use gauss jordan elimination method tofind solution set of4x+5y+3z=23.3x+2y+z=10.2x+5y-z=9
there are several handy calculators for this. You can see the details at
http://www.gregthatcher.com/Mathematics/GaussJordan.aspx
x=3/2,y=4,z=1/5
To use the Gauss-Jordan elimination method to find the solution set for the given system of equations, we will follow these steps:
Step 1: Set up an augmented matrix for the system of equations.
Step 2: Perform row operations to reduce the augmented matrix to row-echelon form.
Step 3: Continue row operations to convert the matrix to reduced row-echelon form.
Step 4: Interpret the transformed matrix to determine the solution set.
Let's begin by setting up the augmented matrix for the given system of equations:
[ 4 5 3 | 23 ]
[ 3 2 1 | 10 ]
[ 2 5 -1 | 9 ]
Now, we will perform the row operations to reduce the augmented matrix to row-echelon form. The goal is to get zeros below the leading coefficients in each row.
Step 1: Swap rows if needed to make the first entry of the first row, 4 in this case, non-zero. Since the first entry is already non-zero, we do not need to swap any rows.
[ 4 5 3 | 23 ]
[ 3 2 1 | 10 ]
[ 2 5 -1 | 9 ]
Step 2: Multiply the first row by a scalar and subtract it from the second row to eliminate the first entry in the second row.
Row 2 = Row 2 - (3/4) * Row 1
[ 4 5 3 | 23 ]
[ 0 -1/4 -5/4 | -55/4 ]
[ 2 5 -1 | 9 ]
Step 3: Multiply the first row by a scalar and subtract it from the third row to eliminate the first entry in the third row.
Row 3 = Row 3 - (1/2) * Row 1
[ 4 5 3 | 23 ]
[ 0 -1/4 -5/4 | -55/4 ]
[ 0 -5/2 -7/2 | -5/2 ]
Step 4: Multiply the second row by a scalar and subtract it from the third row to eliminate the second entry in the third row.
Row 3 = Row 3 - (5/2) * Row 2
[ 4 5 3 | 23 ]
[ 0 -1/4 -5/4 | -55/4 ]
[ 0 0 -9/4 | 11/4 ]
Now, we have row-echelon form.
Next, we will continue the row operations to transform the matrix into reduced row-echelon form.
Step 5: Multiply the third row by a scalar to make the leading coefficient of the third row equal to 1.
Row 3 = (4/9) * Row 3
[ 4 5 3 | 23 ]
[ 0 -1/4 -5/4 | -55/4 ]
[ 0 0 1 | -11/9 ]
Step 6: Multiply the third row by the necessary scalar values and subtract them from the previous rows to reduce the entries above the leading 1 in the third row to zero.
Row 2 = Row 2 + (5/4) * Row 3
Row 1 = Row 1 - 3 * Row 3
[ 4 5 0 | 10 ]
[ 0 -1/4 0 | -33/4 ]
[ 0 0 1 | -11/9 ]
Step 7: Multiply the second row by a scalar to make the leading coefficient of the second row equal to 1.
Row 2 = (-4) * Row 2
[ 4 5 0 | 10 ]
[ 0 1 0 | 33/4 ]
[ 0 0 1 | -11/9 ]
Step 8: Multiply the second row by the necessary scalar values and subtract it from the first row to reduce the entry above the leading 1 in the second row to zero.
Row 1 = Row 1 - 5 * Row 2
[ 4 0 0 | -21/4 ]
[ 0 1 0 | 33/4 ]
[ 0 0 1 | -11/9 ]
The matrix is now in reduced row-echelon form.
Interpreting the transformed matrix, we can deduce the solution set of the given system of equations:
4x = -21/4 --> x = (-21/4) / 4 = -21/16
y = 33/4
z = -11/9
Therefore, the solution set for the given system of equations is:
{(x, y, z) = (-21/16, 33/4, -11/9)}.