use gauss jordan elimination method tofind solution set of4x+5y+3z=23.3x+2y+z=10.2x+5y-z=9
run-on sentences are bad not only in English.
4x+5y+3z=23
3x+2y+z=10
2x+5y-z=9
add the bottom two
add the top and 3 times the bottom, then divide by 10
5x+7y = 19
x+2y = 5
subtract 5 times the bottom from the top
-3y = -6
y = 2
so x = 1 and z = 3
To use the Gauss-Jordan elimination method to find the solution set of the system of linear equations:
1) Write the augmented matrix of the system.
2) Perform row operations to transform the augmented matrix into row-echelon form or reduced row-echelon form.
3) Interpret the resulting matrix to obtain the solution.
Let's go through these steps to solve the given system of equations:
Step 1: Write the augmented matrix
The augmented matrix of the system is obtained by writing the coefficients of the variables and the constant terms in a matrix form.
```
[ 4 5 3 | 23 ]
[ 3 2 1 | 10 ]
[ 2 5 -1 | 9 ]
```
The leftmost column represents the coefficients of `x`, the middle column represents the coefficients of `y`, the rightmost column represents the coefficients of `z`, and the rightmost column represents the constant terms.
Step 2: Perform row operations
The goal is to transform the augmented matrix into row-echelon form or reduced row-echelon form. We can achieve this by performing row operations.
R2 -> R2 - (3/4) * R1 # Multiply R1 by (3/4) and subtract from R2
R3 -> R3 - (2/4) * R1 # Multiply R1 by (2/4) and subtract from R3
The modified augmented matrix becomes:
```
[ 4 5 3 | 23 ]
[ 0 -11/4 -19/4 | -19 ]
[ 0 15/4 -7/4 | -5 ]
```
Next, multiply R2 by -4/11 to make the coefficient of y in R2 equal to 1.
```
[ 4 5 3 | 23 ]
[ 0 1 19/11 | 19/11 ]
[ 0 15/4 -7/4 | -5 ]
```
Now, perform Row 1 × (15/4), Row 2 - (15/4) × Row 1, and Row 3 - (15/4) × Row 1 to make the coefficients of x in Rows 2 and 3 zero.
```
[ 4 0 77/4 | 103/4 ]
[ 0 1 19/11 | 19/11 ]
[ 0 0 -64/11 | -64/11 ]
```
Next, multiply R3 by -11/64 to make the coefficient of z in R3 equal to 1.
```
[ 4 0 77/4 | 103/4 ]
[ 0 1 19/11 | 19/11 ]
[ 0 0 1 | 1 ]
```
Perform Row 2 - (19/11) × Row 3 and Row 1 - (77/4) × Row 3 to make the coefficients of z in Rows 1 and 2 zero.
```
[ 4 0 0 | 7/4 ]
[ 0 1 0 | 0 ]
[ 0 0 1 | 1 ]
```
Step 3: Interpret the resulting matrix
The matrix is now in reduced row-echelon form. The last column of the matrix represents the values of `x`, `y`, and `z`, respectively.
Therefore, the solution set of the given system of equations is:
x = 7/4
y = 0
z = 1