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Question 1202812: Solve the system of linear equations, using the Gauss-Jordan elimination method. (If there is no solution, enter NO SOLUTION. If there are infinitely many solutions, express your answer in terms of the parameters t and/or s.)
x + 2y + z = −2
−2x − 3y − z = 3
4x + 8y + 4z = −8
(x, y, z) =
I'm very confused to how I would start to do this. Please give steps then a final answer so that I can check my work. Please and thank you.
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
You can't check your work by looking at my solution, because we can easily take very different paths to the answer.
Here is one way to get to the solution to the system of equations.
The original matrix, directly from the system of equations:
I can see immediately that this system is not going to have a unique solution, because the 3rd row of the matrix is exactly 4 times the 1st row. One of the operations we can perform is to multiply one row by a constant, so the first thing I would do in this example is multiply the 3rd row by 1/4:
Now I am ready to start on the standard sequence of steps to find the solution.
First step: get "1" in row 1 column 1 if it is not already there. Row 1 column 1 is already 1, so we don't need to do anything here.
Next step: Use the "1" in row 1 column 1 to get "0" in every other row in column 1 by multiplying the first row by a constant and adding it to another row.
Get "0" in row 2 column 1 by multiplying row 1 by 2 and adding it to row 2; get a "0" in row 3 column 1 by multiplying row 1 by -1 and adding it to row 3.
Note the row of all "0" in row 3 is because in the previous matrix rows 3 and 1 were identical.
Next step: get "1" in row 2 column 2 if it is not already there. Again that "1" is already there, so there is no work to do here.
Next step: Use the "1" in row 2 column 2 to get "0" in every other row in column 2 by multiplying the second row by a constant and adding it to another row.
Get a "0" in row 1 column 2 by multiplying row 2 by -2 and adding it to row 1.
In a general problem, the next step would be to get a "1" in row 3 column 3. But since the third row is all "0", we have gone as far as we can performing operations on the matrix.
The matrix we finish with tells us....
row 1: x-z = 0
row 2: y+z = -1
From this we get an infinite set of solutions using a parameter t:
(1) x = t
(2) x-z = 0; t-z = 0; so z = t
(3) y+z = -1; y+t = -1; so y = -t-1
The family of solutions is defined by (x,y,z) = (t,-t-1,t) where t is any constant.
To check the solution, we substitute these parametric expressions for x, y, and z in each of the original equations.
(1) x+2y+z = (t)+(-2t-2)+t = -2 correct
(2) -2x-3y-z = (-2t)+(3t+3)-t = 3 correct
(3) 4x+8y+4z = (4t)+(-8t-8)+(4t) = -8 correct
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