solve using the gauss-jordan method.
x-2y+4z=7
-x+3y-z=-1
2x+y=-2
*I get through the first two columbs, but can't get the third one to the one and zero.
http://www.irs.gov/pub/irs-pdf/i1040tt.pdf
Step 2: Enter the values for your matrix, and press the "Solve" button.
three rows, four columns
Input
+1 -2 +4 +7
-1 +3 -1 -1
+2 +1 +0 -2
Add (1 * row1) to row2
1 -2 4 7
0 1 3 6
2 1 0 -2
Add (-2 * row1) to row3
1 -2 4 7
0 1 3 6
0 5 -8 -16
Add (-5 * row2) to row3
1 -2 4 7
0 1 3 6
0 0 -23 -46
Divide row3 by -23
1 -2 4 7
0 1 3 6
0 0 1 2
Add (-3 * row3) to row2
1 -2 4 7
0 1 0 0
0 0 1 2
Add (-4 * row3) to row1
1 -2 0 -1
0 1 0 0
0 0 1 2
Add (2 * row2) to row1
1 0 0 -1
0 1 0 0
0 0 1 2
http://www.gregthatcher.com/Mathematics/GaussJordan.aspx
thanks, wasn't too sure whether I could divide by -23 or not. Found the website, was really helpfull =)
To solve the given system of equations using the Gauss-Jordan method, we need to perform a series of row operations to transform the augmented matrix into reduced row echelon form. Let's go step by step:
Step 1: Write the augmented matrix corresponding to the system of equations:
```
[ 1 -2 4 | 7 ]
[ -1 3 -1 | -1 ]
[ 2 1 0 | -2 ]
```
Step 2: Perform row operations to transform the matrix into triangular form. We'll focus on making zeros below the first entry in the first column.
Multiply the second row by 1 and add it to the first row:
```
[ 0 1 3 | 6 ]
[ -1 3 -1 | -1 ]
[ 2 1 0 | -2 ]
```
Multiply the third row by 2 and subtract it from twice the first row, and add it to the second row:
```
[ 0 1 3 | 6 ]
[ 0 5 5 | 11 ]
[ 2 1 0 | -2 ]
```
Step 3: Now we'll focus on the second column. Make the entries below the second entry zero using row operations:
Multiply the third row by 5 and subtract it from five times the second row:
```
[ 0 1 3 | 6 ]
[ 0 0 -10 | -19 ]
[ 2 1 0 | -2 ]
```
Step 4: To further simplify the matrix, scale the second row by -1/10:
```
[ 0 1 3 | 6 ]
[ 0 0 1 | 19/10 ]
[ 2 1 0 | -2 ]
```
Step 5: Make the entry above the second entry zero using row operations:
Multiply the second row by -3 and add it to three times the first row:
```
[ 0 1 0 | -1/10 ]
[ 0 0 1 | 19/10 ]
[ 2 1 0 | -2 ]
```
Multiply the second row by -1 and add it to the second row:
```
[ 0 1 0 | -1/10 ]
[ 0 0 1 | 0 ]
[ 2 1 0 | -2 ]
```
Step 6: Next, we'll focus on the first column. Make the entries above and below the first entry zero using row operations:
Multiply the first row by -2 and add it to twice the third row:
```
[ 0 1 0 | -1/10 ]
[ 0 0 1 | 0 ]
[ 0 -1 0 | 6/5 ]
```
Step 7: Simplify the matrix by scaling the third row by -1:
```
[ 0 1 0 | -1/10 ]
[ 0 0 1 | 0 ]
[ 0 1 0 | -6/5 ]
```
Step 8: Finally, make the entry above the second entry zero using row operations:
Multiply the second row by -1 and add it to the first row:
```
[ 0 1 0 | 1/10 ]
[ 0 0 1 | 0 ]
[ 0 1 0 | -6/5 ]
```
Multiply the second row by -1 and add it to the third row:
```
[ 0 1 0 | 1/10 ]
[ 0 0 1 | 0 ]
[ 0 0 0 | -1/2 ]
```
Step 9: The augmented matrix is now in reduced row echelon form. Convert it back to the system of equations:
```
1x + 0y + 0z = 1/10
0x + 0y + 1z = 0
0x + 0y + 0z = -1/2
```
Therefore, the solution to the system of equations is:
```
x = 1/10
y = 0
z = 0
```
The third column being different from zero might be due to a calculation error. Please double-check the steps or share your working if you still have issues.