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 + x - y + z = - 2
2w - x + 2y - z = 7
- w + 2x + y + 2z = - 1

Accepted Answer

hey everyone here we are asked to use Gaussian elimination to solve the given system of equations. So the first step here is to write out our given system into an augmented matrix. So if we recall an augmented matrix gets this bar to represent the equal sign. And on the left hand side we have all of the coefficients of each given variable from the equations. So we have the variables W, x, y and Z. So for our first row we look at our first equation and we get the values one negative three negative two negative two and negative one. Then for our second row we have 32 negative 61 and one. And finally for our third row we have negative 2645 and one. So now from here we need to reduce this augmented matrix to the upper triangle matrix by using elementary row operations. So first we want to begin at the first column and looking at the first term, we have positive one which is what we want in the diagonal. So we can move down to the sec term which is three. We want this three to become zero. So we need to perform an operation here. So we will have row to be replaced by the result of negative three times row one plus row two because then we'll have negative three plus three giving us zero. So now we have to rewrite our matrix. The first row remains the same. So we have one, negative three negative two negative two and negative one. And now our second row becomes 0, 11, 07 and four. And then our third row is the same as before. So we have negative 2645 and one. And now we can move down to the last term of the first column which is -2. We want this term to be zero. So Will three be replaced by the result of two times world one plus row three. And now we can again rewrite our matrix. So the first two rows are the same as our previous matrix. So we have one negative three negative two negative two and negative one. Then we have 0 11, 07 and four. And our third row becomes 0001 negative one. And now we are all done with our first columns. We can move on to the second column and we notice that this diagonal term which is at road to so 11 here we want it to be equal to positive one. So what we have to do is have row to be replaced by row two divided by 11. And so from this we now have our final reduced matrix. So the first row is still the same with one, negative three, negative two negative two and negative one. And now our second row becomes 0107 divided by 11 and four divided by 11. And then our third row is the same as before. So we have 0001, negative one. And now from here we're done reducing. So we can now recall that the left hand side terms here are the W, X, y and Z coefficients. And this bar is the equal sign. So we can now write three new equations from this matrix beginning with our first row, we get the equation w minus three, X minus two, Y minus two, Z is equal to negative one. And then from the second row we get the equation X plus seven divided by 11 times Z is equal to four divided by 11. And then from our third row we get a direct solution where we are told that Z is equal to negative one. And now we can use this value for Z and plug it into our second equation here to solve for X. So we have X plus 7/11 times Z, which is negative one is equal to 4/11. And now, just quickly simplifying, we find that X is equal to one and now we can use this value for X and the value for Z and substitute them into the first equation. So doing this, We have W -3 times X, which is one minus times y minus two times Z which is negative, one is equal to negative one. And now simplifying, we find that w is equal to two, Y and we know that Y is simply equal to y. So now we have a solution for each of our variables. And now from here we just need to write this as a set. So our final solution set is two Y, one Y and negative, one, which leaves us with answer choice. B. 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 + x - y + z = - 2 2w - x + 2y - z = 7 - w + 2x + y + 2z = - 1

Key Concepts

Gaussian Elimination

Row Operations

Augmented Matrix

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