Find an LU factorization of the matrices in Exercise 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.

\[\left[ {\begin{array}{*{20}{c}}{\bf{2}}&{ - {\bf{4}}}&{\bf{2}}\\{\bf{1}}&{\bf{5}}&{ - {\bf{4}}}\\{ - {\bf{6}}}&{ - {\bf{2}}}&{\bf{4}}\end{array}} \right]\]

Let \(A = \left[ {\begin{array}{*{20}{c}}2&{ - 4}&2\\1&5&{ - 4}\\{ - 6}&{ - 2}&4\end{array}} \right]\).

Apply row operation to reduce the matrix into an upper triangular matrix.

At row 3, multiply row 1 by 3 and add it to row 3, i.e., \({R_3} \to {R_3} + 3{R_1}\).

\(\left[ {\begin{array}{*{20}{c}}2&{ - 4}&2\\1&5&{ - 4}\\0&{ - 14}&{10}\end{array}} \right]\)

At row 2, divide row 1 by 2 and subtract it from row 2, i.e., \({R_2} \to {R_2} - \frac{1}{2}{R_1}\).

\(\left[ {\begin{array}{*{20}{c}}2&{ - 4}&2\\0&7&{ - 5}\\0&{ - 14}&{10}\end{array}} \right]\)

At row 3, multiply row 2 by 2 and add it to row 3, i.e., \({R_3} \to {R_3} + 2{R_2}\).

\(\left[ {\begin{array}{*{20}{c}}2&{ - 4}&2\\0&7&{ - 5}\\0&0&0\end{array}} \right]\)

Using matrix \(A\), matrix Lcan be written as

\(\left[ {\begin{array}{*{20}{c}}2&0&0\\1&7&0\\{ - 6}&{ - 14}&1\end{array}} \right]\).

Divide column 1 by 2 and column 2 by 7.

\(\left[ {\begin{array}{*{20}{c}}1&0&0\\{\frac{1}{2}}&1&0\\{ - 3}&{ - 2}&1\end{array}} \right]\)

The product of lower and upper triangular matricescan be written as

\(LU = \left[ {\begin{array}{*{20}{c}}1&0&0\\{\frac{1}{2}}&1&0\\{ - 3}&{ - 2}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2&{ - 4}&2\\0&7&{ - 5}\\0&0&0\end{array}} \right]\).

So, the product \(LU\) is \(\left[ {\begin{array}{*{20}{c}}1&0&0\\{\frac{1}{2}}&1&0\\{ - 3}&{ - 2}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2&{ - 4}&2\\0&7&{ - 5}\\0&0&0\end{array}} \right]\).