91Ó°ÊÓ

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

The permuted lower triangular matrix of \(A\) is \[\left[ {\begin{array}{*{20}{c}}1&0&0\\{ - 1}&1&0\\2&0&1\end{array}} \right]\], and theupper triangular matrix is \[\left[ {\begin{array}{*{20}{c}}4&3&{ - 5}\\0&{ - 2}&2\\0&0&2\end{array}} \right]\].

The inverse of the lower triangular can be calculated as shown below:

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

For row 3, multiply row 1 by 2 and subtract it from row 3, i.e., \({R_3} \to {R_3} - 2{R_1}\).

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

For row 2, add rows 2 and 1, i.e., \({R_2} \to {R_2} + {R_1}\).

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

So, the value of \({L^{ - 1}}\) is \(\left[ {\begin{array}{*{20}{c}}1&0&0\\1&1&0\\{ - 2}&0&1\end{array}} \right]\).

Theinverse of the upper triangular matrix can be calculated as shown below:

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

Divide row 2 by \( - 2\) and row 3 by 2.

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

At row 2, add rows 3 and 2, i.e., \({R_2} \to {R_2} + {R_3}\).

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

At row 1, multiply row 3 by 5 and add it to row 1, i.e., \({R_1} \to {R_1} + 5{R_3}\).

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

At row 1, multiply row 2 by 3 and subtract it from row 1, i.e., \({R_1} \to {R_1} - 3{R_2}\).

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

At row 1, divide row 1 by 4, i.e., \({R_1} \to \frac{1}{4}{R_1}\).

\(\left[ {\begin{array}{*{20}{c}}U&I\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}\,\,\,\,\begin{array}{*{20}{c}}{\frac{1}{4}}&{\frac{3}{8}}&{\frac{1}{4}}\\0&{ - \frac{1}{2}}&{\frac{1}{2}}\\0&0&{\frac{1}{2}}\end{array}} \right]\)

So, the inverse of the upper triangular matrix is \(\left[ {\begin{array}{*{20}{c}}{\frac{1}{4}}&{\frac{3}{8}}&{\frac{1}{4}}\\0&{ - \frac{1}{2}}&{\frac{1}{2}}\\0&0&{\frac{1}{2}}\end{array}} \right]\).

Substitute values in the equation \({A^{ - 1}} = {U^{ - 1}}{L^{ - 1}}\).

\(\begin{array}{c}{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{\frac{1}{4}}&{\frac{3}{8}}&{\frac{1}{4}}\\0&{ - \frac{1}{2}}&{\frac{1}{2}}\\0&0&{\frac{1}{2}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0&0\\1&1&0\\{ - 2}&0&1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{\frac{1}{4} + \frac{3}{8} - \frac{1}{2}}&{\frac{3}{8}}&{\frac{1}{4}}\\{0 - \frac{1}{2} - 1}&{ - \frac{1}{2}}&{\frac{1}{2}}\\{ - 1}&0&{\frac{1}{2}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{\frac{1}{8}}&{\frac{3}{8}}&{\frac{1}{4}}\\{ - \frac{3}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\\{ - 1}&0&{\frac{1}{2}}\end{array}} \right]\end{array}\)

So, the value of \({A^{ - 1}}\) is \(\left[ {\begin{array}{*{20}{c}}{\frac{1}{8}}&{\frac{3}{8}}&{\frac{1}{4}}\\{ - \frac{3}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\\{ - 1}&0&{\frac{1}{2}}\end{array}} \right]\).