Answer: The solution to the system of equations is x = -21/2, y = -21/2, and z = -6.
Explanation:
To solve this system of equations, we can use the technique of Gaussian elimination or row reduction. Here are the steps:
Write the system of equations in augmented matrix form:
[ -2 2 -2 | 10 ]
[ -3 1 -1 | 5 ]
[ 2 -3 1 | -1 ]
Apply row operations to the matrix to transform it into row echelon form. The goal is to create zeros below the diagonal elements in the first column.
Multiply the first row by -1/2 and add it to the second row to eliminate the first column in the second row.
[ -2 2 -2 | 10 ]
[ 0 2/3 -1/3| -5 ]
[ 2 -3 1 | -1 ]
Multiply the first row by -1 and add it to the third row to eliminate the first column in the third row.
[ -2 2 -2 | 10 ]
[ 0 2/3 -1/3| -5 ]
[ 0 -7 3 | -21 ]
Multiply the second row by 3/2 and add it to the third row to eliminate the second column in the third row.
[ -2 2 -2 | 10 ]
[ 0 2/3 -1/3| -5 ]
[ 0 0 2 | -12 ]
The matrix is now in row echelon form. We can use back substitution to solve for the variables. Starting from the last row, we can solve for z, then substitute that value into the second equation to solve for y, and finally substitute those values into the first equation to solve for x.
From the third row, we have:
2z = -12
Therefore, z = -6.
Substituting z = -6 into the second row, we have:
(2/3)y - (1/3)(-6) = -5
Simplifying, we get:
(2/3)y + 2 = -5
Therefore, y = -21/2.
Substituting z = -6 and y = -21/2 into the first row, we have:
-2x + 2(-21/2) - 2(-6) = 10
Simplifying, we get:
-2x - 11 = 10
Therefore, x = -21/2.
Therefore, the solution to the system of equations is x = -21/2, y = -21/2, and z = -6.