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. $\begin{cases}2w + x - y = 3 \w - 3x + 2y = -4 \3w + x - 3y + z = 1 \w + 2x - 4y - z = -2\end{cases}$

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 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 given very from the system of equations. So for our first row we look at our first given equation and we get the values one negative one, negative 20 is equal to 17. And then for our second row we have three negative two, negative 10 and 85. And then for our third row we have one negative 213 and negative 34 finally four. Our fourth row we have negative two and 51. So now from here we need to reduce this augmented matrix into the upper triangle matrix using elementary row operations. So beginning with our first column we see that our first term is correctly positive one. So we can move to the next term which is three and we want this three to become a zero. So our first operation will be to have ro to become the result of negative three times row one plus row two. And with this we can rewrite our matrix to the reduced results. So row one still remains the same as one negative one, negative 2017. But our second row becomes 0150 and 34. And then our last two rows remain the same. So we have one negative 213 negative 34. And for our 4th row we have 341, and 51. So now we can continue to move down our first column to the next term which is one and we want this one to become zero. So our next operation we will have row three become the result of negative one times row one plus row three. And so once again we rewrite this result in our matrix. So row one continues to be the same as one, negative one, negative 2017, row two is still 34. And now row three becomes zero, negative 133 and negative 51. And then our last row is still 341, two and 51. So now once again we move down to the last term in column one which is three and we want three to become zero. So our next operation we will have row four become the result of negative three times row one plus row four. And now once again we rewrite the matrix including this result here. So our first row is still one negative one, negative 20 and 17. Our second row is still 0150 34. And our third row is zero negative 133 negative 51. And finally our fourth row is now 077, negative two and zero. And so now we can move to our second column while we see that at row two we have a positive one as we should. So we can move to the next term which is negative one. We want this negative one to become zero. So we'll have row three become the result of row three plus row two. And so now we can rewrite our matrix one is still one, negative one, negative 20 and 17. Row two is still 0150 and 34. And now row three is becomes 0083 and negative 17. And our final row is still unchanged. So we have 077, negative 20. And once again we moved down column to to the last term which is seven. We want this seven to become zero. So we'll have row four become the result of negative seven times row two plus row four. And with this we can rewrite our matrix. So for row one we still have one negative one, negative 20 and 17 4, row two. We have 0150, 34 4, row three. We have 0083 and negative 17. And now finally our final row is now becomes 00, negative 28 negative two and negative 238. And now we are all done with column two. So we can move on to our third column at the diagonal here at row three, we have eight. Well we want eight to become one. So all we have to do here is divide this row by eight. So we'll have row three become row three divided by eight. Once again we rewrite our matrix. So our first row remains as one negative one, negative 20 and 17. Our second row is still 34. Our third row now becomes 001, 3/8 and negative 17 8th. And then our fourth row is still the same. We have 00, negative 28 negative two and negative 238. And now from here we can move down our A third column once again. So we go to the next term which is -28. We want this negative 28 to become zero. So we'll have row four undergo the operation where we take 28 times row three and then add it to row four. Doing this, we can rewrite our matrix. So for row one we continue to have one, negative one, negative 20 and 17 for row two, we still have 0150 and four, row three. We have 001, 3/8 and negative 17 8th. And now for row four it now becomes 000, 17 halves and then negative 595 halves. And now we are all done with column three. So we can move to column four and take a look at our last term which is 17 halves. We want this 17 halves to become positive one, so we will take row four and have it become the result of this reciprocal here, which is to 17th times row four. So from here we now have our final reduced augmented matrix. So our matrix is now one negative one, negative 2017 for the first row. And then for the second row we have 34. And then we have 001, 3/8 and negative 17 8th. And finally our fourth row becomes 001, Oh sorry 0001, And now with this we have our final reduced matrix. So now if we recall the left hand side terms here or the w x, Y and d coefficients in this bar again is an equal sign. So we can now take each of these rows and write four new equations. So from the first row we get the equation, w minus x minus two, Y is equal to 17. Then from our second row we have X plus five, Y is equal to three Y. Or sorry 34. And then from our third row we get the equation, Y plus 3/8 times Z is equal to negative 17 8th. And then from our fourth row we get a direct solution for Z, we find that Z is equal to negative 35 and now we can use this solution this value for Z and substituted in our third row equation. So we have here we have Y plus 3/8 times Z, which is negative, 35 is equal to negative 8th. And now simplifying we find that y is equal to 11 and now we can use this value for Y and substitute it in the second equation here to find our value for X. So doing this, we have X plus five times Y which is 11 Is equal to 34. And now simplifying, we have a solution for X where X is equal to negative 21 and now we can use this value for X and the value for Y and substitute it into our first equation here to solve for W. So doing this, we have W -X which is negative Times Y, which is 11 is equal to 17. And now simplifying we see that W is equal to 18 and now we have found the solution for each variable. So all we have to do is rewrite this as a solution set. So with this our final answer is the solution set of 18, negative 21 11 and negative 35 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. $\begin{cases}2w + x - y = 3 \\w - 3x + 2y = -4 \\3w + x - 3y + z = 1 \\w + 2x - 4y - z = -2\end{cases}$

Key Concepts

Systems of Linear Equations

Gaussian Elimination

Row Operations and Row-Echelon Form

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