Express the system as an augmented matrix and solve using Gaussian elimination.
x=+2y+z=3
2y+3z =2
-x+2z = 1
hop on over to
http://www.gregthatcher.com/Mathematics/GaussJordan.aspx
and you can see all the steps involved in the solution
Thank you.
To solve the system of equations using Gaussian elimination, we need to express the system as an augmented matrix and then perform row operations to transform it into row-echelon form.
First, let's represent the system of equations as an augmented matrix. Each row of the matrix corresponds to the coefficients of the variables in each equation, and the last column represents the constant terms.
The augmented matrix is:
[ 1 2 1 | 3 ]
[ 0 2 3 | 2 ]
[-1 0 2 | 1 ]
Now, let's perform row operations to transform the matrix into row-echelon form:
1. R2 = R2 - 2R1
[ 1 2 1 | 3 ]
[ 0 -2 1 | -4 ]
[-1 0 2 | 1 ]
2. R3 = R3 + R1
[ 1 2 1 | 3 ]
[ 0 -2 1 | -4 ]
[ 0 2 3 | 4 ]
3. R3 = R3 + R2
[ 1 2 1 | 3 ]
[ 0 -2 1 | -4 ]
[ 0 0 4 | 0 ]
Now, the matrix is in row-echelon form. Let's perform back substitution to solve for the variables:
From the third row, we have: 0z = 0, which indicates that z can have any value.
From the second row, we have: -2y + z = -4
Simplifying, we get: -2y = -4 - z
From the first row, we have: x + 2y + z = 3
Substituting the value of -2y from the second row, we get: x - 4 - z + z = 3
Simplifying, we get: x = 7 - z
Therefore, the solution to the system of equations is:
x = 7 - z
y = (-4 - z)/2
z can have any value.