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.

3. \(A = \left[ {\begin{array}{*{20}{c}}2&{ - 1}&2\\{ - 6}&0&{ - 2}\\8&{ - 1}&5\end{array}} \right],{\mathop{\rm b}\nolimits} = \left[ {\begin{array}{*{20}{c}}1\\0\\4\end{array}} \right]\)

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

Short answer

\(x = \left( { - 1,3,3} \right)\) and \(y = \left[ {\begin{array}{*{20}{c}}1\\3\\3\end{array}} \right]\)

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

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

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 4 and subtract it from row three.

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

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

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

The arithmetic values take place only in column four.

Thus, \(y = \left[ {\begin{array}{*{20}{c}}1\\3\\3\end{array}} \right]\).

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:

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

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

At row two, multiply row three by 4 and subtract it from row two. At row one, multiply row three by 2 and subtract it from row one.

\[ \sim \left[ {\begin{array}{*{20}{c}}2&{ - 1}&0&{ - 5}\\0&{ - 3}&0&{ - 9}\\0&0&1&3\end{array}} \right]\]

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

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

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

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

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

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

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