Solve each system, using the method indicated.
5x + 2y = -10
3x...

Question

Solve each system, using the method indicated. 
5x + 2y = -10 
3x - 5y = -6 
(Gauss-Jordan)

Accepted Answer

Hello, today we're gonna be solving the system of linear equations using the Gauss Jordan method. So we're currently given a system of two equations. We have three X plus Y is equal to negative 12 and we have X minus two, Y is equal to negative four. What we're going to do is we're going to set up an augmented matrix now because we're given the elements of X and Y, we're going to be setting up an augmented two by two matrix. Now, what's gonna go on the right hand side is going to be the current solutions to the system that's going to be negative 12 and negative four. And on the left hand side, we're going to be writing the coefficients for each of the terms. So in the first row, the first equation given to us is three X plus Y. So X has a coefficient of three, Y has a coefficient of one. So our first row is going to be three and one. Then in the second row, we have the second equation which is X minus two Y X has a coefficient of one, which is going to give us one in the second row and Y has a coefficient of negative two, which is going to give us negative two. So now that we have the augmented matrix, our goal is to get the identity matrix on the left hand side and whatever we are left with on the right hand side, after getting the identity matrix is going to be the solution to the system. So the trick to this is you want to first get a one in the upper left element, we have a one below it. So we can start by swapping the positions of row one and row two. So the first step is to swap row one with row two, that is going to give us the matrix of one negative two augmented with negative four and augmented with negative 12. Now, what we want to go ahead and do is we want to get this lower left element to be zero. In order to do that, we can go ahead and multiply the first row by negative three, then add it to the second row. So we're going to be changing the second row by multiplying the first row by negative three, then adding it to the second row. So the first row is gonna stay unchanged. We have one negative two augmented with negative four, but one times negative three is going to give us negative three. Add that to three in the second row and we have zero negative three. Times negative two is going to give us positive six and positive six plus one is going to give us positive seven, negative four times negative three is going to give us positive 12 and 12 minus 12 is going to give us zero. Now, what we wanna do is we want to go ahead and get the lower right element to be one. In order to do that, we can multiply row two or divide row two by seven. Or in other words, that's the same thing as saying as multiplying it by 1/7. So we can rewrite the matrix as the unchanged first row, which is one negative two, negative four and 010. Finally, we want to get the element above the lower right one to be zero. In order to do that, we can multiply the second row by two and add it to the first row. So row one is going to equal to two times row two plus row one. The second row is being unchanged in this step. So we have 01 augmented with zero and zero times two is going to give us zero plus one is going to give us one, one times two is positive two minus two is zero and zero times two is zero minus four is going to be nine minus four. So what we now have is the identity matrix on the left hand side. And that means that negative four and zero is going to be the solution or more specifically X is going to equal to negative four and Y is going to equal to zero. And this can be represented as negative four comma zero, which is going to be our solution. And with that being said, the answer to this problem is going to be D. So I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

Solve each system, using the method indicated. 5x + 2y = -10 3x - 5y = -6 (Gauss-Jordan)

Key Concepts

Systems of Linear Equations

Gauss-Jordan Elimination

Augmented Matrix

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