chiqititaa
New member
- Joined
- Mar 12, 2024
- Messages
- 1
Hi all!
So I'm trying to wrap my head around gaussian elimination and found myself stuck on the very first problem I encountered, I was hoping to get some help to clear it up.
Solve the system of equations:
[math]\begin{cases} x_1+2x_2-x_3 = 6\\ 2x_1-x_2+x_3 = 1\\ -x_1+x_2-2x_3 = 3 \end{cases}[/math]
After writing this as a augmented matrix I multiply row 1 with -2 and 1, and add that to row 2 and 3 respectively, giving me:
[math]\left( \begin{array}{ccc|c} 1 & 2 & -3 & 6 \\ 0 & -5 & 3 & -11\\ 0 & 3 & -3 & 9 \end{array} \right)[/math]
I then multiply row 2 with -1/5 and add it to itself to make -5 (middle, row 2) into 1. I also multiply row 3 with -1 and add it to itself to make 3 (middle, row 3) into 0. This gives me the matrix:
[math]\left( \begin{array}{ccc|c} 1 & 2 & -1 & 6\\ 0 & 1 & -3/5 & -11/5\\ 0 & 0 & 3 & -9 \end{array} \right)[/math]
When trying to solve for x3 though, I get x3=-3 when it should be x3=-2. I would be very grateful for any help, thank you!
So I'm trying to wrap my head around gaussian elimination and found myself stuck on the very first problem I encountered, I was hoping to get some help to clear it up.
Solve the system of equations:
[math]\begin{cases} x_1+2x_2-x_3 = 6\\ 2x_1-x_2+x_3 = 1\\ -x_1+x_2-2x_3 = 3 \end{cases}[/math]
After writing this as a augmented matrix I multiply row 1 with -2 and 1, and add that to row 2 and 3 respectively, giving me:
[math]\left( \begin{array}{ccc|c} 1 & 2 & -3 & 6 \\ 0 & -5 & 3 & -11\\ 0 & 3 & -3 & 9 \end{array} \right)[/math]
I then multiply row 2 with -1/5 and add it to itself to make -5 (middle, row 2) into 1. I also multiply row 3 with -1 and add it to itself to make 3 (middle, row 3) into 0. This gives me the matrix:
[math]\left( \begin{array}{ccc|c} 1 & 2 & -1 & 6\\ 0 & 1 & -3/5 & -11/5\\ 0 & 0 & 3 & -9 \end{array} \right)[/math]
When trying to solve for x3 though, I get x3=-3 when it should be x3=-2. I would be very grateful for any help, thank you!