Question

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.

9. \(\left[ {\begin{array}{*{20}{c}}3&{ - 1}&2\\{ - 3}&{ - 2}&{10}\\9&{ - 5}&6\end{array}} \right]\)

Short answer

The LU factorization of the matrices is \[L = \left[ {\begin{array}{*{20}{c}}1&0&0\\{ - 1}&1&0\\3&{\frac{2}{3}}&1\end{array}} \right]\]and \(U = \left[ {\begin{array}{*{20}{c}}3&{ - 1}&2\\0&1&{ - 4}\\0&0&{ - 8}\end{array}} \right]\).

Step-by-step solution

Apply the row operation to determine matrix U

Consider the matrix \[A = \left[ {\begin{array}{*{20}{c}}3&{ - 1}&2\\{ - 3}&{ - 2}&{10}\\9&{ - 5}&6\end{array}} \right]\].

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

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

At row two, add rows one and two. At row three, multiply row one by 3 and subtract it from row three.

At row two, multiply row two by \( - \frac{1}{3}\).

\[ \sim \left[ {\begin{array}{*{20}{c}}3&{ - 1}&2\\0&1&{ - 4}\\0&{ - 2}&0\end{array}} \right]\]

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

Thus, \(U = \left[ {\begin{array}{*{20}{c}}3&{ - 1}&2\\0&1&{ - 4}\\0&0&{ - 8}\end{array}} \right]\).

Determine matrix L

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

Divide the first column of \[\left[ {\begin{array}{*{20}{c}}3&{ - 1}&2\\{ - 3}&{ - 2}&{10}\\9&{ - 5}&6\end{array}} \right]\]by pivot entry 3. And at each pivot column, divide the entries by the pivot, as shown below:

\[\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{\frac{3}{3}}\\{ - \frac{3}{3}}\\{\frac{9}{3}}\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}{\frac{{ - 3}}{{ - 3}}}\\{\frac{{ - 2}}{{ - 3}}}\end{array}} \right]\left[ {\frac{{ - 8}}{{ - 8}}} \right]\\\,\,\, \downarrow \,\,\,\,\,\,\,\, \downarrow \,\,\,\,\,\, \downarrow \\\left[ {\begin{array}{*{20}{c}}1&{\begin{array}{*{20}{c}}{}&{}\end{array}}&{}\\{ - 1}&1&{}\\3&{\frac{2}{3}}&1\end{array}} \right]\end{array}\]

Place the result in matrix L:

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

Thus, the LU factorization of the matrices is \[L = \left[ {\begin{array}{*{20}{c}}1&0&0\\{ - 1}&1&0\\3&{\frac{2}{3}}&1\end{array}} \right]\]and \(U = \left[ {\begin{array}{*{20}{c}}3&{ - 1}&2\\0&1&{ - 4}\\0&0&{ - 8}\end{array}} \right]\).