x-y+z=3
2x-y+z=4
x+2y-z=-1
solve in Gaus-Jordan method
To solve the given system of equations using the Gauss-Jordan method, we'll perform row operations to transform the augmented matrix into its reduced row echelon form. Here are the steps:
Step 1: Write the augmented matrix
The augmented matrix represents the coefficients of the variables and the constants. The original system can be represented as:
[1 -1 1 | 3]
[2 -1 1 | 4]
[1 2 -1 | -1]
Step 2: Perform row operations to create zeros below the main diagonal
We can eliminate the non-zero entries below the main diagonal by performing row operations.
R2 = R2 - 2R1
R3 = R3 - R1
This gives us:
[1 -1 1 | 3]
[0 1 -1 | -2]
[0 3 -2 | -4]
Step 3: Perform row operations to create zeros above the main diagonal
We can eliminate the non-zero entries above the main diagonal by performing additional row operations.
R3 = R3 - 3R2
This gives us:
[1 -1 1 | 3]
[0 1 -1 | -2]
[0 0 1 | -2]
Step 4: Perform row operations to create leading ones in each row
To create leading ones (1s) in each row, we divide each corresponding row by the leading coefficient.
R1 = R1 + R2 - R3
This gives us:
[1 -1 0 | 3]
[0 1 0 | -2]
[0 0 1 | -2]
Step 5: Read the solution
The solution to the system of equations can be read directly from the augmented matrix:
x = 3
y = -2
z = -2
Therefore, the solution to the given system of equations using the Gauss-Jordan method is:
x = 3, y = -2, and z = -2.
To solve a system of equations using the Gauss-Jordan method, you will need to perform row operations to eliminate variables and eventually obtain the values of the variables.
Let's begin by representing the given system of equations in an augmented matrix form:
1 -1 1 | 3
2 -1 1 | 4
1 2 -1 | -1
Step 1: Perform row operations to eliminate the x-coefficient in the second and third rows.
R2 = R2 - 2R1 (Replace R2 with the result)
R3 = R3 - R1 (Replace R3 with the result)
The augmented matrix becomes:
1 -1 1 | 3
0 1 -1 | -2
0 3 -2 | -4
Step 2: Perform row operations to eliminate the x-coefficient in the third row.
R3 = R3 - 3R2 (Replace R3 with the result)
The augmented matrix becomes:
1 -1 1 | 3
0 1 -1 | -2
0 0 1 | 2
Step 3: Perform row operations to obtain a diagonal matrix.
R1 = R1 + R2 (Replace R1 with the result)
R2 = R2 + R3 (Replace R2 with the result)
The augmented matrix becomes:
1 0 0 | 1
0 1 0 | 0
0 0 1 | 2
Now, the augmented matrix is in row-echelon form.
Step 4: Back-substitute to find the values of the variables.
From the augmented matrix, we obtain the following equations:
x = 1
y = 0
z = 2
Therefore, the solution to the given system of equations is x = 1, y = 0, and z = 2.