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

Gauss-Jordan Elimination

A systematic method to solve linear systems by converting the augmented matrix of the system into reduced row-echelon form through the application of elementary row operations. This method reveals the solution directly, whether it is unique, infinite, or nonexistent.

Systems of Linear Equations

These are collections of equations involving several variables that need to be solved simultaneously. They are a central topic in linear algebra and can be represented compactly in matrix form for systematic analysis and solution.

Elementary Row Operations

Operations including row-switching, scalar multiplication of a row, and adding a multiple of one row to another, which are the tools used to simplify matrices to an equivalent form that is easier to solve.

Infinite Solutions and Parameterization

When a system does not have a unique solution, it often has infinitely many solutions. This occurs when there are free parameters in the reduced matrix form, allowing the solution to be expressed in terms of one or more arbitrary variables.

Matrix Methods in Linear Algebra

These techniques, including Gauss-Jordan elimination, allow for a structured and algorithmic approach to solving systems, providing clarity on the consistency and nature of the solutions while utilizing the power of matrices to represent complex linear interrelationships.

Transcript

00:01 All right.

00:02 So today we're going to do a problem about the gauss jordan method for solving a system of linear equations.

00:11 And so the gauss jordan method takes a system of linear equations and converts it into an augmented matrix.

00:18 And that augmented matrix contains all of the coefficients of the variables in rows as a matrix.

00:26 And then it has a final column of sort of the solutions to those equations, the numbers that are on the right -hand side of the equal sign that does not contain any variables.

00:42 So once you have your augmented matrix, the ways you can manipulate that matrix to try to get it into row echelon form, which remember is a upper triangular matrix where each row begins with a one, is you have three options with your augmented matrix.

01:01 You can swap two rows, you can multiply one row by a scalar, or you can add the multiple of one row to another.

01:11 And so we're going to solve the equations given in this problem using this method.

01:18 So firstly, let's get our problem down.

01:22 So we have 2x plus y minus c plus 3w equals 0.

01:35 It's equation 1.

01:37 We have 3x minus 2y plus c minus 4 w is negative 24.

01:53 We have x plus y minus c plus w equals 2.

02:03 And we have x minus y plus 2 z minus 5 w minus 5 w equals negative 16 so i'm going to just warn you right now when i was trying to work through the solution to this problem i made a couple of mistakes it's really easy to just try to quickly do the math and you make a mistake and it propagates through your solution and it's very easy to make mistakes and then get the wrong solution.

02:41 So i want you to, as you're going through this method, make sure every time you do an operation, one of these three bullets, make sure you check that answer.

02:50 Make sure that everything actually is what, it follows the operation that you specified.

02:58 And you should be good to go as long as you're very careful and you don't try to do too much at once.

03:04 All right.

03:05 So first step is to conduct.

03:06 The system of equations to an augmented matrix.

03:10 So really this means to put all of the coefficients over here.

03:15 And now if you saw another system of equations that didn't have one of the variables in it, you would just put a zero for that variable.

03:24 So i'm doing, i'm just filling it up right now.

03:27 You know, i mean, you've seen the equations.

03:29 I'm just filling it up.

03:32 How's your day? hope you're doing well.

03:34 Hope you're safe out there, you know.

03:37 Hope math is fun.

03:39 I like math.

03:40 Good stuff.

03:43 And there we go.

03:45 And remember, because this is an augmented matrix, we're going to add this final column that contains sort of this last row over here, or last column.

03:55 Zero, negative 24, 2, and negative 16.

04:03 That's our augmented matrix.

04:05 It's augmented by the sort of solutions.

04:08 Okay.

04:09 So, let's get to it.

04:14 So the first thing that i'm going to do is i am going to move row three up to row one.

04:26 I want this to be sort of my main row.

04:29 So i'm going to rewrite this perhaps elsewhere.

04:42 Let me erase some things really quick.

04:43 Oh gosh, this eraser is not really what i was hoping.

04:50 Do do -do -do -do -do -do -do -do -do -do something like that.

05:01 Making a little more space.

05:02 Okay, so we're now going to convert this to new matrix.

05:12 It's an equivalent matrix.

05:14 Imagine just solving this by hand, but now we're just working with the numbers rather than the letters.

05:20 So this is equivalent to putting the first row up here, 1, 1, negative 1, 1 and 2.

05:29 And now that i've put this up here, i'm going to use this first equation and use it to try to eliminate the first column of the other three rows.

05:43 That means i want to try to get the other rows to zero.

05:46 So how do i do that? well, if i multiply the first row by, or this new first row, by negative 2, and add it to add it to, the original first row over here, i can eliminate the first variable.

06:05 So i would get negative 2 times 1, which is plus 2 is 0.

06:13 And we have negative 2 times 1, plus 1 is negative 1.

06:20 Negative 2 times negative 1 is 2 minus 1 is 1.

06:26 Negative 2 times 1 is negative 2 plus 3 is negative 1.

06:37 And negative 2 times 2 is negative 4 plus 0 is negative 4.

06:44 We do that same process for the other two rows.

06:47 You see how we can imagine we multiplied this little side over here by negative 2.

06:52 We multiplied the row by negative 2, and then we added it to this row.

06:56 We got rid of one of its coefficients.

07:01 Okay, so let's do it for the other rows.

07:04 You try to do it on paper in front of yourself, and i'm going to do it up here as well, so you make sure that your answers are correct and you don't make the time -consuming mistakes that i did.

07:14 Okay, so now i'm going to multiply the first row by negative 3, and that cancels out the first term with the second row.

07:26 Negative 5, 4, negative 7, negative 30, and the final row.

07:41 It's just a negative 1 multiply.

07:44 I've got 0, negative 2, negative 6 and negative 18.

07:52 All right, sick.

07:55 So we have this new matrix now that has zeros for the first coefficient of all the rows except for the first one.

08:03 So we're on our way to making this upper triangular matrix.

08:06 That's what we want.

08:06 We want these guys to also be zero.

08:10 All right.

08:11 So let's keep going.

08:14 I'm going to make some more space over here.

08:16 We don't need this original matrix anymore because we're, you know, we're just modifying this one.

08:21 So now the next thing i want to do is i don't, i'm not a huge fan of all these negatives in the leading coefficients.

08:31 I would rather begin my row with one so that, you know, if i have that all set up in row echelon form, i can simply, you know, get the answer for the last variable and plug it into the previous equation, which solves for the next variable and plug it into the previous equation.

08:47 That's why we want these numbers right here to be ones.

08:52 Makes our solving of the equations relatively easy.

08:57 So let's get to that point.

08:59 So what i'm going to do next is i'm going to multiply the bottom three rows by negative 1.

09:05 And so that's pretty trivial.

09:07 So i get rid of these negatives, and then i add negatives here.

09:14 Boom, boom, boom...

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