Solve each system of equations using matrices. Use Gaussian el...

Question

Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.

${\begin{matrix} x + 2 y + 3 z = - 5 \\ 2 x + y + z = 1 \\ x + y - z = 8 \end{matrix}$

Accepted Answer

Welcome back. I'm so glad you're here. We've got a system of equations here and I'm going to demonstrate, we've got two choices here, but I'm going to demonstrate with Gauss Jordan elimination, how to solve for x, y and Z. So first let's write this system of equations as an augmented matrix, meaning we're going to look at the coefficients and copy those over. So for the first row, excess coefficient is one wise coefficient is one. Zis coefficient is one and the constant term that they equal is one. Second row, X is three, Y is two, Z is one, constant is four, third row X is six, Y is negative one, Z is three and they equal to there's our augmented matrix and we're going to be transforming this in Gauss Jordan elimination so that we've got a diagonal row of ones with zeros above and below them and those will equal our X value, our y value, and our Z value. I don't want to cut that off. There we go. So the um we're going to, the process will be going through, is that first we're going to go column by column. So the first column, we're going to get our one in place and then we'll use that one to turn the other numbers in the column into zeros when we finish, the first column will proceed to the 2nd and 3rd and then we'll have our answers in the fourth column. If that was all you needed to know, go for it, we'll catch it at the end with the answer if you need to see me go step by step, we'll be doing that. It will take a little bit but we'll get there. So we start off looking for a one in that first row, first column. And we've already got it. That one is done for us. Great. Now we can move on to replacing this three with zero. So all right, all the steps over here that we're going to go through, we're going to be replacing row two and in order to do that, we work with the road that has the one in it. So that's the first row and we're going to multiply that row by negative three. So negative three times row one and add that to our row two. So for step one, we're replacing row two, which means we can copy over rows one and three. And I'm not going to write the brackets for the whole time just to save a little bit of time and space. So row two will begin with the multiplication and addition. So negative three times one is negative three plus three is the zero that we desire negative three times one is negative three plus two is negative one, negative three times one plus one is negative two and negative three times one plus four is positive one. Now let's get a zero into this first column, third row where the sixes. So we're going to replace row three. We'll do that by working with the row that has a one in the desired spot which is row one. And we're going to take row one and we're going to multiply it by negative six. So -6 times row one plus row three. Because we are replacing row three we can copy over rows one and two. So 1111 and zero negative one negative 21, proceed with the multiplication. In addition negative six times one is negative six plus six is the zero that we wanted negative six times one is negative six plus negative one is negative seven negative six times one is negative six plus three is negative three negative six times one is negative six plus two is negative four. Step two is done. Now we're looking at the second column and we want to get a one Into this 2nd row. And in order to do that, replacing road to we can just multiply road to by -1. So -1 times row two. And because we're replacing road to we can copy over rose one and three. And now beginning with the multiplication negative one times 00 negative one times negative one is the one that we want negative one times negative two is two negative one times one is negative one. Great. We've got the one in the position that we can now use it to get zeros for the rest of the column. So let's start with row one for column two. So step four we're going to take row two And we'll multiply it by negative one and add it two, Row 1, which will replace row one. First, step four, we're replacing row one will copy over rows two and 3012 negative 10, negative seven, negative three negative four. And begin with the multiplication and addition negative one times zero is zero plus one is one negative one times one is negative one plus one gives us the zero that we want negative one times two is negative two plus one is negative one and negative one times negative one is one plus one is two. We've replaced our role one. Now let's get a zero into that spot where we've got a negative seven. Step five, we'll be replacing row three. We'll do that by working with the row that has a one in the desired spot. And let's multiply that rho times seven. So seven times row two plus the replacement row row three for step five. Because we're replacing row three, we copy over rows one and 210 negative 012 negative one. Now, with the multiplication in addition seven times 00 plus zero is 07 times one is seven plus negative seven gives us the zero. We want there seven times two is 14 plus negative three gives us 11 and seven times negative one is negative seven plus a negative four gives us a negative 11. Great. We've got the first two columns done. We are making progress. Now let's put a one in our desired location in the third column. So we're replacing that 11 And we can do that by taking our row three. So we're going to replace row three And we're going to divide it by 11. So row three divided by 11. Because we're replacing row three, we can copy over rows one and 210 negative 12, 012 negative one. And now dividing row three by 11, 0, divided by 11, 00, divided by 11 to 0. 11, divided by 11 is one -11 divided by 11 is -1. We're getting there just two more to go. Let's replace this negative one at the top of the third column. This will be our step seven And we're replacing row one and we will do that by working with the row that has the one in the desired spot which is row three. And this time we just have to add it to row one. So row three plus row one will get us the zero in the desired spot. We're replacing, row one. So we'll copy over rows two and three. Row 21012 negative one and row three is 001, negative one. Begin with our addition zero plus one is 10 plus zero is 01 plus negative one is zero and negative one plus two is one one spot left. Guys, we're almost there. Let's get this two out of here. So for our two for step number eight, we're replacing row two. We'll do that by working with the row that has the one in the desired spot. Let's multiply this by negative two. So negative two times row three and add that to row two. Because we're replacing row two, we can copy over rose one and 31001 and 001 negative one. And now with the multiplication and addition negative two times zero is zero plus zero is zero, negative two times zero is zero plus one is one, negative two times one is negative two plus two is the zero that we're looking for and negative two times negative one is two plus negative one is one. And that's it guys. It is in the proper format, which means the one here is X. The one in the second row is Y. And the negative one in the third row is Z. We look at our answer choices and that matches with answer choice. D. Well done. We'll catch on the next one

Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
${\begin{matrix} x + 2 y + 3 z = - 5 \\ 2 x + y + z = 1 \\ x + y - z = 8 \end{matrix}$

Key Concepts

Systems of Linear Equations

Gaussian Elimination with Back-Substitution

Gauss-Jordan Elimination

Watch next