91Ó°ÊÓ

Let \[A = \left[ {\begin{array}{*{20}{c}}1&4&{ - 1}&5\\3&7&{ - 2}&9\\{ - 2}&{ - 3}&1&{ - 4}\\{ - 1}&6&{ - 1}&7\end{array}} \right]\].

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

At row 4, add rows 1 and 4, i.e., \({R_4} \to {R_4} + {R_1}\).

\[\left[ {\begin{array}{*{20}{c}}1&4&{ - 1}&5\\3&7&{ - 2}&9\\{ - 2}&{ - 3}&1&{ - 4}\\0&{10}&{ - 2}&{12}\end{array}} \right]\]

At row 3, multiply row 1 by 2 and add it to row 3, i.e., \({R_3} \to {R_3} + 2{R_1}\).

\[\left[ {\begin{array}{*{20}{c}}1&4&{ - 1}&5\\3&7&{ - 2}&9\\0&5&{ - 1}&6\\0&{10}&{ - 2}&{12}\end{array}} \right]\]

At row 2, multiply row 1 by 3 and subtract it from row 2, i.e., \({R_2} \to {R_2} - 3{R_1}\).

\[\left[ {\begin{array}{*{20}{c}}1&4&{ - 1}&5\\0&{ - 5}&1&{ - 6}\\0&5&{ - 1}&6\\0&{10}&{ - 2}&{12}\end{array}} \right]\]

At row 4, multiply row 2 by 2 and add it to row 4, i.e., \({R_4} \to {R_4} + 2{R_2}\).

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

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

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

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

\(\left[ {\begin{array}{*{20}{c}}1&0&0&0\\3&{ - 5}&0&0\\{ - 2}&5&1&0\\{ - 1}&{10}&0&1\end{array}} \right]\).

Divide column 2 by \( - 5\).

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

The product of lower and upper triangular matrices can be written as

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

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