Solve the system of equations by finding a row echelon form for the augmented matrix.
x-2y+z=8
2x+y-3z=-9
-3x+y+3z=5
there are lots of online calculators, such as this one, which shows all the operations.
http://www.dougbabcock.com/matrix.php
Let's write the system of equations as an augmented matrix:
[ 1 -2 1 | 8 ]
[ 2 1 -3 | -9 ]
[ -3 1 3 | 5 ]
To find the row echelon form, we'll perform row operations to zero out the entries below the main diagonal of the matrix.
1. Multiply the first row by 2 and add it to the second row:
[R2 = R2 - 2R1]
[ 1 -2 1 | 8 ]
[ 0 5 -5 | -25 ]
[ -3 1 3 | 5 ]
2. Multiply the first row by -3 and add it to the third row:
[R3 = R3 + 3R1]
[ 1 -2 1 | 8 ]
[ 0 5 -5 | -25 ]
[ 0 -5 6 | 29 ]
3. Multiply the second row by -1 and add it to the third row:
[R3 = R3 + R2]
[ 1 -2 1 | 8 ]
[ 0 5 -5 | -25 ]
[ 0 0 1 | 4 ]
Now, the augmented matrix is in row echelon form. Let's write it back as a system of equations:
x - 2y + z = 8
5y - 5z = -25
z = 4
To solve for the variables, we'll do back-substitution:
From the third equation, we have z = 4. Substituting this back into the second equation, we get:
5y - 5(4) = -25
5y - 20 = -25
5y = -5
y = -1
Substituting the values of y = -1 and z = 4 into the first equation, we get:
x - 2(-1) + 4 = 8
x + 2 + 4 = 8
x + 6 = 8
x = 2
Therefore, the solution to the system of equations is x = 2, y = -1, and z = 4.
To solve the system of equations by finding the row echelon form for the augmented matrix, we can follow these steps:
Step 1: Write down the augmented matrix for the system of equations.
The augmented matrix is obtained by writing the coefficients of the variables and the constants on the right-hand side of the equations. The augmented matrix for the given system of equations is:
[1 -2 1 | 8]
[2 1 -3 | -9]
[-3 1 3 | 5]
Step 2: Perform row operations to transform the matrix into row echelon form.
To do this, we will use row operations such as swapping rows, multiplying a row by a constant, and adding rows together.
The aim is to obtain zeros below the main diagonal of the matrix.
Step 3: Start with the first column, perform row operations to make all elements below the first element in the first column equal to zero.
Perform the following operations:
Row 2 = Row 2 - 2 * Row 1 (Multiply Row 1 by 2 and subtract it from Row 2)
Row 3 = Row 3 + 3 * Row 1 (Multiply Row 1 by 3 and add it to Row 3)
The new matrix after performing these operations is:
[1 -2 1 | 8]
[0 5 -5 | -25]
[0 -5 6 | 29]
Step 4: Continue with the second column, perform row operations to make all elements below the second element in the second column equal to zero.
Perform the following operation:
Row 3 = Row 3 + Row 2 (Add Row 2 to Row 3)
The new matrix after performing this operation is:
[1 -2 1 | 8]
[0 5 -5 | -25]
[0 0 1 | 4]
Step 5: Continue with the third column, perform row operations to make all elements above the third element in the third column equal to zero.
Perform the following operation:
Row 1 = Row 1 - Row 3 (Subtract Row 3 from Row 1)
Row 2 = Row 2 + 5 * Row 3 (Multiply Row 3 by 5 and add it to Row 2)
The new matrix after performing these operations is:
[1 -2 0 | 4]
[0 5 0 | 0]
[0 0 1 | 4]
Step 6: Now we have a row echelon form. Solve for the variables.
From the row echelon form, we can directly deduce the values of the variables:
x = 4
y = 0
z = 4
So, the solution to the system of equations is x = 4, y = 0, and z = 4.