use a matrix to solve each system
2x + 7y +2z = -16
-5x - 8y = 17
x - 2y - 8z = -13
Look here:
https://www.wolframalpha.com/input/?i=%7B%7B2%2C7%2C2%7D%2C%7B-5%2C-9%2C0%7D%2C%7B1%2C-2%2C-8%7D%7D*%7B%7Bx%7D%2C%7By%7D%2C%7Bz%7D%7D+%3D+%7B%7B-16%7D%2C%7B17%7D%2C%7B-13%7D%7D
To do it by hand, you'll need the inverse matrix here:
https://www.wolframalpha.com/input/?i=inverse+%7B%7B2%2C7%2C2%7D%2C%7B-5%2C-9%2C0%7D%2C%7B1%2C-2%2C-8%7D%7D
To use a matrix to solve this system of equations, we can form an augmented matrix and then use row operations to transform it into row echelon form or reduced row echelon form. Here are the steps to solve the system using a matrix:
1. Write down the coefficients of the variables and the constants from each equation. Label the variables in the same order in each equation. Place them in a matrix called the coefficient matrix, and the constants in a separate column matrix called the constant matrix. The augmented matrix will be formed by combining the coefficient matrix and the constant matrix.
Coefficient matrix:
[ 2 7 2 ]
[ -5 -8 0 ]
[ 1 -2 -8 ]
Constant matrix:
[ -16 ]
[ 17 ]
[ -13 ]
Augmented matrix:
[ 2 7 2 | -16 ]
[ -5 -8 0 | 17 ]
[ 1 -2 -8 | -13 ]
2. Perform row operations to transform the augmented matrix into row echelon form or reduced row echelon form. The goal is to create zeros below the leading entries (the leftmost non-zero entry) in each row.
Using row operations, we can start by multiplying the first row by 5 and adding it to the second row:
[ 2 7 2 | -16 ]
[ 0 9 10 | -21 ]
[ 1 -2 -8 | -13 ]
Next, we can multiply the first row by -1/2 and add it to the third row:
[ 2 7 2 | -16 ]
[ 0 9 10 | -21 ]
[ 0 -11 -12 | 5 ]
Now, multiply the second row by 2/3 to simplify the coefficient in the (2,2) position:
[ 2 7 2 | -16 ]
[ 0 6 20 | -14 ]
[ 0 -11 -12 | 5 ]
Next, multiply the second row by 11/9 and add it to the third row:
[ 2 7 2 | -16 ]
[ 0 6 20 | -14 ]
[ 0 0 38 | -13 ]
3. Now, we can convert the augmented matrix into reduced row echelon form by dividing the third row by 38:
[ 2 7 2 | -16 ]
[ 0 6 20 | -14 ]
[ 0 0 1 | -13/38 ]
4. We can now read off the solutions from the resulting matrix. The variables correspond to the columns of the coefficient matrix, and the values are given by the entries in the last column.
From the last row, we have z = -13/38.
From the second row, we have 6y + 20z = -14. Substituting z = -13/38, we can solve for y:
6y + 20(-13/38) = -14
6y - 260/38 = -14
6y - 130/19 = -14
6y = -14 + 130/19
6y = (-14 * 19 + 130) / 19
6y = (-266 + 130) / 19
6y = -136 / 19
y = -136 / (6 * 19)
y = -136/114
y = -68/57
From the first row, we have 2x + 7y + 2z = -16. Substituting y = -68/57 and z = -13/38, we can solve for x:
2x + 7(-68/57) + 2(-13/38) = -16
2x - 476/57 - 26/38 = -16
2x - (476 * 38)/(57 * 38) - (26 * 57)/(38 * 57) = -16
2x - 18112/2166 - 1482/2166 = -16
2x = (-16 * 2166 + 18112 + 1482) / 2166
2x = (-34656 + 18112 + 1482) / 2166
2x = (-1462) / 2166
x = (-731) / 1083
So the solution to the system is x = -731/1083, y = -68/57, and z = -13/38.