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.

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

Short answer

The LU factorization of the matrices is \(L = \left[ {\begin{array}{*{20}{c}}1&0\\{ - \frac{3}{2}}&1\end{array}} \right]\)and \(U = \left[ {\begin{array}{*{20}{c}}2&5\\0&{\frac{7}{2}}\end{array}} \right]\).

Step-by-step solution

Apply row operation to determine matrix U

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

Place the first pivot column of \(\left[ {\begin{array}{*{20}{c}}2&5\\{ - 3}&{ - 4}\end{array}} \right]\) in the first column of L after dividing the column by top pivot entry 2.

At row two, multiply row one by \(\frac{3}{2}\) and add it to row two to produce matrix U.

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

Determine matrix L

Divide the first column of \(\left[ {\begin{array}{*{20}{c}}2&5\\{ - 3}&{ - 4}\end{array}} \right]\)by pivot entry 2.

And divide row two of U by \(\frac{7}{2}\), as shown below:

\(\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{\frac{2}{2}}\\{ - \frac{3}{2}}\end{array}} \right]\,\left[ {\frac{{\frac{7}{2}}}{{\frac{7}{2}}}} \right]\\\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\, \downarrow \\\left[ {\begin{array}{*{20}{c}}1&{}\\{\frac{{ - 3}}{2}}&1\end{array}} \right]\end{array}\)

Place the result in matrix L.

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

Thus, the LU factorization of the matrices is \(L = \left[ {\begin{array}{*{20}{c}}1&0\\{ - \frac{3}{2}}&1\end{array}} \right]\)and \(U = \left[ {\begin{array}{*{20}{c}}2&5\\0&{\frac{7}{2}}\end{array}} \right]\).