solve using gauss jordan
10y − 4z + w = 1
x + 4y − z + w = 2
3x + 2y + z + 2w = 5
−2x − 8y + 2z − 2w = −4
x − 6y + 3z = 1
To solve the system of equations using Gauss-Jordan elimination, we can write the augmented matrix for the given system:
⎡ 0 10 -4 1 | 1 ⎤
⎢ 1 4 -1 1 | 2 ⎥
⎢ 3 2 1 2 | 5 ⎥
⎢ -2 -8 2 -2 |-4 ⎥
⎣ 1 -6 3 0 | 1 ⎦
Now, let's perform the row operations step-by-step to transform the matrix into row-echelon form:
1. First, we switch to row 1 and row 5 to have a non-zero entry in the first column by interchanging them:
⎡ 1 -6 3 0 | 1 ⎤
⎢ 1 4 -1 1 | 2 ⎥
⎢ 3 2 1 2 | 5 ⎥
⎢ -2 -8 2 -2 |-4 ⎥
⎣ 0 10 -4 1 | 1 ⎦
2. Next, we perform row 2 - row 1, row 3 - 3 * row 1, and row 4 + 2 * row 1 to eliminate the x variable from rows 2, 3, and 4:
⎡ 1 -6 3 0 | 1 ⎤
⎢ 0 10 -4 1 | 1 ⎥
⎢ 0 20 -8 2 | 2 ⎥
⎢ 0 -4 8 -2 | 2 ⎥
⎣ 0 10 -4 1 | 1 ⎦
3. Now, we perform row 3 - 2 * row 2, and row 4 + row 2 to eliminate the x variable from rows 3 and 4 further:
⎡ 1 -6 3 0 | 1 ⎤
⎢ 0 10 -4 1 | 1 ⎥
⎢ 0 0 0 0 |-2 ⎥
⎢ 0 6 4 -1 | 3 ⎥
⎣ 0 10 -4 1 | 1 ⎦
4. Next, we perform row 4 - row 2 and row 2 - row 1 to eliminate the y variable further:
⎡ 1 -6 3 0 | 1 ⎤
⎢ 0 4 0 -3 | 4 ⎥
⎢ 0 0 0 0 |-2 ⎥
⎢ 0 6 4 -1 | 3 ⎥
⎣ 0 10 -4 1 | 1 ⎦
5. Finally, we perform row 5 - 2 * row 2, row 4 - (3/2) * row 2, and row 1 - 2 * row 2:
⎡ 1 -6 3 0 | 1 ⎤
⎢ 0 4 0 -3 | 4 ⎥
⎢ 0 0 0 0 |-2 ⎥
⎢ 0 0 4 2 | 1 ⎥
⎣ 0 0 -4 7 |-7 ⎦
Now, we have reached row-echelon form. To transform this matrix into reduced row-echelon form (also known as Gauss-Jordan form), we continue with the following steps:
6. We perform row 4 / 4 to make the leading coefficient of the fourth row equal to 1:
⎡ 1 -6 3 0 | 1 ⎤
⎢ 0 4 0 -3 | 4 ⎥
⎢ 0 0 0 0 |-2 ⎥
⎢ 0 0 1 1/2 |1/4⎥
⎣ 0 0 -4 7 |-7 ⎦
7. Next, we perform row 2 + (3/2) * row 4, and row 1 + 6 * row 4 to eliminate the z variable from rows 2 and 1:
⎡ 1 -6 0 3/2 | 4/3⎤
⎢ 0 4 0 -2/3 |10/3⎥
⎢ 0 0 0 0 |-2/7⎥
⎢ 0 0 1 1/2 |1/4⎥
⎣ 0 0 -4 7 |-7 ⎦
8. Finally, we perform row 1 + (1/2) * row 3 to eliminate the w variable from row 1:
⎡ 1 -6 0 0 | 13/3⎤
⎢ 0 4 0 0 | 14/3⎥
⎢ 0 0 0 0 |-2/7⎥
⎢ 0 0 1 1/2 | 1/4⎥
⎣ 0 0 -4 7 | -7 ⎦
The reduced row-echelon form of the augmented matrix corresponds to the system of equations:
1x - 6y = 13/3
4y = 14/3
z + (1/2)w = 1/4
-4z + 7w = -7
To obtain the solution from the system:
1. From the second equation, we can solve for y:
4y = 14/3
y = (14/3) / 4
y = 14/12
y = 7/6
2. Substituting the value of y in the first equation:
1x - 6(7/6) = 13/3
x - 7 = 13/3
x = (13/3) + 7
x = 26/3
3. From the third equation, we can solve for z:
z + (1/2)w = 1/4
z = 1/4 - (1/2)w
4. From the fourth equation, we can solve for w:
-4z + 7w = -7
7w = -7 + 4z
w = (-7 + 4z) / 7
w = (-1 + (4/7)z) / 7
Thus, the solution to the system of equations is:
x = 26/3,
y = 7/6,
z = 1/4 - (1/2)w,
w = (-1 + (4/7)z) / 7
I see more equations (5) than unknowns (4). The problem could be overdetermined.
The last equation is a simple combination of the third and fourth. So there is a solution, but you don't need the x − 6y + 3z = 1 equation to get it.
For more about Gauss Jordan, see
http://aubuchon.jsc.vsc.edu/finiteNotes7SystemsUnique.htm