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.

8. \(\left[ {\begin{array}{*{20}{c}}6&9\\4&5\end{array}} \right]\)

Short answer

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

Step-by-step solution

Apply the row operation to determine matrix U

Place the first pivot column of \(\left[ {\begin{array}{*{20}{c}}6&9\\4&5\end{array}} \right]\) in the first column of L after dividing the column by top pivot entry 6. Then at row two, multiply row one by \(\frac{2}{3}\) and subtract it from row two to produce matrix U.

\[A = \left[ {\begin{array}{*{20}{c}}6&9\\4&5\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}6&9\\0&{ - 1}\end{array}} \right] = U\]

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 \(\left[ {\begin{array}{*{20}{c}}6&9\\4&5\end{array}} \right]\)by pivot entry 6.

Also, divide row two of matrix U by \( - 1\), as shown below.

\[\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{\frac{6}{6}}\\{\frac{4}{6}}\end{array}} \right]\,\left[ {\frac{{ - 1}}{{ - 1}}} \right]\\\,\,\, \downarrow \,\,\,\,\, \downarrow \\\left[ {\begin{array}{*{20}{c}}1&{}\\{\frac{2}{3}}&1\end{array}} \right]\end{array}\]

Place the result in L.

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

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