My name is Michael and I'm in the 11th grade. I have a Math question that I can't solve. The problem is system of equations that I need to do in augmented matrix form, find the row echelon form, and solve it by using back substitution.
2x + 3y + 7z =13
3x + 2y - 5z = -22
5x+ 7y - 3z = -28
Please respond. Thank you!
Hi Michael,
There are many ways to arrive at a row echelon form. I try do do as much as I can without introducing fractions.
Multiply the first row by -1 and add it to the second row and then multiply the first row by -2 and add it to the third row. Hence
| 2 | 3 | 7 | 13 |
| 3 | 2 | -5 | -33 |
| 5 | 7 | -3 | -28 |
becomes
| 2 | 3 | 7 | 13 |
| 1 | -1 | -12 | -35 |
| 1 | 1 | -17 | -54 |
Now multiply the second row by -1 and add it to the third row and then multiply the second row by -2 and add it to the first row. The matrix then becomes
| 0 | 5 | 31 | 83 |
| 1 | -1 | -12 | -35 |
| 0 | 2 | -5 | -19 |
Next multiply the third row by -2 and add it to the first row to get
| 0 | 1 | 41 | 121 |
| 1 | -1 | -12 | -35 |
| 0 | 2 | -5 | -19 |
Multiply the first row by -2 and add to the third row to get
| 0 | 1 | 41 | 121 |
| 1 | -1 | -12 | -35 |
| 0 | 0 | -87 | -261 |
Finally interchange the first and second rows and multiply the third row by -87 to get the row echilon form
| 1 | -1 | -12 | -35 |
| 0 | 1 | 41 | 121 |
| 0 | 0 | 1 | 3 |
Thus z = 3. Back substitute to get y and x and then VERIFY that you have the correct solution.
Penny