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Q5.3-2E

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 $${A^{\bf{4}}}$$.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 required value is $${A^4} = \frac{1}{{16}}\left( {\begin{array}{*{20}{c}}{151}&{90}\\{ - 225}&{ - 134}\end{array}} \right)$$.

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 $$P = \left( {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 3}&5\end{array}} \right)$$ 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 $${P^{ - 1}}$$.

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

## Step 3: Find $${A^{\bf{4}}}$$

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

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