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Q5.3-1E

Expert-verified
Found in: Page 267

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

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