In Exercises 1 - 24, use Gaussian Eliminaion to find the complete...

Question

In Exercises 1 - 24, use Gaussian Eliminaion to find the complete solution to each system of equations, or show that none exists.

8x + 5y + 11z = 30
- x - 4y + 2z = 3
2x - y + 5z = 12

Accepted Answer

Welcome back. I am so glad you're here. We are given this system of equations and we are asked to find the complete solution for it using Gaussian elimination. Will you recall from previous lessons that when we're given a system of equations we can create a matrix from it using the coefficients and constant terms. So copying over the first equation, just copying its coefficients and constant term to the first row will have to to negative negative eight. And that makes up our first row of this matrix and now continuing with the second equation, we have 30 negative 21 negative 15 and then our third equation, the third row 6, 10, negative 72 negative 20. Now we recall previous lessons on Gaussian elimination, that what we want is to turn this matrix into a matrix that has a one in the upper left hand corner and then zeros below it. And then we're making a diagonal of ones. So we'll have something in this second column and then a one and then a zero below it. The third column will have something and something and then a one and then that final column will give us something something and then rz solution what Z will equal in that last spot. And then we'll use substitution to go back and solve because we know what Z is will solve for y and then solve for X. Alright, so that's what we're doing. The first step is to get a one in this upper left hand position here in order to do that, we're going to take row one and I'll write our steps down here on the left hand side. We're going to replace row one and we're going to divide everything by two or multiply it by one half. So we're going to take one half of rho one because we're replacing row one, Rose two and three will stay the same. So I'm going to copy those down. We have row 230 negative negative 15. I'm not going to copy all the brackets and everything because we're going to do this a bunch of times and row 36, 10 negative 72 negative 20. Excuse me. And so now doing the work on row 12 times one half is one. That's what we want two times one half is one, negative 20 times one half is negative 10 and negative eight times one half is negative four. And there is our new matrix. What we want to do next is get a zero in the first column below the one. In order to get a zero there, we're going to do some work on the row that has the one in the desired position in this column. So that's rho one to replace row two. We're going to take row one and in this case we're going to multiply it by negative three, negative three times row one and then add that to row two. So because we're replacing road too, we can copy over rose one and 311 negative negative four And 6, 10 -72 -20. Now doing the work for row two, negative three times one is negative three plus three is the zero that we want negative three times one is negative three plus zero is negative three negative three times negative 10 is a positive 30 minus is going to be a positive nine negative three times negative four is positive 12 minus 15 is negative three. Okay, we have got that zero where we want it in the second row, first column. Now we want a zero in the third row, first column and we're replacing row three to get the zero there. We're going to do the work on the road that has the one in the desired spot. So that's rho one. We're going to take row one this time times negative six And add that to row three. Because we're replacing row three, we can copy down rows one and 211 negative 10 negative 40 negative negative three. And now row three negative six times one is negative six plus six. Is the zero that we want negative six times one is negative six plus 10 is a positive four, -6 times negative 10 is positive, 60 -72 is negative 12 negative six times negative four is positive 24 minus 20 is a positive four. Okay, That first column is done. Now we want a one in the second row. 2nd column. So we're replacing road to and to get that to be a zero. Excuse me, a positive one. We're going to multiply it by a negative so it will be positive and we'll multiply it by its reciprocal one third. So we'll take row two times a negative one third. Because we're replacing road to we can copy over rose one in -10 -404 -124. Now taking road two times negative one third, zero times negative one third is still zero negative three times negative one third is a positive 19 times negative one third is negative three and negative three times negative one third. That's a positive one. Okay, we have got a column to halfway done. Now let's finish off Column two and we'll replace that four with a zero. So we're replacing row three and to get a zero there we do the work on the row that has the one in the desired spot. So we're going to take row two times a negative four And add that to row three because we're replacing row three, we can copy over rows one and negative 10 negative 401, negative 31. Now replacing row three negative four times zero is zero plus zero is still zero negative four times one is negative four plus four is zero. Excuse me. That doesn't look like a zero negative four times negative three is a positive 12 plus negative 12. Well that's zero and negative four times one is negative four plus four is zero. Hang on a minute. Look at that. Look at our row three. We recall that this is our system of equations, coefficients. So that's zero, X plus zero, Y plus zero, Z equals the constant term zero. And yeah, that is just true across the board. So we recall from previous lessons that by definition our answer here is infinite solutions. Well done, we'll catch you on the next one.

In Exercises 1 - 24, use Gaussian Eliminaion to find the complete solution to each system of equations, or show that none exists. 8x + 5y + 11z = 30 - x - 4y + 2z = 3 2x - y + 5z = 12

Key Concepts

Gaussian Elimination

Row Operations

Augmented Matrix

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