Question

In Exercises \({\bf{3}}\) and \({\bf{4}}\), use the factorization \(A = PD{P^{ - {\bf{1}}}}\) to compute \({A^k}\) where \(k\) represents an arbitrary positive integer.

4. \(\left( {\begin{array}{*{20}{c}}{ - 2}&{12}\\{ - 1}&5\end{array}} \right) = \left( {\begin{array}{*{20}{c}}3&4\\1&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 1}&4\\1&{ - 3}\end{array}} \right)\)

Short answer

The required answer is \({A^k} = \left( {\begin{array}{*{20}{c}}{4 - 3\left( {{2^k}} \right)}&{ - 12 + 3\left( {{2^{k + 2}}} \right)}\\{1 - {2^k}}&{ - 3 + {2^{k + 2}}}\end{array}} \right)\).

Step-by-step solution

Write the Diagonalization Theorem

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 \(A\).

Find the inverse of the invertible matrix

Consider the given equation form \(\left( {\begin{array}{*{20}{c}}{ - 2}&{12}\\{ - 1}&5\end{array}} \right) = \left( {\begin{array}{*{20}{c}}3&4\\1&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 1}&4\\1&{ - 3}\end{array}} \right)\).

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

\({A^n} = P{D^n}{P^{ - 1}}\).

Compare the given equation form with \(A = PD{P^{ - 1}}\).

\[\left( {\begin{array}{*{20}{c}}{ - 2}&{12}\\{ - 1}&5\end{array}} \right) = \left( {\begin{array}{*{20}{c}}3&4\\1&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 1}&4\\1&{ - 3}\end{array}} \right)\]

Therefore,

\[\begin{array}{c}A = \left( {\begin{array}{*{20}{c}}{ - 2}&{12}\\{ - 1}&5\end{array}} \right)\\P = \left( {\begin{array}{*{20}{c}}3&4\\1&1\end{array}} \right)\\D = \left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right)\\{P^{ - 1}} = \left( {\begin{array}{*{20}{c}}{ - 1}&4\\1&{ - 3}\end{array}} \right)\end{array}\]

Find \({A^k}\)

\[\begin{array}{c}{A^k} = P{D^k}{P^{ - 1}}\\ = \left( {\begin{array}{*{20}{c}}3&4\\1&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{2^k}}&0\\0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 1}&4\\1&{ - 3}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{3\left( {{2^k}} \right) + 0}&{0 + 4}\\{{2^k} + 0}&{0 + 1}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 1}&{ - 3}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{3\left( {{2^k}} \right)}&4\\{{2^k}}&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 1}&4\\1&{ - 3}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 3\left( {{2^k}} \right) + 4}&{3\left( {{2^k}} \right)\left( 4 \right) - 12}\\{ - {2^k} + 1}&{4\left( {{2^k}} \right) - 3}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{4 - 3\left( {{2^k}} \right)}&{ - 12 + 3\left( {{2^{k + 2}}} \right)}\\{1 - {2^k}}&{ - 3 + {2^{k + 2}}}\end{array}} \right)\end{array}\]

Thus, \({A^k} = \left( {\begin{array}{*{20}{c}}{4 - 3\left( {{2^k}} \right)}&{ - 12 + 3\left( {{2^{k + 2}}} \right)}\\{1 - {2^k}}&{ - 3 + {2^{k + 2}}}\end{array}} \right)\).