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Linear Algebra and its Applications
Found in: Page 267
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

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

1. \(P{\bf{ = }}\left( {\begin{array}{*{20}{c}}{\bf{5}}&{\bf{7}}\\{\bf{2}}&{\bf{3}}\end{array}} \right)\), \(D{\bf{ = }}\left( {\begin{array}{*{20}{c}}{\bf{2}}&{\bf{0}}\\{\bf{0}}&{\bf{1}}\end{array}} \right)\)

The required value is \({A^4} = \left( {\begin{array}{*{20}{c}}{226}&{ - 525}\\{90}&{ - 209}\end{array}} \right)\).

See the step by step solution

Step by Step Solution

Step 1: 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 \(A\).

Step 2: Find the inverse of the invertible matrix

Consider the matrix \(P = \left( {\begin{array}{*{20}{c}}5&7\\2&3\end{array}} \right)\) and \(D = \left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right)\).

\[\begin{array}{P^{ - 1}} = \frac{1}{{5 \times 3 - 7 \times 2}}\left( {\begin{array}{*{20}{c}}3&{ - 7}\\{ - 2}&5\end{array}} \right)\\ = \frac{1}{{15 - 14}}\left( {\begin{array}{*{20}{c}}3&{ - 7}\\{ - 2}&5\end{array}} \right)\\ = \frac{1}{1}\left( {\begin{array}{*{20}{c}}3&{ - 7}\\{ - 2}&5\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}3&{ - 7}\\{ - 2}&5\end{array}} \right)\end{array}\]

Step 3: Find \({{\bf{4}}^{{\bf{th}}}}\)the power of the diagonal matrix

\[\begin{array}{l}D = \left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right)\\{D^2} = \left( {\begin{array}{*{20}{c}}{{2^2}}&0\\0&{{1^2}}\end{array}} \right)\\{D^3} = \left( {\begin{array}{*{20}{c}}{{2^3}}&0\\0&{{1^3}}\end{array}} \right)\\{D^4} = \left( {\begin{array}{*{20}{c}}{{2^4}}&0\\0&{{1^4}}\end{array}} \right)\\{D^4} = \left( {\begin{array}{*{20}{c}}{16}&0\\0&1\end{array}} \right)\end{array}\]

Step 4: Find \({A^{\bf{4}}}\)

Consider the matrix \(P = \left( {\begin{array}{*{20}{c}}5&7\\2&3\end{array}} \right)\) and \(D = \left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right)\).

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

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

Then we get:

\[\begin{array}{A^4} = P{D^4}{P^{ - 1}}\\ = \left( {\begin{array}{*{20}{c}}5&7\\2&3\end{array}} \right)\left( {\begin{array}{*{20}{c}}{16}&0\\0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}3&{ - 7}\\{ - 2}&5\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{80}&7\\{32}&3\end{array}} \right)\left( {\begin{array}{*{20}{c}}3&{ - 7}\\{ - 2}&5\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{226}&{ - 525}\\{90}&{ - 209}\end{array}} \right)\end{array}\]

Thus, \({A^4} = \left( {\begin{array}{*{20}{c}}{226}&{ - 525}\\{90}&{ - 209}\end{array}} \right)\).

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