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- Jan 26, 2012

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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 \(\displaystyle \left( \begin{array}{ccc} 2 & -1 & 2 \\ -6 & 0 & -2 \\ 8 & -1 & 5 \end{array} \right)\).

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 \(\displaystyle \left( \begin{array}{ccc} 2 & -1 & 2 \\ 0 & -3 & 4 \\ 0 & 0 & 1 \end{array} \right)\).

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: \(\displaystyle \left( \begin{array}{ccc} 1 & 0 & 0 \\ a & 1 & 0 \\ b & c & 1 \end{array} \right)\).

Finally if we use the fact that \(\displaystyle \left( \begin{array}{ccc} 1 & 0 & 0 \\ a & 1 & 0 \\ b & c & 1 \end{array} \right) \left( \begin{array}{ccc} 2 & -1 & 2 \\ 0 & -3 & 4 \\ 0 & 0 & 1 \end{array} \right) = \left( \begin{array}{ccc} 2 & -1 & 2 \\ -6 & 0 & -2 \\ 8 & -1 & 5 \end{array} \right)\)

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!