In Exercises 1 - 24, use Gaussian Elimination to find the comp...

Question

In Exercises 1 - 24, use Gaussian Elimination to find the complete solution to each system of equations, or show that none exists. $\begin{cases}w - 3x + y - 4z = 4 \-2w + x + 2y = -2 \3w - 2x + y - 6z = 2 \-w + 3x + 2y - z = -6\end{cases}$

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 given system into an augmented matrix. So if we recall this matrix gets a bar to represent the equal sign into the left hand side. We place each coefficient for each variable from the given equations. So for our first row we look at the first equation and we have the values three negative two negative 11 and 10. For our second role we have negative one to negative 10 and negative five for our third row we get 23 negative 14 and 20. And for our final role we have negative four negative two negative 12 and negative 15. So now we need to work to reduce this augmented matrix into the upper triangle matrix using elementary row operations. So the first step here is to work to get our first value this three, we want it to be positive one. Well one way to do this, we can just swap row one with row two, so we'll have row one, interchange with row two and now we'll have negative one as our first value. So rewriting our matrix, our first row becomes negative one to negative 10 negative five. And our second row becomes three negative two negative 11 and 10. And now our last two roles remain unchanged. So we have 23 negative 14 and 20. And lastly we have negative four negative two negative 12 and negative 15. So now from here we want this negative one. This first term we want it to be positive one. So all we have to do is have roll one be equal to negative one times the entire world one. So we'll have roll one is equal to negative role one. And now once again we rewrite the matrix. So four row one we now have one negative 210 and five. And our last three rows remain unchanged. So we have three negative two negative and 10. Then we have 23 negative 14 and 20. And lastly negative four negative two negative 12 and negative 15. So now we can continue to work down our first column to the next term which is three and we want this term to become zero. So we can take row two and have it be equal to negative three times row one plus row two. And so now we can again rewrite our matrix so our row one remains the same. We have one negative 2105 and now our second row changes and we have negative 41 and negative five. Then our last two rows are unchanged. So we have 23 negative 14 and 20. And lastly we have negative four negative two negative 12 and negative 15. So now we can continue to move down our first column to the next term which is two and we want this to to become zero. So we can have row three B equal to negative two times row one plus row three. Now once again we rewrite our matrix. So for row one it remains the same and we have one negative 2105. Row two is also the same with 04, negative 41 negative five. And now row three becomes 07, negative 34 and 10. And our fourth row is still negative four, negative two, negative 12 and negative 15. And so we can continue to move down column one to the last term, which is negative four, which we want to become zero. So we'll have row four B equal to four times row one plus row four. So with this we can rewrite our new matrix, Our first row remains the same. We have 1 -2, 105. And then for the second row it also remains unchanged with 04 - -5. Our third row is still 07, negative 34 and 10. And now our last row becomes zero negative 10, 3 to 5. So now we can move to our second column and we move to the second row because we want this term to be one. So we take this positive four which we want to be one which means we must have row to be equal to rho two divided by four. And so with this we write down our reduced matrix. So for our first row we have one negative 2105. Now our second row is 01, negative 1 1/4 and negative 5/4. Our last two rows are still the same with 07, negative 34 and 10. And lastly zero negative 10, 3 to 5. So now once again we move down column two and we take the next term which is seven, we want seven to be zero. So we'll have row three B equal to set negative seven times row two plus row three. Now we can rewrite our matrix, Row one is still the same with 1 -2105. Row two is still 01 negative 1, 1/4 and negative 5/4. And now row three becomes 004, 9/4 and then 75 4th and then our last row is still zero negative 10, 3 to 5. So now moving down column two to our last term which is negative 10, we want negative 10 to become zero. So we have row four B equal to 10 times row one plus row four. And now we can rewrite our matrix. So for rule one we have 1 -2105. For row two we have 01 negative 1/4 and negative 5/4 four row three. We have 004 9/4 and 75 4th and now four row four we have 00 negative seven and then we have four or sorry nine halves and negative 15 halves. And so now we can move to our third column and we go to the third row which has four and we want this four to be become a positive one. So all we do is have row three B equal to rho three divided by four. And now once again rewriting our Matrix we have for our first row one negative 2105. For our second row we have 01 negative 1/4 and negative 5/4. Our third row now becomes 9/16 and 16th and now our fourth remains the same. So we have 00 negative 79 halves and negative 15 halves. So now we can move down our third column to the last term which is negative seven. We want this negative seven to become zero. So we have row four the equal to seven times row two plus row four. We then rewrite our matrix Row one is still -2105, row two remains as 01 negative 1 1/4 and negative 5/4 row three remains as 001, 9/16 and 16th. And now our third row becomes 135 divided by 16. And finally 405 divided by 16. So now the last step for reducing here is to make this last term at column four, row four a positive one. So we'll have rope four B equal to the reciprocal here which is 16 divided by times row four. So with this we can write our final reduced matrix. So our first row remains as 1 -2, 105. We then have the second roll with -1, 1/4 and -5/4. Our third row is then 9/16 and 75 16th. And that leaves us with our final roll which will become 0001 and three. So now with this we have finished reducing our matrix here. So all that is left is to recall that this left hand side here, all of these terms are the w, x, Y and z coefficient. And so now we can write four new equations and then solve them for our solutions. So beginning with row one we get our first equation which will be w minus two, X plus y is equal to five. From our second row we get our second equation which is x minus Y plus 1/ Z is equal to negative 5/4. And now from our third row we get our third equation which is why 9/16 times Z is equal to 75 16th. And the fourth row gives us a direct solution, it tells us that Z is equal to three. So now we can substitute Z into the third row equation and then solve for Y. So here we have Y plus 9/16 times Z which is three is equal to 75 16th. And now simplifying, we get that Y is equal to three. And now we can take this value for Y and the value for C. And substitute them into the second equation to solve for X. So we have x minus three plus 14 times three is equal to negative 5/4. And then solving this equation we see that X is equal to one. And now we can take the value for X and the value for Y and substitute them into the first equation to find our solution for W. So we have W -2 times X. Which is one why? Which is three is equal to five. And now solving this we see that W is equal to four. And so now we have found the solution for each variable. So we can just write them as our final solution set. So we have our final solution set of 4, 1 three which leaves us with answer choice. A thanks for watching. I hope you found this video helpful and I'll see you again next time

In Exercises 1 - 24, use Gaussian Elimination to find the complete solution to each system of equations, or show that none exists. $\begin{cases}w - 3x + y - 4z = 4 \\-2w + x + 2y = -2 \\3w - 2x + y - 6z = 2 \\-w + 3x + 2y - z = -6\end{cases}$

Key Concepts

Systems of Linear Equations

Gaussian Elimination

Row Operations and Consistency of Systems

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