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Hi all,
I realize there might not be a "best method" but I want to ask if anyone has any improvements to the method taught in my class. I've looked at the Wikipedia page on this already.
Let's use the matrix [math]\left( \begin{array}{ccc} 2 & -1 & 2 \\ -6 & 0 & -2 \\ 8 & -1 & 5 \end{array} \right)[/math].
My teacher said to get the upper triangular matrix, $U$, we should simply find an echelon form of this matrix.
The quickest one for this appears to be [math]\left( \begin{array}{ccc} 2 & -1 & 2 \\ 0 & -3 & 4 \\ 0 & 0 & 1 \end{array} \right)[/math].
Now we need to find the lower triangular matrix, $L$. The way I was told to do this is by setting up another matrix in the following form: [math]\left( \begin{array}{ccc} 1 & 0 & 0 \\ a & 1 & 0 \\ b & c & 1 \end{array} \right)[/math].
Finally if we use the fact that [math]\left( \begin{array}{ccc} 1 & 0 & 0 \\ a & 1 & 0 \\ b & c & 1 \end{array} \right) \left( \begin{array}{ccc} 2 & -1 & 2 \\ 0 & -3 & 4 \\ 0 & 0 & 1 \end{array} \right) = \left( \begin{array}{ccc} 2 & -1 & 2 \\ -6 & 0 & -2 \\ 8 & -1 & 5 \end{array} \right)[/math]
then we should be able to solve for $a,b,c$. Is there a better method that anything uses or is this one just fine?
Thank you!
I realize there might not be a "best method" but I want to ask if anyone has any improvements to the method taught in my class. I've looked at the Wikipedia page on this already.
Let's use the matrix [math]\left( \begin{array}{ccc} 2 & -1 & 2 \\ -6 & 0 & -2 \\ 8 & -1 & 5 \end{array} \right)[/math].
My teacher said to get the upper triangular matrix, $U$, we should simply find an echelon form of this matrix.
The quickest one for this appears to be [math]\left( \begin{array}{ccc} 2 & -1 & 2 \\ 0 & -3 & 4 \\ 0 & 0 & 1 \end{array} \right)[/math].
Now we need to find the lower triangular matrix, $L$. The way I was told to do this is by setting up another matrix in the following form: [math]\left( \begin{array}{ccc} 1 & 0 & 0 \\ a & 1 & 0 \\ b & c & 1 \end{array} \right)[/math].
Finally if we use the fact that [math]\left( \begin{array}{ccc} 1 & 0 & 0 \\ a & 1 & 0 \\ b & c & 1 \end{array} \right) \left( \begin{array}{ccc} 2 & -1 & 2 \\ 0 & -3 & 4 \\ 0 & 0 & 1 \end{array} \right) = \left( \begin{array}{ccc} 2 & -1 & 2 \\ -6 & 0 & -2 \\ 8 & -1 & 5 \end{array} \right)[/math]
then we should be able to solve for $a,b,c$. Is there a better method that anything uses or is this one just fine?
Thank you!