In Exercises 8–11, use Gaussian elimination to find the comple...

Question

In Exercises 8–11, use Gaussian elimination to find the complete solution to each system, or show that none exists.

Accepted Answer

Welcome back. I'm so glad you're here. We have got this system of equations and we are going to use Gaussian elimination to figure out the solution. So first let's start by writing the augmented matrix that's just taking the coefficients of everything. And so for X that's one for why? That is one. And for Z, that's negative three and that equals five, continuing with the second row three for X, negative five for Y and 12 for Z. And then that equals negative six for that constant term. And for our X in the third row, it's too wise to Z is negative six and our constant is 10. Alright, We've got the augmented matrix and for Gaussian elimination, our goal is to have a diagonal row of ones with zeros underneath them. And in order to do that, we want to first get a one in the upper left position and hey look cool, it's already there. Alright, so now let's work on getting zeros in the column below that one. So we will start off by taking the one, we're going to use the one in the upper left position and we're going to use that to cancel out to get that three into a zero. So we're working on this position which means we're going to replace row two and we're going to do so by taking row one and multiplying it by negative three negative three times row one and then adding that to row two. So that's the step we're going to take for our number one for our first step. And because we're replacing row two, it means we can copy down rose one and three. So we can copy down 11, negative 35. And the row three is two to negative 6, 10. And now to replace row two we take negative three times one is negative three plus three is zero. There's r zero. We wanted. Now just working across the row negative three times one is again negative three plus negative five is negative eight negative three times negative three is positive nine plus 12 positive nine plus 12 gives us 21 and negative three times five is negative 15 plus are negative six is going to give us negative 21. Cool. Now we want to work on getting a zero in this position, the first column of the third row. So for step number two, we're replacing row three. We'll do that by working with our one in that first row. So we're going to take the one. We're going to multiply it by negative two. So negative two times that whole first row. Then that gets added to row three. So for step number two, we want to replace row three so we can just copy over rows one and two. So 11 negative 350 negative 21 negative 21. Great. Now we take negative two times our first row. So negative two times one, negative two plus two is zero. We got our zero that we wanted continuing across the road negative two times one is negative two plus two. Well, hey, a bonus zero again for us, negative two times negative three. That is positive six plus negative six is another zero. Okay. Wasn't wanting wanting a zero there. We want to one there eventually. But let's keep going. And the final, the fourth column of our first row negative two times five is negative 10 plus positive 10. Hey, that's another zero. By definition, when we have zero, X plus zero, Y plus zero, Z equals zero. There are infinitely many solutions so that one's done. Well done. We'll catch you on the next one.

In Exercises 8–11, use Gaussian elimination to find the complete solution to each system, or show that none exists.

Key Concepts

Gaussian Elimination

Row Echelon Form

Existence of Solutions

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