Acan be written in the form \(A = LU\), where Lis an \(m \times m\) lower triangular matrix with 1s on the diagonal, and U is an \(m \times n\) echelon form of A.
When \(A = LU\), the equation \(Ax = b\) can be written as \(L\left( {Ux} \right) = {\mathop{\rm b}\nolimits} \).
Write y for \(Ux\) and find x by solving the pair of equations
\(\begin{array}{l}Ly = b,\\Ux = y.\end{array}\)
It is given that \[L = \left[ {\begin{array}{*{20}{c}}1&0&0&0\\2&1&0&0\\{ - 1}&0&1&0\\{ - 4}&3&{ - 5}&1\end{array}} \right],{\rm{ }}U = \,\left[ {\begin{array}{*{20}{c}}1&{ - 2}&{ - 4}&{ - 3}\\0&{ - 3}&1&0\\0&0&2&1\\0&0&0&1\end{array}} \right],{\rm{ }}b = \left[ {\begin{array}{*{20}{c}}1\\7\\0\\3\end{array}} \right]\].
The matrix \(\left[ {\begin{array}{*{20}{c}}L&b\end{array}} \right]\) is shown below:
\(\left[ {\begin{array}{*{20}{c}}L&b\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0&0&0&1\\2&1&0&0&7\\{ - 1}&0&1&0&0\\{ - 4}&3&{ - 5}&1&3\end{array}} \right]\)
Perform an elementary row operation to produce the row-echelon form of the matrix.
At row two, multiply row one by 2 and subtract it from row two. At row three, multiply row one by 1 and subtract it from row three. At row four, multiply row one by 4 and add it to row four.
\( \sim \left[ {\begin{array}{*{20}{c}}1&0&0&0&1\\0&1&0&0&5\\0&0&1&0&1\\0&3&{ - 5}&1&7\end{array}} \right]\)
At row four, multiply row two by 3 and subtract it from row four.
\( \sim \left[ {\begin{array}{*{20}{c}}1&0&0&0&1\\0&1&0&0&5\\0&0&1&0&1\\0&0&{ - 5}&1&{ - 8}\end{array}} \right]\)
At row four, multiply row three by 5 and add it to row four.
\( \sim \left[ {\begin{array}{*{20}{c}}1&0&0&0&1\\0&1&0&0&5\\0&0&1&0&1\\0&0&0&1&{ - 3}\end{array}} \right]\)
The arithmetic values take place only in column five.
Thus, \(y = \left[ {\begin{array}{*{20}{c}}1\\5\\1\\{ - 3}\end{array}} \right]\).
