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 are going to use Gaussian elimination to find the solution or identify if there's no solution for this system of equations. Gaussian elimination means that we're going to get a final matrix that has a diagonal row of ones and then zeros below those ones and then just any numbers. I'll put dots here in these other spaces. And then what we do is we've identified this final value, the X sub four in the fourth row and we'll use back substitution to go back and find the solutions for X and three X two and X sub one. So let's start off by writing our matrix for our given system of equations. Looking at the coefficients of row one, we will write one negative two negative four, negative five and two Row 2s coefficients, 1 -1 -4 - and zero row threes coefficients to negative four. Oh, nice of them to leave us a space there. So there's zero for that. Except three value in row three negative eight and eight. Sometimes they don't leave a space and you got to watch for that row four. We've got two negative four negative eight negative 10 and four. Great there is are given matrix. Usually you'll see it written with these brackets but I'm not going to write the brackets going forward or this vertical line here just to save a little bit of space. So we'll start off by trying to get a one in the first row in the desired position. And hey look, we did it, it's already there. So now let's work on that first column and we'll get a zero in the second row. So to get a zero in that row, working on row two, we're going to work with the road that has the one in the desired spot. And that's row one. We're going to multiply row one times negative one. So negative row one and then add that to row two to replace our road to because we're replacing road to, we can copy down Rose 13 and four. Let's do that one, negative two, negative four, negative 52 skipping road to to negative 40, negative 88 and row four to negative four, negative eight, negative 10 and four. Now we'll do our multiplication and addition. So multiplying in row one negative one, times one is negative one and then adding that to row two. We've got the zero that we're looking for negative one times negative two is two plus negative one is one negative one times negative four is four plus negative four is zero negative one, times negative five is five plus negative two is three negative one, times two is negative two plus zero is negative two. And we've got row two replaced with zero in the proper location. So now let's keep working down the first column and get zero to replace the two in row three. So first step two, we're going to replace row three and do so by working with the row that has the one in the desired place which is row one. This time when won't supply it? By negative two And then add that to row three. Because we are replacing row three, we can copy over rows 12 and four. So we've got one negative two, negative four, negative zero, 03, -2, skipping row 32, negative four, negative eight, negative 10 and four in row four. Now we will do our multiplication in addition, negative two times one is negative two plus two is the zero that we're looking for negative two. Times negative two is four plus negative four is a bonus zero negative two times negative four is eight plus zero is eight, negative two times negative five is 10 plus negative eight is two. The negative two times two is negative four plus eight is four. Great. We've got that step done. Now let's get a zero in that final row, row four. For the first column. Again, we'll do that by working with the row that has the one in the desired spot. And again, we will multiply it by negative two, row one, times negative two. And then adding that to row four because we're replacing row four, we can copy down rows 12 and three. So that gives us one, negative two, negative four, negative 0103, negative two 0, 2, 4. And now doing our multiplication. In addition we've got negative two times one is negative two plus two is zero. There's a zero that we're looking for negative two times negative two is four plus negative four. Well that's also zero bonus negative two times negative four is eight plus negative eight. Well there is an extra bonus. Zero negative two times negative five. That's 10 plus 10. Or plus negative 10 is another zero. Let's see what happens with this last one negative two times two is negative four plus four. That is zero. We've got zeros across the bottom row which means that zero x ones plus zero x two's plus zero X three's plus zero X four's equals zero. And we know that by definition from previous lessons we have learned that this system has infinitely many solutions. 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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