Solving a 3x3 System of Equations

In summary, Gauss-Jordan reduction is a method used to solve systems of linear equations. It involves manipulating a matrix by using elementary row operations to transform it into reduced row echelon form (rref), which allows for easy solution of the system. The key is to eliminate as many leading terms as possible by adding, subtracting, and multiplying rows. By following this method, the solution to the system of equations can be easily read off from the rref matrix.
  • #1
Amaz1ng
42
0

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:
 
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  • #2
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?

[tex]
\left[
\begin{array}{ccc|c}
1 & 1 & 1 & 2\\
0 & 5 & 2 & 12\\
0 & -4 & 8 & 0\\
\end{array}
\right]

\left[
\begin{array}{ccc|c}
1 & 1 & 1 & 2\\
0 & 20 & 8 & 48\\
0 & -20 & 40 & 0\\
\end{array}
\right]

\left[
\begin{array}{ccc|c}
1 & 1 & 1 & 2\\
0 & 0 & 48 & 48\\
0 & 1 & -2 & 0\\
\end{array}
\right]

\left[
\begin{array}{ccc|c}
1 & 1 & 1 & 2\\
0 & 0 & 1 & 1\\
0 & 1 & -2 & 0\\
\end{array}
\right]

[/tex]

Edit: I fixed it.

Amaz1ng said:
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 a2,1. So a good way to start is by considering R2 and R3, and trying to eliminate either entry a2,2 or a3,2.

If you're not familiar with the notation I'm using; ai,j corresponds to the entry in the ith row and jth column of matrix A.
 
Last edited:
  • #3
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:
 
  • #4
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:
[tex]
\left[
\begin{array}{ccc|c}
0 & 2 & 3 & 8\\
2 & 3 & 1 & 5\\
1 & -1 & -2 & -5\\
\end{array}
\right]
[/tex]

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 a1,1) 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:

[tex]
\left[
\begin{array}{ccc|c}
0 & 2 & 3 & 8\\
2 & 3 & 1 & 5\\
1 & -1 & -2 & -5\\
\end{array}
\right]

\left[
\begin{array}{ccc|c}
1 & -1 & -2 & -5\\
2 & 3 & 1 & 5\\
0 & 2 & 3 & 8\\
\end{array}
\right]

\left[
\begin{array}{ccc|c}
1 & -1 & -2 & -5\\
0 & 5 & 5 & 15\\
0 & 2 & 3 & 8\\
\end{array}
\right]

\left[
\begin{array}{ccc|c}
1 & -1 & -2 & 5\\
0 & 1 & 1 & 3\\
0 & 2 & 3 & 8\\
\end{array}
\right]

\left[
\begin{array}{ccc|c}
1 & -1 & -2 & 5\\
0 & 1 & 1 & 3\\
0 & 0 & 1 & 2\\
\end{array}
\right]

[/tex]

What I did after changing R1 and R3 was I subtracted 2*R1 from R2, this eliminated the leading 2 in a2,1.
I then multiplied R2 by [tex] \frac {1}{5} [/tex] After that I subtracted 2*R2 from R3, which eliminated the leading 2 in a3,2.

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?
 

Related to Solving a 3x3 System of Equations

1. What is a 3x3 system of equations?

A 3x3 system of equations is a set of three equations with three variables in each equation. These equations are typically solved simultaneously to determine the values of the variables that satisfy all three equations.

2. How do you solve a 3x3 system of equations?

To solve a 3x3 system of equations, you can use the elimination method or the substitution method. In the elimination method, you manipulate the equations to eliminate one variable at a time until you are left with one equation and one variable. In the substitution method, you solve for one variable in terms of the other two and then substitute that into the other equations to solve for the remaining variables.

3. What is the importance of solving a 3x3 system of equations?

Solving a 3x3 system of equations is important in many fields of science and engineering. It allows us to determine the relationships between multiple variables and find solutions that satisfy all given equations. These solutions can then be used to make predictions and solve real-world problems.

4. Can a 3x3 system of equations have more than one solution?

Yes, a 3x3 system of equations can have one unique solution, an infinite number of solutions, or no solution at all. This depends on the equations and the relationships between the variables. For example, if the equations are parallel, there will be no solution, and if the equations are the same, there will be an infinite number of solutions.

5. How do you check if a set of values is a solution to a 3x3 system of equations?

To check if a set of values is a solution to a 3x3 system of equations, you can substitute the values into each equation and see if they satisfy all three equations. If they do, then the values are a solution to the system of equations. You can also graph the equations and see if the values lie on the intersection point of all three lines.

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