Answer:
(x1, x2, x3) = (4, 5, -2)
Explanation:
You want to solve this system of equations using Gauss elimination:
- x -y +2z = -5
- 3x +4y +15z = 2
- 2x -y +z = 1
Gauss elimination
The method of Gauss elimination eliminates the variables from the equations one by one until the final equation has only one variable. Essentially, the coefficient matrix is put in row-echelon (upper triangular) form.
The solution to the final equation is then back-substituted into the other equations. That process is repeated until all variable values have been found.
Eliminate x
Subtract 3 times the first equation from the second, and 2 times the first equation from the third.
x - y + 2z = -5
0x +7y +9z = 17
0x +y -3z = 11
Eliminate y
It is convenient to swap the last two equations so the coefficient of y on the diagonal is 1.
x - y + 2z = -5
0x +y -3z = 11
0x +7y +9z = 17
Now, we will subtract 7 times the second equation from the third:
x -y +2z = -5
0x +y -3z = 11
0x +0y +30z = -60
Find z
The solution to the last equation is ...
z = -60/30 = -2
Find y
Substituting for z in the second equation, we find y to be ...
y -3(-2) = 11
y = 5 . . . . . . . . subtract 6
Find x
Substituting for z and y in the first equation, we find x to be ...
x -(5) +2(-2) = -5
x = 4 . . . . . . . . . add 9
The solution to the system of equations is (x1, x2, x3) = (4, 5, -2).
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Additional comment
It is painful to write subscripts on numerous instances of a variable, so we have used x, y, z for x1, x2, x3. The working is the same.
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