Use the Gauss-Jordan method to solve each system of equations. For systems in two variables with

Use the Gauss-Jordan method to solve each system of...

Key Concepts

Parameterization of Solutions

Parameterization of solutions involves expressing the general solution of a system of linear equations in terms of one or more parameters, typically corresponding to the free variables identified in the RREF of the matrix. This method is essential when a system has infinitely many solutions, as it provides a compact and general description of all possible solutions.

Free Variables

Free variables are variables in a system of linear equations that are not leading variables in the reduced row echelon form of the matrix. They can take on arbitrary values, which is often the case in systems that have infinitely many solutions. Recognizing free variables allows one to express the general solution set in terms of these arbitrary parameters.

Gauss-Jordan Elimination

Gauss-Jordan elimination is a systematic method for solving systems of linear equations. It works by applying a series of row operations to transform the augmented matrix of the system into a reduced row echelon form, from which the solutions can be directly read. This process helps in simplifying the system and identifying whether it has a unique solution, infinitely many solutions, or no solution at all.

Systems of Linear Equations

A system of linear equations is a collection of one or more linear equations involving the same set of variables. The goal is to find the values of the variables that satisfy all the equations simultaneously. Such systems can be represented in matrix form, which facilitates the application of algebraic methods like Gauss-Jordan elimination for finding solutions.

Reduced Row Echelon Form (RREF)

Reduced row echelon form is a special form of a matrix achieved through row operations where each leading entry is 1, all entries in the column above and below the leading 1 are zero, and each leading 1 appears to the right of the previous row’s leading 1. This form is particularly useful because it makes the structure of the system transparent, clearly showing the relationships between variables and indicating any free variables.

Transcript

00:01 The concept involved in this problem is to solve a system of linear equations by using the gauss jordan method.

00:09 So to use the gauss -jordan method, need to set up an augmented matrix using the coefficients and the constants in the system.

00:20 So using the first equation, setting up our first row, the coefficients will give us 1, negative 1, 2, 1 then we'll have our vertical bar and 4 okay second row 0 1 1 0 and 3 then our next row we'll have 0 0 1 and negative 1 and then 2 and then the last row would be 1 negative 1 0 and the constant 0 so we will use the gastroard method in a series of row operations to solve this system.

01:15 Now we'll take quite a few row operations.

01:18 My first thing that i like to do when i work with row operations is try to produce as many zeros as i can.

01:33 And also work towards getting the first column towards 1 ,001.

01:42 So i'm going to go through a couple steps.

01:44 I will get that column.

01:45 But the first thing i'm going to do on this one is i'm going to take row two and i'm going to add it to row one.

02:01 Okay so that will produce a new matrix that will look like this.

02:09 Okay, my new row one would be one, zero, three, one, prickle bar, and seven.

02:27 Okay, so i'm going to leave rows two, three, and four, as they are for this step.

02:47 Okay, then my next step i'm going to take row 1 times a negative 1 and add that to row 4 and that will produce a new row 4.

03:09 So come down here, so i'll keep row 1 and row 2 and row 3 and i will produce a new row 4 by taking row 1 times a negative 1 and adding it to row 4.

03:50 Okay, so this new row 4 is going to be 0, 0, i'm sorry, negative 1, negative 3, negative 1, and negative 7.

04:17 Okay, so i've accomplished really my first goal is to get the 1 and the 0 stand in the first column, and i did produce another in the first row in one of my operations.

04:31 Okay, so my next step is i'm going to add rows two.

04:38 I'm going to add row two to row four.

04:47 Because what i'm going to try to go for is to get, well, i don't have to get that bottom row into the form of zero zero zero one.

04:56 So that's kind of what i'm going for now.

05:04 So row one's going to stay the same and row 2 and row 3 are going to stay the same and i'm producing a new row 4 by adding row 2 to row 4 so give me a 0 0 0 0 0 negative 2 negative 1 and negative 4 okay then i'm going to take row 3 times 2 and add that to row four.

06:21 And that'll produce a zero in this slot.

06:25 And then i'll be on my way to figure out what w is equal to.

06:31 Okay, so again rows 1 and 2 are going to stay as they are.

06:50 So row 1 and row 2 stay as they are and i'm taking row 3 times 2 and add it to row 4.

07:01 So row 3 will stay what it is in my new row 4 is going to be 0 0 0 0 0 0.

07:22 So i'm on my way to my first solution for a variable because now i can take row 3 times the negative 1 3rd.

07:36 Not doing anything else with it but times negative 1 3rd.

07:42 Okay so we're getting some of once we find out one variable the others pretty well will start falling in place quickly.

07:55 Okay so rows 1 2 and 3 are going to stay as they are and it's row 4 that i'm going to change.

08:15 Okay so row 4 multiply everything in it times a negative 1 it's going to produce me 0 -0 -0 of positive 1 in 0.

08:29 So now we have figured out that w is equal to 0.

08:37 Okay, so that's going to be to our advantage that we've found out that w is equal to 1.

08:44 I mean 0.

08:46 As soon as you get one answer, you pretty well have it made.

08:51 Okay, so now i'm going to do a couple things.

08:54 I am going to take row 4 and i'm going to add that to row 3.

09:06 Okay so let's do that...

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