Question

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.

3. \(\left( {\begin{array}{*{20}{c}}a&0\\{3\left( {a - b} \right)}&b\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0\\3&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}a&0\\0&b\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&0\\{ - 3}&1\end{array}} \right)\)

Short answer

The required value is \({A^k} = \left( {\begin{array}{*{20}{c}}{{a^k}}&0\\{3{a^k} - 3{b^k}}&{{b^k}}\end{array}} \right)\).

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 \(\left( {\begin{array}{*{20}{c}}a&0\\{3\left( {a - b} \right)}&b\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0\\3&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}a&0\\0&b\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&0\\{ - 3}&1\end{array}} \right)\).

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

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

Therefore,

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

Find \({A^k}\)

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

Thus, \({A^k} = \left( {\begin{array}{*{20}{c}}{{a^k}}&0\\{3{a^k} - 3{b^k}}&{{b^k}}\end{array}} \right)\).