Solve each system, using the method indicated.
x - z = -3
y + z...

Question

Solve each system, using the method indicated. 
x - z = -3 
y + z = 6 
2x - 3z = -9 
(Gauss-Jordan)

Accepted Answer

Hello. Today we're gonna be using the Gauss Jordan elimination method to solve the system of linear equations. So the system of linear equations are given to us is X minus Z is equal to negative one Y plus Z is equal to four and four, X minus five. Z is going to equal to negative seven. In order to use the Gauss Jordan method, we can go ahead and set up this system of equations as an augmented matrix. So we're going to make a three by three matrix since we have X Y and Z, but we're gonna be augmenting it with the given solutions. So what's gonna go in this right hand column is going to be the solutions for the system of equations? And the solutions are going to be negative 14 and negative seven. Now what goes in the left hand column is going to be the coefficient of the given of the given equations. So in the first equation, we have X and negative Z, the coefficient of X is going to be one. So the first element in the first row is one, we are not given a Y term in the first equation. So we're gonna represent that as zero and the coefficient of Z in the first equation is negative one. So we're going to be writing negative one for Z. Then we're gonna repeat this process for the next two rows. In the second equation, we are not given an X value. So we're gonna represent that with the coefficient of zero. The coefficient of Y in the second equation is one. So we're gonna write down one and the coefficient of Z in the second equation is one. So we're going to represent that with one. Finally, in the third equation, the coefficient of four of X is four. So we're going to represent that with four, we are not given a Y term. So we're gonna represent that with zero and the coefficient of Z is negative five. So we're gonna write down negative five. Now, the goal is to go ahead and get the left hand side into the form of an identity matrix or in other words, we want 100010 and 001. And whatever is gonna be on the right hand side of that augmented matrix is going to be the solutions to the system of equations. So a trick to approaching this is to get a lead one or get ones in the diagonal of the matrix. Currently, we have a one in the first element, but we want to get this number 4 to 0. What we can go ahead and do is we can let row three equal to negative four times row one and then add it to row three. So multiplying one by negative four is going to give us negative four and adding that to the first element of the third row is going to give us zero. So before we do that, actually, let's go ahead and write down the first two rows. Since we're not changing those. In this step, we have 10, negative one augmented with negative one, then we have 011 with four, no, negative four times one in the first row is going to give us negative four. And if we add that to the first element in the third row, that's going to give us zero negative four times zero in the first row is going to give us zero and zero plus zero in the third row is going to give us zero a negative one in the third in the first row times negative four is going to give us positive four and positive four minus five in the third row is going to give us negative one. And we have to repeat that for the augmented value. So negative four times negative one in the first row is going to give us positive four and positive four minus seven is going to give us negative three. So now we want to make sure that our second column is in the form of 010. Currently, it's in that form. So now we want to try to change this last column to be in order to do that, let's go ahead and get a leading one in the third row. In order to do that, we can take row three and set it equal to negative one times row three. Doing this is going to change our matrix to give us 10, negative one augmented with negative one 0114 and 00, positive one, positive three. Now what we want to go ahead and do is get all the elements above this last one to be zero. What we can do is for row two, we can multiply row three by negative one and then add that to row two that is going to give us 10, negative one augmented with negative one. And since we're multiplying row three by negative one, that is going to give us zero plus zero, which is 00 plus one, which is one and one times negative one, which is negative one plus one, which is zero. Then we're gonna do negative one times positive three, which is negative three, add that to four and that is going to give us one, then we have 001 and three. And finally to get this last element to zero, we can just go ahead and add row three to row one. So row one is going to equal to row three plus row one. And that is going to give us the matrix of three minus one, which is two 0101 and 0013. So what we currently have is the identity matrix for a three by three system. And that means that this is going to be the solution to our system of equations. So if we write that going from left to right, this is in the form of 21 and three, this is where the first element represents the X value. The second element represents the Y value and the third element represents the Z value. So that means that X is going to equal to two Y is going to equal to one and Z is going to equal to three. And this is going to be the solution to the system of equations. And with that being said, the answer to this problem is going to be C 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. x - z = -3 y + z = 6 2x - 3z = -9 (Gauss-Jordan)

Key Concepts

Systems of Equations

Gauss-Jordan Elimination

Augmented Matrix

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