Answer:
(2, 7, 1)
Explanation:
We have three equations, and using Gauss-Jordan Elimination, we can solve for x, y, and z
3x + y - 2z = 11
4x - 2y + z = -5
x + 5y - 4z = 33
We can start by taking out the z from all rows except one. To do this, we can work with the second row. I chose the second row because -5 is small and easy to add up with other numbers, and z has no coefficient in this row.
We can add 2 times the second row to the first row and 4 times the second row to the third row to get
11x - 3y = 1
4x - 2y + z = -5
17x -3y = 13
We then have the first and third rows having two variables. Since the y coefficients are the same, we can eliminate the y by adding the negative of the first row to the third row. Our result is then
11x - 3y = 1
4x - 2y + z = -5
6x = 12
From the third row, we can gather that x= 2. We can then plug that into the first row to get
22 -3y = 1
subtract 22 from both sides
-3y = -21
divide both sides by -3
y = 7
We can then plug our x and y values into the second row to get
4(2) - 2(7) + z = -5
8 - 14 + z = -5
-6 + z = -5
add 6 to both sides
z = 1
Our answer is thus (2, 7, 1)