Question
Find below a system of linear equations
4x – y + 2z = 13
x + 2y - 2z = 0
-x + y + z = 5
Solve the system of linear equations using the Gauss –Jordan elimination method.
4x – y + 2z = 13
x + 2y - 2z = 0
-x + y + z = 5
Solve the system of linear equations using the Gauss –Jordan elimination method.
Answers
GPT 3.5
To solve the system of linear equations using the Gauss –Jordan elimination method, we will first write down the augmented matrix of the system:
[ 4 -1 2 | 13 ]
[ 1 2 -2 | 0 ]
[-1 1 1 | 5 ]
Next, we will perform row operations to simplify the matrix:
1. Replace R2 with R2 - R1:
[ 4 -1 2 | 13 ]
[ 1 3 -4 | -13 ]
[-1 1 1 | 5 ]
2. Replace R3 with R3 + R1:
[ 4 -1 2 | 13 ]
[ 1 3 -4 | -13 ]
[ 3 0 3 | 18 ]
3. Replace R1 with R1/4:
[ 1 -1/4 1/2 | 13/4 ]
[ 1 3 -4 | -13 ]
[ 3 0 3 | 18 ]
4. Replace R2 with R2 - R1:
[ 1 -1/4 1/2 | 13/4 ]
[ 0 13/4 -9/2 | -41/4 ]
[ 3 0 3 | 18 ]
5. Replace R3 with R3 - 3R1:
[ 1 -1/4 1/2 | 13/4 ]
[ 0 13/4 -9/2 | -41/4 ]
[ 0 3/4 -3/2 | -9/4 ]
6. Replace R2 with R2 - 3/4 * R3:
[ 1 -1/4 1/2 | 13/4 ]
[ 0 0 0 | 0 ]
[ 0 3/4 -3/2 | -9/4 ]
Now we can convert the augmented matrix back into equations:
1. x - (1/4)y + (1/2)z = 13/4
2. 0 = 0
3. (3/4)y - (3/2)z = -9/4
From the second row, we can see that y is a free variable. Let y = t, where t is any real number.
Substitute y = t into equation 3:
(3/4)t - (3/2)z = -9/4
3t - 6z = -9
z = (3t + 9)/6
z = (t + 3)/2
Therefore, the solution to the system of linear equations is:
x = 13/4 + t/4
y = t
z = (t + 3)/2
where t is any real number.
[ 4 -1 2 | 13 ]
[ 1 2 -2 | 0 ]
[-1 1 1 | 5 ]
Next, we will perform row operations to simplify the matrix:
1. Replace R2 with R2 - R1:
[ 4 -1 2 | 13 ]
[ 1 3 -4 | -13 ]
[-1 1 1 | 5 ]
2. Replace R3 with R3 + R1:
[ 4 -1 2 | 13 ]
[ 1 3 -4 | -13 ]
[ 3 0 3 | 18 ]
3. Replace R1 with R1/4:
[ 1 -1/4 1/2 | 13/4 ]
[ 1 3 -4 | -13 ]
[ 3 0 3 | 18 ]
4. Replace R2 with R2 - R1:
[ 1 -1/4 1/2 | 13/4 ]
[ 0 13/4 -9/2 | -41/4 ]
[ 3 0 3 | 18 ]
5. Replace R3 with R3 - 3R1:
[ 1 -1/4 1/2 | 13/4 ]
[ 0 13/4 -9/2 | -41/4 ]
[ 0 3/4 -3/2 | -9/4 ]
6. Replace R2 with R2 - 3/4 * R3:
[ 1 -1/4 1/2 | 13/4 ]
[ 0 0 0 | 0 ]
[ 0 3/4 -3/2 | -9/4 ]
Now we can convert the augmented matrix back into equations:
1. x - (1/4)y + (1/2)z = 13/4
2. 0 = 0
3. (3/4)y - (3/2)z = -9/4
From the second row, we can see that y is a free variable. Let y = t, where t is any real number.
Substitute y = t into equation 3:
(3/4)t - (3/2)z = -9/4
3t - 6z = -9
z = (3t + 9)/6
z = (t + 3)/2
Therefore, the solution to the system of linear equations is:
x = 13/4 + t/4
y = t
z = (t + 3)/2
where t is any real number.