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.2x + y - z = 23x + 3y - 2z = 3

Accepted Answer

hey everyone here we are asked to use Gaussian elimination to solve the given system of equations. So the first step is to set up our system in an augmented matrix. So first if we recall the augmented matrix gets this bar to represent the equal sign. And on the left hand side we place the X, y and z. The coefficient from the equation. So starting with our first row, we look at our first given equation and we have the values three negative 24 is equal to seven. And for our second row we get the values from our second equation which are one negative 15 is equal to 24. And now we can use elementary row operations to start reducing this matrix. So one thing we can do to get a one as our first value, we can simply swap the two rows so we can let ro one be interchanged with row two. And so doing this, our new matrix is now row one being one, negative one, negative five is equal to four and our second row is now three negative 24 is equal to seven. And now we can see that we could easily get a zero instead of this three Here if we take this one from row one and multiply it by negative three. So with this information we can let row to be equal to negative three times row one plus row two. So quickly solving this we have negative three times row one which gives us negative three positive 15 and negative 12 and then our row two is still the same as three negative 24 is equal to seven and now adding them together, we see that row two will now be negative three plus three is 03 plus negative two is 1, 15 plus four is 19 is equal to negative 12 plus seven gives us negative five. And now writing this into our newly reduced matrix, we have roll one being unchanged from before. So we have one negative one, negative five is equal to four and our road to is now 01 19 is equal to negative five. And now we have reduced these this matrix as much as possible so we can now evaluate it and get two new equations. So recalling the left hand side here, these values are the x, y and z coefficients. And this bar is the equal sign. So from our first row we see that we have x minus y minus five, Z is equal to four. And from our second row we get the equation y plus 19, Z is equal to negative five. And now we can get an equation for y by transposing this 19 Z to the right hand side. So we see that y is equal to negative 19 Z minus five. And now we can use this equation for y and plug into our first equation that we got from the reduced matrix. So doing this, we have x minus y which is this quantity of negative 19 Z minus five minus five, Z is equal to four. And now we can get an equation for X by simplifying and rearranging this equation to get X in terms of Z. So first we can factor in this negative sign. So we have x minus negative 19, gives us plus 19, Z -5 gives us plus -5, Z is equal to four. And now we can add the Z terms together. So we have X plus 14, Z plus five is equal to four. And now we can subtract five from both sides as well as subtract 14 Z from both sides so that we get X alone. So we finally have a solution here where X is equal to negative 14 Z minus one. And so now we have a solution for Y and we have a solution for X. And we also know that Z is simply equal to Z. So we can write our final solution set. So the solution set for this given system of equations is negative 14. Z minus one negative 19 Z minus five. And Z which is our answer here for 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 Eliminaion to find the complete solution to each system of equations, or show that none exists.2x + y - z = 23x + 3y - 2z = 3

Key Concepts

Gaussian Elimination

Row Echelon Form

Consistency of a System

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