Use gauss Jordan elimination method to find solution set of
4x+5y+3z=23
3x+2y+z=10
2x+5y-z=9
enter your data at this site for all the details:
http://www.gregthatcher.com/Mathematics/GaussJordan.aspx
To solve the given system of linear equations using the Gauss-Jordan elimination method, we will perform row operations on an augmented matrix until it is in reduced row-echelon form. This will allow us to determine the solution set.
Step 1: Initialize the augmented matrix
Start by setting up the augmented matrix for the given system of equations:
[ 4 5 3 | 23 ]
[ 3 2 1 | 10 ]
[ 2 5 -1 | 9 ]
Step 2: Apply row operations
Now, we will perform row operations to transform the augmented matrix into reduced row-echelon form.
R2 = R2 - (3/4)R1
R3 = R3 - (1/2)R1
The resulting matrix will be:
[ 4 5 3 | 23 ]
[ 0 -5/4 -1/4 | -4/4 ]
[ 0 5/2 -7/2 | -5/2 ]
Next, we will apply row operations to eliminate the remaining non-zero entries below and above the leading entries in each row.
R1 = R1 + (5/4)R2
R3 = R3 + (5/2)R2
The matrix becomes:
[ 4 0 2 | 21 ]
[ 0 -5/4 -1/4 | -4/4 ]
[ 0 0 -4/2 | -6/2 ]
After that, we need to convert the leading entries to 1 by dividing the corresponding rows:
R1 = (1/4)R1
R2 = (-4/5)R2
R3 = (-2/4)R3
The matrix now looks like:
[ 1 0 1/2 | 21/4 ]
[ 0 1 1/5 | 4/5 ]
[ 0 0 1 | 3 ]
Finally, we perform additional row operations to eliminate the entries above the leading 1s:
R1 = R1 - (1/2)R3
R2 = R2 - (1/5)R3
This gives us the final matrix:
[ 1 0 0 | 3/4 ]
[ 0 1 0 | 1 ]
[ 0 0 1 | 3 ]
Step 3: Interpret the results
The reduced row-echelon form of the augmented matrix represents the solution set of the original system of linear equations. From the final matrix, we can determine the solution:
x = 3/4,
y = 1,
z = 3.
Therefore, the solution set of the given system of equations is {(3/4, 1, 3)}.