Find a LU factorization of the matrices in Exercises 7-16 (with L unit lower triangular). Note that MATLAB will usually produce a permuted LU factorization because it uses partial pivoting for numerical accuracy.

10. \(\left[ {\begin{array}{*{20}{c}}{ - 5}&3&4\\{10}&{ - 8}&{ - 9}\\{15}&1&2\end{array}} \right]\)

Consider the matrix \[A = \left[ {\begin{array}{*{20}{c}}{ - 5}&3&4\\{10}&{ - 8}&{ - 9}\\{15}&1&2\end{array}} \right]\].

Perform an elementary row operation to produce the row echelon form of the matrix.

Mark the pivot columns in matrix A and the row-echelon form of A.

At row two, multiply row one by 2 and add it to row two. At row three, multiply row one by 3 and add it to row three.

At row three, multiply row two by 5 and add it to row three.

Thus, \[U = \left[ {\begin{array}{*{20}{c}}{ - 5}&3&4\\0&{ - 2}&{ - 1}\\0&0&9\end{array}} \right]\].

The first pivot column of L is the first column of Adivided by the top pivot entry.

Divide the first column of \[\left[ {\begin{array}{*{20}{c}}{ - 5}&3&4\\{10}&{ - 8}&{ - 9}\\{15}&1&2\end{array}} \right]\]by top pivot entry \( - 5\). And at each pivot column, divide the entries by the pivot, as shown below:

\[\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{ - \frac{5}{5}}\\{ - \frac{{10}}{5}}\\{ - \frac{{15}}{5}}\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}{\frac{{ - 2}}{{ - 2}}}\\{\frac{{10}}{{ - 2}}}\end{array}} \right]\left[ {\frac{9}{9}} \right]\\\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\, \downarrow \\\left[ {\begin{array}{*{20}{c}}1&{\begin{array}{*{20}{c}}{}&{}\end{array}}&{}\\{ - 2}&1&{}\\{ - 3}&{ - 5}&1\end{array}} \right]\end{array}\]

Place the result in matrix L:

\[L = \left[ {\begin{array}{*{20}{c}}1&0&0\\{ - 2}&1&0\\{ - 3}&{ - 5}&1\end{array}} \right]\]

Thus, the LU factorization of the matrices is \[L = \left[ {\begin{array}{*{20}{c}}1&0&0\\{ - 2}&1&0\\{ - 3}&{ - 5}&1\end{array}} \right]\]and \[U = \left[ {\begin{array}{*{20}{c}}{ - 5}&3&4\\0&{ - 2}&{ - 1}\\0&0&9\end{array}} \right]\].