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 + 2y + 3z = 5
y - 5z = 0

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 system of equations in an augmented matrix. Now if we recall an augmented matrix has this bar here to represent the equal sign. And on the left hand side we place the X. Y. And the coefficients from each equation. So starting with our first row, we take the values from the first equation. So we have two negative 79 is equal to negative four. And for our second row we look at the second equation, so we have 013 is equal to zero. And now from here we just have to use elementary row operations to work to reduce this matrix. So one thing I notice is that we can easily change this to this first value of the matrix into a one by dividing the row by one. So we can let row one be equal to rho one divided by two. And from here we can reduce this matrix. So our one is now two divided by two, which again gives one negative seven divided by two. So we have negative seven halves and then we have nine halves is equal to negative two. And our second row just stays the same. So we have 0130. And now this matrix is reduced as much as possible. So we can now evaluate it and see we have two equations. So recalling that the left hand side are the X, Y and the coefficients. Our first row gives us the equation x minus seven halves, Y plus nine. Z is equal to negative two. And our second equation or second row gives us the equation, Y plus three, Z is equal to zero. So now we can simply solve this second equation here for Y. In terms of Z. So we see that Y is equal to negative three Z. After we transpose this three Z to the right hand side. And so now we can plug in this expression for Y into our first equation. So from here we have x minus seven halves Y. Which we found as negative three. Z. Z is equal to -2. And now we can just simply arrange this equation here to find our expression for X. So doing this we can first add these Z terms together. So we end up getting X plus 15. Z is equal to negative two. And now we can subtract 15 Z from both sides to get X alone. So we see that X is equal to negative 15, Z minus two. So now we have our expression our solution for Y and we have our solution for X. And we know that our solution for Z is just Z is simply equal to Z. So now we our final solution set which we found as negative 15 Z minus two, negative three. Z. And Z which is our answer here for choice. The 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 + 2y + 3z = 5 y - 5z = 0

Key Concepts

Gaussian Elimination

Row Echelon Form

Systems of Linear Equations

Watch next