Question: In Exercises \({\bf{1}}\) and \({\bf{2}}\), let \(A = PD{P^{ - {\bf{1}}}}\) and compute \({A^{\bf{4}}}\).

2. \(P{\bf{ = }}\left( {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 3}&5\end{array}} \right)\), \(D{\bf{ = }}\left( {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{0}}\\{\bf{0}}&{\frac{{\bf{1}}}{{\bf{2}}}}\end{array}} \right)\)

The Diagonalization Theorem:An \(n \times n\) matrix \(A\) is diagonalizable if and only if \(A\) has \(n\) linearly independent eigenvectors. As \(A = PD{P^{ - 1}}\) which has \(D\) a diagonal matrix if and only if the columns of \(P\) are \(n\) linearly independent eigenvectors of \(A\).

Consider the matrix \(P = \left( {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 3}&5\end{array}} \right)\)and \(D = \left( {\begin{array}{*{20}{c}}1&0\\0&{\frac{1}{2}}\end{array}} \right)\).

As it is given that \(A = PD{P^{ - 1}}\)than by using the formula for \({n^{th}}\) power we get:

\[{A^n} = P{D^n}{P^{ - 1}}\].

Compute \({P^{ - 1}}\).

\[\begin{array}{P^{ - 1}} = \frac{1}{{2 \times 5 - \left( { - 3} \right) \times \left( { - 3} \right)}}\left( {\begin{array}{*{20}{c}}5&3\\3&2\end{array}} \right)\\ = \frac{1}{{10 - 9}}\left( {\begin{array}{*{20}{c}}5&3\\3&2\end{array}} \right)\\ = \frac{1}{1}\left( {\begin{array}{*{20}{c}}5&3\\3&2\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}5&3\\3&2\end{array}} \right)\end{array}\]

\[\begin{array}{A^4} = P{D^4}{P^{ - 1}}\\ = \left( {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 3}&5\end{array}} \right){\left( {\begin{array}{*{20}{c}}1&0\\0&{\frac{1}{2}}\end{array}} \right)^4}\left( {\begin{array}{*{20}{c}}5&3\\3&2\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 3}&5\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{{\left( 1 \right)}^4}}&0\\0&{{{\left( {\frac{1}{2}} \right)}^4}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}5&3\\3&2\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 3}&5\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&0\\0&{\frac{1}{{16}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}5&3\\3&2\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{2 + 0}&{0 - \frac{3}{{16}}}\\{ - 3 + 0}&{0 + \frac{5}{{16}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}5&3\\3&2\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}2&{ - \frac{3}{{16}}}\\{ - 3}&{\frac{5}{{16}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}5&3\\3&2\end{array}} \right)\end{array}\]

Further,

\[\begin{array}{c}{A^4} = \left( {\begin{array}{*{20}{c}}2&{ - \frac{3}{{16}}}\\{ - 3}&{\frac{5}{{16}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}5&3\\3&2\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{10 - \frac{9}{{16}}}&{6 - \frac{6}{{16}}}\\{ - 15 + \frac{{15}}{{16}}}&{ - 9 + \frac{{10}}{{16}}}\end{array}} \right)\\ = \frac{1}{{16}}\left( {\begin{array}{*{20}{c}}{151}&{90}\\{ - 225}&{ - 134}\end{array}} \right)\end{array}\]

Thus, \({A^4} = \frac{1}{{16}}\left( {\begin{array}{*{20}{c}}{151}&{90}\\{ - 225}&{ - 134}\end{array}} \right)\).