91Ó°ÊÓ

It is known that Acan be written in the form \(A = LU\), where Lis a \(m \times m\) lower triangular matrix with 1s on the diagonal. Also, 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, \(L = \left[ {\begin{array}{*{20}{c}}1&0&0\\{ - 1}&1&0\\2&{ - 5}&1\end{array}} \right]{\rm{, }}U = \,\left[ {\begin{array}{*{20}{c}}3&{ - 7}&{ - 2}\\0&{ - 2}&{ - 1}\\0&0&{ - 1}\end{array}} \right]{\rm{, }}b = \left[ {\begin{array}{*{20}{c}}{ - 7}\\5\\2\end{array}} \right]\).

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

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

At row two, multiply row one by 1 and add it to row two. At row three, multiply row one by 2 and subtract it from row three.

\( \sim \left[ {\begin{array}{*{20}{c}}1&0&0&{ - 7}\\0&1&0&{ - 2}\\0&{ - 5}&1&{16}\end{array}} \right]\)

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

\( \sim \left[ {\begin{array}{*{20}{c}}1&0&0&{ - 7}\\0&1&0&{ - 2}\\0&0&1&6\end{array}} \right]\)

The arithmetic values take place only in last column.

Thus, \(y = \left[ {\begin{array}{*{20}{c}}{ - 7}\\{ - 2}\\6\end{array}} \right]\).

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

\(\left[ {\begin{array}{*{20}{c}}U&y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3&{ - 7}&{ - 2}&{ - 7}\\0&{ - 2}&{ - 1}&{ - 2}\\0&0&{ - 1}&6\end{array}} \right]\)

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

At row three, multiply row three by \( - 1\).

\[ \sim \left[ {\begin{array}{*{20}{c}}3&{ - 7}&{ - 2}&{ - 7}\\0&{ - 2}&{ - 1}&{ - 2}\\0&0&1&{ - 6}\end{array}} \right]\]

At row two, multiply row three by 1 and add it to row two. At row one, multiply row three by 2 and add it to row 1.

\[ \sim \left[ {\begin{array}{*{20}{c}}3&{ - 7}&0&{ - 19}\\0&{ - 2}&0&{ - 8}\\0&0&1&{ - 6}\end{array}} \right]\]

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

\[ \sim \left[ {\begin{array}{*{20}{c}}3&{ - 7}&0&{ - 19}\\0&1&0&4\\0&0&1&{ - 6}\end{array}} \right]\]

At row one, multiply row two by \(7\) and add it to row one.

\[ \sim \left[ {\begin{array}{*{20}{c}}3&0&0&9\\0&1&0&4\\0&0&1&{ - 6}\end{array}} \right]\]

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

\[ \sim \left[ {\begin{array}{*{20}{c}}1&0&0&3\\0&1&0&4\\0&0&1&{ - 6}\end{array}} \right]\]

Thus, \(x = \left( {3,4, - 6} \right)\).

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

\(\left[ {\begin{array}{*{20}{c}}A&b\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3&{ - 7}&{ - 2}&{ - 7}\\{ - 3}&5&1&5\\6&{ - 4}&0&2\end{array}} \right]\)

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

At row two, multiply row one by 1 and add it to row two. At row three, multiply row one by 2 and subtract it from row three.

\( \sim \left[ {\begin{array}{*{20}{c}}3&{ - 7}&{ - 2}&{ - 7}\\0&{ - 2}&{ - 1}&{ - 2}\\0&{10}&4&{16}\end{array}} \right]\)

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

\( \sim \left[ {\begin{array}{*{20}{c}}3&{ - 7}&{ - 2}&{ - 7}\\0&{ - 2}&{ - 1}&{ - 2}\\0&0&{ - 1}&6\end{array}} \right]\)

Thus, the row reduction is the same as\(\left[ {\begin{array}{*{20}{c}}U&y\end{array}} \right]\), yielding the same result.