Q4E For the matrices A in Exercises ... [FREE SOLUTION]

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Linear Algebra With Applications

Otto Bretscher

$Math Studyset 91影视 Explanations$ Math

5th Edition 路 442 pages

ISBN: 9780321796974

Chapter 7: Q4E (page 355)

For the matrices A in Exercises 1 through 12, find closed formulas for $A^{t}$ , where t is an arbitrary positive integer. Follow the strategy outlined in Theorem 7.4.2 and illustrated in Example 2. In Exercises 9 though 12, feel free to use technology.
4. $A = [\begin{matrix} 4 & - 2 \\ 1 & 1 \end{matrix}]$

Short Answer

Expert verified

The closed formulas for $A^{t} = [\begin{matrix} - 2 + 2 脳 3^{t} & 2^{t - 1} - 2 脳 3^{t} \\ - 2 + 3^{t} & 2^{t + 1} - 3^{t} \end{matrix}]$

Step by step solution

Finding the diagonalization matrix and eigenvalues.

The given matrix is $A = [\begin{matrix} 4 & - 2 \\ 1 & 1 \end{matrix}]$ .We need to find the eigenvalue for the given matrix.

$\det (A - 位 I) = 0 |\begin{matrix} 4 - 位 & - 2 \\ 1 & 1 - 位 \end{matrix}| = 0 (4 - 位) (1 - 位) + 2 = 0 位^{2} - 5 位 + 6 = 0$

$(位 - 2) (位 - 3) = 0 位_{1} = 2, 位_{2} = 3$

Since, there are two distinct real eigenvalues, the $2 脳 2$ matrix is diagonalizable and its diagonalization will be

$B = [\begin{matrix} 2 & 0 \\ 0 & 3 \end{matrix}]$

$B^{t} = [\begin{matrix} 2^{t} & 0 \\ 0 & 3^{t} \end{matrix}]$

Finding the characteristic equation for the eigenvalues.

Now first, we can solve for $位 = 2$

$(A - 2 I) x = 0 [\begin{matrix} 2 & - 2 \\ 1 & - 1 \end{matrix}] \begin{matrix} x_{1} \\ x_{2} \end{matrix} = \begin{matrix} 0 \\ 0 \end{matrix} x_{1} - x_{2} = 0 E_{2} = s p a n [\begin{matrix} 1 \\ 1 \end{matrix}]$

Now, we can solve for $位 = 3$

$(A - 2 I) x = 0 [\begin{matrix} 1 & - 2 \\ 1 & - 2 \end{matrix}] \begin{matrix} x_{1} \\ x_{2} \end{matrix} = \begin{matrix} 0 \\ 0 \end{matrix} x_{1} - 2 x_{2} = 0 E_{2} = S p a n [\begin{matrix} 2 \\ 1 \end{matrix}]$

So,

$S = [\begin{matrix} 1 & 2 \\ 1 & 1 \end{matrix}]$

Finding the closed formulas for At

Now we can find the $S^{- 1}$ matrix.

$S = [\begin{matrix} 1 & 2 \\ 1 & 1 \end{matrix}], S^{- 1} = [\begin{matrix} - 1 & 2 \\ 1 & - 1 \end{matrix}]$

Finally now, we can calculate,

$A^{t} = S B^{t} S^{- 1} = [\begin{matrix} 1 & 2 \\ 1 & 1 \end{matrix}] 脳 [\begin{matrix} 2^{t} & 0 \\ 0 & 3^{t} \end{matrix}] 脳 [\begin{matrix} - 1 & 2 \\ 1 & - 1 \end{matrix}] = [\begin{matrix} - 2 + 2 脳 3^{t} & 2^{t + 1} - 2 脳 3^{t} \\ - 2 + 3^{t} & 2^{t + 1} - 3^{t} \end{matrix}]$

Hence the closed formulas for $A^{t} = [\begin{matrix} - 2 + 2 脳 3^{t} & 2^{t - 1} - 2 脳 3^{t} \\ - 2 + 3^{t} & 2^{t + 1} - 3^{t} \end{matrix}]$

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91影视

Short Answer

Step by step solution

Finding the diagonalization matrix and eigenvalues.

Finding the characteristic equation for the eigenvalues.

Finding the closed formulas for At

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91影视

For the matrices A in Exercises 1 through 12, find closed formulas for At, where t is an arbitrary positive integer. Follow the strategy outlined in Theorem 7.4.2 and illustrated in Example 2. In Exercises 9 though 12, feel free to use technology.4.A=[4-211]

Short Answer

Step by step solution

Finding the diagonalization matrix and eigenvalues.

Finding the characteristic equation for the eigenvalues.

Finding the closed formulas for At

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

Calculus

Decision Maths

Geometry

Mechanics Maths

Probability and Statistics

Study anywhere. Anytime. Across all devices.