Solve the system of equations by finding the reduced row-echelon form of the augmented matrix for the system of equations.
x + y - z = -2
2x - y + 3z = 9
x - 4y - 2z = 1
There is a handy calculator here, to check your work:
www.math.odu.edu/~bogacki/cgi-bin/lat.cgi?c=rref
3,4,6
To solve the system of equations, we can start by transforming the augmented matrix for the system into reduced row-echelon form using the Gauss-Jordan elimination method.
The augmented matrix for the system is:
[ 1 1 -1 | -2 ]
[ 2 -1 3 | 9 ]
[ 1 -4 -2 | 1 ]
To eliminate the x-coefficient in the second and third rows, we will use the first row as our pivot row and eliminate the x term from the second and third rows by multiplying the first row by the appropriate constants and subtracting it from the corresponding rows.
First, we'll eliminate the x-term in the second row:
2 * [ 1 1 -1 | -2 ] - [ 2 -1 3 | 9 ] = [ 0 3 -5 | 5 ]
Next, we'll eliminate the x-term in the third row:
[ 1 -4 -2 | 1 ] - [ 1 1 -1 | -2 ] = [ 0 -5 -1 | 3 ]
Now, to eliminate the y-term in the third row, we can multiply the second row by 5/3 and subtract it from the third row:
(5/3) * [ 0 3 -5 | 5 ] - [ 0 -5 -1 | 3 ] = [ 0 0 -10/3 | 10/3 ]
We can now simplify this row further by dividing the entire row by -10/3:
(3/10) * [ 0 0 -10/3 | 10/3 ] = [ 0 0 1 | -1 ]
Now that we have the reduced row-echelon form, we can back-substitute to find the values of x, y, and z.
From the third row, we can conclude that z = -1.
From the second row, we have 0y + 3z = 5. Substituting z = -1, we get 0y + 3(-1) = 5, which simplifies to -3 = 5. This is not a true statement, so there is no solution to the system of equations.
Thus, the system of equations is inconsistent and has no solution.
To solve the system of equations by finding the reduced row-echelon form of the augmented matrix, we first need to set up the augmented matrix for the system of equations.
The system of equations can be written as:
x + y - z = -2 (Equation 1)
2x - y + 3z = 9 (Equation 2)
x - 4y - 2z = 1 (Equation 3)
To set up the augmented matrix, we write down the coefficients of the variables and the constants from each equation. The augmented matrix is formed by arranging these coefficients and constants in a matrix form.
The augmented matrix for the given system of equations is:
[1 1 -1 -2]
[2 -1 3 9]
[1 -4 -2 1]
Now, we will perform row operations to find the reduced row-echelon form of the augmented matrix.
Step 1: Swap rows if necessary to bring a non-zero entry in the top left position.
Looking at the first column, the first entry in the second row is already non-zero, so there is no need to swap rows.
Step 2: Perform row operations to make all entries below the leading coefficient in the first column equal to zero.
Multiply the first row by -2 and add it to the second row, and multiply the first row by -1 and add it to the third row to eliminate the 1's below the leading 1 in the first column.
The updated augmented matrix is:
[1 1 -1 -2]
[0 -3 5 13]
[0 -5 -1 -1]
Step 3: Perform row operations to make all leading coefficients equal to 1.
Multiply the second row by -1/3 to make the leading coefficient in the second row equal to 1.
The updated augmented matrix is:
[1 1 -1 -2]
[0 1 -5/3 -13/3]
[0 -5 -1 -1]
Step 4: Perform row operations to make all entries above and below the leading 1's equal to zero.
Multiply the second row by -1 and add it to the first row to eliminate the 1 above the leading 1 in the second row.
The updated augmented matrix is:
[1 0 2/3 4/3]
[0 1 -5/3 -13/3]
[0 -5 -1 -1]
Step 5: Perform row operations to make all entries above and below the leading 1's equal to zero.
Multiply the second row by 5 and add it to the third row to eliminate the -5 above the leading 1 in the third row.
The updated augmented matrix is:
[1 0 2/3 4/3]
[0 1 -5/3 -13/3]
[0 0 -22/3 -22/3]
Step 6: Perform row operations to make all leading coefficients equal to 1.
Multiply the third row by -3/22 to make the leading coefficient in the third row equal to 1.
The updated augmented matrix is:
[1 0 2/3 4/3]
[0 1 -5/3 -13/3]
[0 0 1 1]
Step 7: Perform row operations to make all entries above and below the leading 1's equal to zero.
Multiply the third row by -2/3 and add it to the first row to eliminate the 2/3 above the leading 1 in the first row.
The updated augmented matrix is:
[1 0 0 2]
[0 1 -5/3 -13/3]
[0 0 1 1]
Step 8: Perform row operations to make all entries above and below the leading 1's equal to zero.
Multiply the third row by 5/3 and add it to the second row to eliminate the -5/3 above the leading 1 in the second row.
The updated augmented matrix is:
[1 0 0 2]
[0 1 0 -4]
[0 0 1 1]
Now that we obtained the reduced row-echelon form of the augmented matrix, we can interpret it as a system of equations:
x = 2
y = -4
z = 1
Therefore, the solution to the system of equations is x = 2, y = -4, z = 1.