Question

1-6: Solve the equation Ax=b by using the LU factorization given for A. In Exercises 1 and 2, also solve \(Ax = b\) by ordinary row reduction.

2. \[A = \left[ {\begin{array}{*{20}{c}}4&3&{ - 5}\\{ - 4}&{ - 5}&7\\8&6&{ - 8}\end{array}} \right],{\mathop{\rm b}\nolimits} = \left[ {\begin{array}{*{20}{c}}2\\{ - 4}\\6\end{array}} \right]\]

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

Short answer

\({\mathop{\rm x}\nolimits} = \left( {\frac{1}{4},2,1} \right)\) and \(y = \left[ {\begin{array}{*{20}{c}}2\\{ - 2}\\2\end{array}} \right]\)

Step-by-step solution

Solve the equation \(Ly = b\) for y

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

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

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

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.

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

The arithmetic values take place only in the last column.

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

Use back-substitution to solve the equation \(Ux = y\) for x

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

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

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

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

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

At row two, multiply row three by 1 and add it to row two. At row one, multiply row three by 5 and add it to row one.

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

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

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

At row one, multiply row two by \( - 3\) and add it to row one.

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

At row one, multiply row one by \(\frac{1}{4}\).

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

Thus, \({\mathop{\rm x}\nolimits} = \left( {\frac{1}{4},2,1} \right)\).

Solve the equation \(Ax = b\)

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

\(\left[ {\begin{array}{*{20}{c}}A&b\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}4&3&{ - 5}&2\\{ - 4}&{ - 5}&7&{ - 4}\\8&6&{ - 8}&6\end{array}} \right]\)

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.

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

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