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.

\[\left[ {\begin{array}{*{20}{c}}{\bf{3}}&{ - {\bf{6}}}&{\bf{3}}\\{\bf{6}}&{ - {\bf{7}}}&{\bf{2}}\\{ - {\bf{1}}}&{\bf{7}}&{\bf{0}}\end{array}} \right]\]

Let \(A = \left[ {\begin{array}{*{20}{c}}3&{ - 6}&3\\6&{ - 7}&2\\{ - 1}&7&0\end{array}} \right]\).

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

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

\(\left[ {\begin{array}{*{20}{c}}3&{ - 6}&3\\6&{ - 7}&2\\0&5&1\end{array}} \right]\)

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

\(\left[ {\begin{array}{*{20}{c}}3&{ - 6}&3\\0&5&{ - 4}\\0&5&1\end{array}} \right]\)

At row 3, subtract row 3 from row 2, i.e., \({R_3} \to {R_3} - {R_2}\).

\(\left[ {\begin{array}{*{20}{c}}3&{ - 6}&3\\0&5&{ - 4}\\0&0&5\end{array}} \right]\)

Using matrix \(A\), matrix Lcan be written as shown below:

\(\left[ {\begin{array}{*{20}{c}}3&0&0\\6&5&0\\{ - 1}&5&5\end{array}} \right]\)

Divide column 1 by 3, column 2 by 5, and column 3 by 5.

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

The product of the lower and upper triangular matrixis

\(LU = \left[ {\begin{array}{*{20}{c}}1&0&0\\2&1&0\\{ - \frac{1}{3}}&1&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}3&{ - 6}&3\\0&5&{ - 4}\\0&0&5\end{array}} \right]\).

So, the product \(LU\) is \(\left[ {\begin{array}{*{20}{c}}1&0&0\\2&1&0\\{ - \frac{1}{3}}&1&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}3&{ - 6}&3\\0&5&{ - 4}\\0&0&5\end{array}} \right]\).