91Ó°ÊÓ

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

\(\begin{array}{l}Ly = b,\\Ux = y.\end{array}\)

Here, \[L = \left[ {\begin{array}{*{20}{c}}1&0&0&0\\{ - 3}&1&0&0\\3&{ - 2}&1&0\\{ - 5}&4&{ - 1}&1\end{array}} \right],{\rm{ }}U = \,\left[ {\begin{array}{*{20}{c}}1&3&4&0\\0&3&5&2\\0&0&{ - 2}&0\\0&0&0&1\end{array}} \right],{\rm{ }}b = \left[ {\begin{array}{*{20}{c}}1\\{ - 2}\\{ - 1}\\2\end{array}} \right]\].

The matrix \(\left[ {\begin{array}{*{20}{c}}L&b\end{array}} \right]\) is shown below:

\(\left[ {\begin{array}{*{20}{c}}L&b\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0&0&0&1\\{ - 3}&1&0&0&{ - 2}\\3&{ - 2}&1&0&{ - 1}\\{ - 5}&4&{ - 1}&1&2\end{array}} \right]\)

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

At row two, multiply row one by 3 and add it to row two. At row three, multiply row one by 3 and subtract it from row three. At row four, multiply row one by 5 and add it to row four.

\( \sim \left[ {\begin{array}{*{20}{c}}1&0&0&0&1\\0&1&0&0&1\\0&{ - 2}&1&0&{ - 4}\\0&4&{ - 1}&1&7\end{array}} \right]\)

At row three, multiply row two by 2 and add it to row three. At row four, multiply row two by 4 and subtract it from row four.

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

At row four, multiply row three by 1 and add it to row four.

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

The arithmetic values take place only in column five.

Thus, \(y = \left[ {\begin{array}{*{20}{c}}1\\1\\{ - 2}\\1\end{array}} \right]\).

The matrix \(\left[ {\begin{array}{*{20}{c}}U&y\end{array}} \right]\) is shown below:

\(\left[ {\begin{array}{*{20}{c}}U&y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&3&4&0&1\\0&3&5&2&1\\0&0&{ - 2}&0&{ - 2}\\0&0&0&1&1\end{array}} \right]\)

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

At row two, multiply row four by 2 and subtract it from row two.

\( \sim \left[ {\begin{array}{*{20}{c}}1&3&4&0&1\\0&3&5&0&{ - 1}\\0&0&{ - 2}&0&{ - 2}\\0&0&0&1&1\end{array}} \right]\)

At row three, multiply row three by \(\frac{{ - 1}}{2}\).

\( \sim \left[ {\begin{array}{*{20}{c}}1&3&4&0&1\\0&3&5&0&{ - 1}\\0&0&1&0&1\\0&0&0&1&1\end{array}} \right]\)

At row two, multiply row three by 5 and subtract it from row two. At row one, multiply row three by 4 and subtract it from row one.

\( \sim \left[ {\begin{array}{*{20}{c}}1&3&0&0&{ - 3}\\0&3&0&0&{ - 6}\\0&0&1&0&1\\0&0&0&1&1\end{array}} \right]\)

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

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

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

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

Thus, \(x = \left( {3, - 2,1,1} \right)\).