₂. Then switch R₂ and R₃. Add R₃ to R₂. Can you take it from there?

<br /> \left[ <br /> \begin{array}{ccc|c} <br /> 1 &amp; 1 &amp; 1 &amp; 2\\ <br /> 0 &amp; 5 &amp; 2 &amp; 12\\ <br /> 0 &amp; -4 &amp; 8 &amp; 0\\ <br /> \end{array} <br /> \right]<br /> →<br /> \left[ <br /> \begin{array}{ccc|c} <br /> 1 &amp; 1 &amp; 1 &amp; 2\\ <br /> 0 &amp; 20 &amp; 8 &amp; 48\\ <br /> 0 &amp; -20 &amp; 40 &amp; 0\\ <br /> \end{array} <br /> \right]<br /> →<br /> \left[ <br /> \begin{array}{ccc|c} <br /> 1 &amp; 1 &amp; 1 &amp; 2\\ <br /> 0 &amp; 0 &amp; 48 &amp; 48\\ <br /> 0 &amp; 1 &amp; -2 &amp; 0\\ <br /> \end{array} <br /> \right]<br /> →<br /> \left[ <br /> \begin{array}{ccc|c} <br /> 1 &amp; 1 &amp; 1 &amp; 2\\ <br /> 0 &amp; 0 &amp; 1 &amp; 1\\ <br /> 0 &amp; 1 &amp; -2 &amp; 0\\ <br /> \end{array} <br /> \right]<br /> <br />

Edit: I fixed it.

Go to position 4, which is 5. Multiply the entire row with that 5 by the reciprocol of 5 (1/5).
Go to position 5 which is -4, needs to be 0. Multiply the first row by 4 then add that result to the row with the 4 in it. So basically: 4(Row1) + Row2 -> New Row 2

D:

The problem with those two steps is that you're not helping yourself at all. Adding R₁ to R₂ only gets you a leading term in a_2,1. So a good way to start is by considering R₂ and R₃, and trying to eliminate either entry a_2,2 or a_3,2.

If you're not familiar with the notation I'm using; a_i,j corresponds to the entry in the i^th row and j^th column of matrix A.

 

[Amaz1ng]

The problem with those two steps is that you're not helping yourself at all.

Yeah the thing is that I'm not learning exactly why I'm doing these things. We're not learning theory, just applications. I'm just given the steps and that's the end of that lol. Luckily my assignment is finished (TI-89 ftw), and I'll just have to ask in class how exactly to do a 3x3 before the test. :shy:

 

[DivisionByZro]

It's not hard. There's not much theory behind reducing a 3x3, 4x4 5x5...9x9... All you have to keep in mind is that you want to eliminate as many leading terms as possible. I'll show you an example:

Let's start with:
<br /> \left[ <br /> \begin{array}{ccc|c} <br /> 0 &amp; 2 &amp; 3 &amp; 8\\ <br /> 2 &amp; 3 &amp; 1 &amp; 5\\ <br /> 1 &amp; -1 &amp; -2 &amp; -5\\ <br /> \end{array} <br /> \right] <br />

What not to do: It's a bad idea here to add R₂ or R₃ to R₁ since that will change the leading zero (in a_1,1) to a 2 or a 1. So we do not change R₁ yet. We can begin by switching R₁ for R₃, we want all leading terms to be in the first few rows, and the leading zeros to be at the bottom:

<br /> \left[ <br /> \begin{array}{ccc|c} <br /> 0 &amp; 2 &amp; 3 &amp; 8\\ <br /> 2 &amp; 3 &amp; 1 &amp; 5\\ <br /> 1 &amp; -1 &amp; -2 &amp; -5\\ <br /> \end{array} <br /> \right] <br /> → <br /> \left[ <br /> \begin{array}{ccc|c} <br /> 1 &amp; -1 &amp; -2 &amp; -5\\ <br /> 2 &amp; 3 &amp; 1 &amp; 5\\ <br /> 0 &amp; 2 &amp; 3 &amp; 8\\ <br /> \end{array} <br /> \right] <br /> → <br /> \left[ <br /> \begin{array}{ccc|c} <br /> 1 &amp; -1 &amp; -2 &amp; -5\\ <br /> 0 &amp; 5 &amp; 5 &amp; 15\\ <br /> 0 &amp; 2 &amp; 3 &amp; 8\\ <br /> \end{array} <br /> \right] <br /> → <br /> \left[ <br /> \begin{array}{ccc|c} <br /> 1 &amp; -1 &amp; -2 &amp; 5\\ <br /> 0 &amp; 1 &amp; 1 &amp; 3\\ <br /> 0 &amp; 2 &amp; 3 &amp; 8\\ <br /> \end{array} <br /> \right]<br /> → <br /> \left[ <br /> \begin{array}{ccc|c} <br /> 1 &amp; -1 &amp; -2 &amp; 5\\ <br /> 0 &amp; 1 &amp; 1 &amp; 3\\ <br /> 0 &amp; 0 &amp; 1 &amp; 2\\ <br /> \end{array} <br /> \right]<br /> <br />

What I did after changing R₁ and R₃ was I subtracted 2*R₁ from R₂, this eliminated the leading 2 in a_2,1.
I then multiplied R₂ by \frac {1}{5} After that I subtracted 2*R₂ from R₃, which eliminated the leading 2 in a_3,2.

Do you see the gist of what is going on here? On most steps, I try to eliminate one of the entries on the bottom left-half of the matrix. I want to create a "triangular" shape so that I can have a rref for my matrix. Basically, I want to be able to just read off the answer.

Can you try to finish the Gauss-Jordan reduction on the matrix that I started?