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

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

It is given that \[L = \left[ {\begin{array}{*{20}{c}}1&0&0\\{\frac{1}{2}}&1&0\\{\frac{3}{2}}&{ - 5}&1\end{array}} \right]\,{\rm{, }}U = \left[ {\begin{array}{*{20}{c}}2&{ - 2}&4\\0&{ - 2}&{ - 1}\\0&0&{ - 6}\end{array}} \right],{\rm{ }}b = \left[ {\begin{array}{*{20}{c}}0\\{ - 5}\\7\end{array}} \right]\].

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

\(\left[ {\begin{array}{*{20}{c}}L&b\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0&0&0\\{\frac{1}{2}}&1&0&{ - 5}\\{\frac{3}{2}}&{ - 5}&1&7\end{array}} \right]\)

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

At row two, multiply row one by \(\frac{1}{2}\) and subtract it from row two. At row three, multiply row one by \(\frac{1}{2}\) and subtract it from row three.

\( \sim \left[ {\begin{array}{*{20}{c}}1&0&0&0\\0&1&0&{ - 5}\\0&{ - 5}&1&7\end{array}} \right]\)

At row three, multiply row two by 5 and add it to row three.

\( \sim \left[ {\begin{array}{*{20}{c}}1&0&0&0\\0&1&0&{ - 5}\\0&0&1&{ - 18}\end{array}} \right]\)

The arithmetic values take place only in column four.

Thus, \(y = \left[ {\begin{array}{*{20}{c}}0\\{ - 5}\\{ - 18}\end{array}} \right]\).

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

\(\left[ {\begin{array}{*{20}{c}}U&y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2&{ - 2}&4&0\\0&{ - 2}&{ - 1}&{ - 5}\\0&0&{ - 6}&{ - 18}\end{array}} \right]\)

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

At row three, multiply row three by \( - \frac{1}{6}\).

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

At row two, multiply row three by \(1\) and add it to row two. At row one, multiply row three by 4 and subtract it from row one.

\[ \sim \left[ {\begin{array}{*{20}{c}}2&{ - 2}&0&{ - 12}\\0&{ - 2}&0&{ - 2}\\0&0&1&3\end{array}} \right]\]

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

\[ \sim \left[ {\begin{array}{*{20}{c}}2&{ - 2}&0&{ - 12}\\0&1&0&1\\0&0&1&3\end{array}} \right]\]

At row one, multiply row two by \(2\) and add it to row one.

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

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

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

Thus, \(x = \left( { - 5,1,3} \right)\).