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.

w + 2x + 3y - z = 7
2x - 3y + z = 4
w - 4x + y = 3

Accepted Answer

hey everyone here we are asked to use Gaussian elimination to solve the given system of equations. So our first step here is to write our system into an augmented matrix. So if we recall this matrix has a bar here to represent the equal sign and to the left we place each coefficient from each given very in relation to each equation. So we have variables W, x, y and Z. So for our first row we will look at our first equation. So we have the values negative 13, negative 23 and three and now four, row two. We have 01, two and negative four. And for our final role we have 23 negative 20 and six. Now all we have to do is reduce this matrix into an upper triangle matrix using elementary row operations. So our first step here would be to look at our first value which is negative one while we want this value to be positive one. So we can let row one be replaced by negative one times row one. So doing this, we rewrite our matrix. So roll one. Now has the values of one, 32, negative three and negative three. And then our last two rows remain the same. We have 0142 and negative four and then we have 23, negative 20 and six. So our next step would be to go down the first column to the next term which is zero. Well we want it to be zero. So we can move to the last term which is to we want this to to also be zero. So we'll have row three become the result of negative two times row one plus row three. And so using this operation we can again rewrite our matrix. Our first row remains as the previous one. So we have one negative 32, negative three and negative three. Row two is also the same. We have 0142 and negative four. And now our third row becomes 09, negative 66 and 12. And now we can move to our second row or sorry, second column and we see that there is a one in the diagonal. So at row two which we would we want it to be. So we can move to the next term which is nine. We want this nine to become a zero. So we'll have row three become the result of negative nine times row two plus row three. And so we can write our final reduced matrix here. So our first row is still the same with one, negative 32, negative three and negative three. Our second row is also the same with 0142 and negative four. And now with our final row we have the values 00, negative 42 negative 12 and 48. So now since we are done reducing if we recall the left hand side here we have the W x, y and z coefficients for each of these terms here and then the bar is the equal sign. So now we can write three new equations with each of these roads. So beginning with our first row we get the equation W minus three, X plus two, Y minus three. Z is equal to negative three. And then from our second row we get the equation X plus four, Y plus two, Z is equal to negative four. And then finally with our third row we get a third equation of negative 42 Y minus 12. Z is equal to 48. And now from here we need to start solving these equations to find the solutions for each variable. So we can begin with our third equation and solve for y. So we can first divide each term by six gives us the equation negative seven, Y minus two. Z is equal to eight. We then can get y alone and simplify the expression. So we are left with, Y is equal to the quantity of negative eight minus two. Z. All divided by seven. And now we can plug in this solution for Y into our second equation that we found. So we have X plus four times Y. Which is we found to be negative eight minus two. Z. All divided by seven. Then we have plus two. Z is equal to negative four. So now we can begin solving for X. In terms of Z and doing this, we can first factor in this four and then add to Z to the numerator. So we have X plus negative 32 minus eight. Z plus 14. Z. All divided by seven is equal to negative four. And now simplifying, we have X plus negative 32 plus six. Z is divided all divided by seven is equal to negative four. And now just transposing this fraction to the right hand side, we are left with X being equal to the quantity of four minus six D divided by seven. And now we can take this solution for X and the solution for Y and substitute it into our first equation to solve for W. In terms of Z. So we have w minus three times X, which is the quantity of four minus six Z divided by seven. Then we have plus two times Y which we found to be the quantity of negative eight minus two. Z divided by seven. Then we have minus three, Z is equal to negative three. And now again we can begin solving for W by first factoring in this three and two we have w minus the quantity here of 12 minus 18. Z. All divided by seven plus quantity of negative 16 minus four Z divided by seven. Then we have minus three. Z is equal to negative three. And now simplifying and rearranging to get W alone, we find that W is equal to the plus one. What we also know that the is simply equal to Z. So now we have a solution for each of the given variables, so all that is left is to write our solutions in a solution set. So our final answer is the solution set of Z plus one, then four minus six Z. All the divided by seven, Then negative 8 -2 Z, all divided by seven and Z. And this leaves us with answer choice. The thanks for watching. I hope you found this video helpful and I'll see you again next time.

In Exercises 1 - 24, use Gaussian Eliminaion to find the complete solution to each system of equations, or show that none exists. w + 2x + 3y - z = 7 2x - 3y + z = 4 w - 4x + y = 3

Key Concepts

Gaussian Elimination

Row Echelon Form

Existence of Solutions

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