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.x + y - 2z = 23x - y - 6z = - 7

Accepted Answer

Hey everyone here we are asked to use Gaussian elimination to solve the given system of equations. So first we need to write our two given equations in an augmented matrix. So if we recall an augmented matrix has this bar to represent the equal sign on the left hand side we have the coefficients for X, Y and Z. Which our our columns for this matrix. And on the right hand side is simply the right hand side of both equations. So our first row from the first equation gives us two negative 34 and negative six. And then from our second equation we get the second row which is negative 53, negative 10 is equal to three. And now from here we need to work to reduce this augmented matrix by using elementary row operations. So the first thing I notice here is we can easily turn this first value of the matrix, this two into a one. If we divide this whole row by by two and so we can let row one be equal to rho one divided by two. This gives us a reduced matrix. So our first row now becomes two divided by two is one, then we have negative three halves, two is equal to negative three, and our second row remains the same. And so now the next thing to notice is we can easily turn this negative five into a zero if we multiply this one, this first row by five, so next we can have row to be equal to five, row one plus row two. And so quickly solving this. Here we multiply row one by five. So we have five negative 15 halves, 10 is equal to negative 15 and our second row is still the same. So we have negative 53, negative 10 is equal to three. And now we can add these rows together and we see that row two is equal to five, plus five is zero, negative 15 halves, plus three, gives us negative nine halves. Then we get another zero from 10 plus negative 10. So we have zero and then finally we have equals to negative plus three, gives us negative 12. So now we can write out our further reduced matrix. So the first row is what we had earlier, so it is one negative three halves, two is equal to negative three and then our new row two is zero negative nine halves, zero and negative 12. So now we've produce this matrix as much as possible. And now from this we have two equations. So if we recall the left hand side are the x, y and z coefficient and this bar is the equal sign. So from our first row we have x minus three halves, Y plus two, Z is equal to negative three. And then we also have negative nine halves, Y is equal to negative 12. And now we can easily solve this second equation to find our value for Y. So dividing both sides by negative nine halves, we see that Y is equal to eight thirds. And now we can plug this value into our first equation that we got from the reduced matrix. So We have X -3/2 y, which is 8/ Z is equal to negative three. And now from this we can simplify and rearrange this equation to express X in terms of Z. So we can first simplify these fractions here. So multiplying, we see that these threes cancel. And so we have x minus eight halves is four plus two. Z is equal to three. And now we can transpose this to Z and the negative four to both sides to the other side. So we have X is equal to negative two, Z plus one. So now we have our solution for Y and we have our our solution for X. And we know that Z is simply we can finally write our answer as the solution set. So we have our final answer of negative two Z plus one eight thirds and Z which is our answer here for choice. C 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.x + y - 2z = 23x - y - 6z = - 7

Key Concepts

Gaussian Elimination

Row Echelon Form

Consistency of a System

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