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

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've got a system of equations and we're going to use Gaussian elimination to figure out what X, Y and Z are. So let's start off by writing this as an augmented matrix, which means we're just going to look at the coefficients of the terms and copy them into the matrix. The first row, excess coefficient is one, wise coefficient is negative one and z's coefficient is three and those equal the constant term six, row two, Xs coefficient is three, Y is negative five, Z is 12 and the equal negative six and row three X is two, Y is negative two, Z is six and the equal 10. We've got our augmented matrix. Now, where are we going with us for Gaussian elimination, we want it to look like a diagonal row of ones with zeros underneath them. And once we have it in that format we'll easily be able to say, okay, well Z equals this third row and then use back substitution to get the Y and the X values. Now we know where we're going, how are we going to get there? We're going to go column by column. We're going to get the one in its place and use that one to get the zeros in the rest of the appropriate places in the column. So looking at column one here, we've already got our one in its appropriate place. Great, we don't have to get that there. And now we can just focus on getting this three into a zero. So we will do that by replacing road to with working with the row one that has the one in the desired spot. We're going to multiply everything there. Times negative three, negative three times row one and then add that to row two. Because we're replacing road too. We can copy down Rose one and three. So one negative 136 and two negative 26 10. I'm not going to write the brackets just to save some time and space now to get road to filled in. We're going to proceed with the multiplication and addition negative three times one is negative three plus three is the zero that we want negative three times negative one is three plus negative five is negative two negative three times three is negative nine plus 12 is positive three negative three times six is negative plus negative six is going to be negative 24. We have got the zero in the spot. We want it. And now let's get a zero where this two is. So we're going to be replacing row three in order to do that. We work with the row that has the one in the desired spot. We're going to multiply that times negative two. So negative two times row one and add that to row three. Because we are replacing row three, we can copy over rows one and 21 negative 1360 negative negative 24. Now proceed with the multiplication. In addition negative two times one is negative two plus two is the zero that we want negative two times negative one is two plus negative two. There's like a bonus zero. Cool negative two times three is negative six plus six will hang on. That is another zero and negative two times six is negative 12 plus 10. That's negative two. And let's hold on just a second here. This is saying that zero X plus zero, Y plus zero Z equals negative two. And I just don't know of a universe where that is true. So the solution does not exist. 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

Consistency of a System

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