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.
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.
1 answer
1 answer