Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

Find a symmetric 2x2 matrix B such that $B^{3} = \frac{1}{5} [\begin{matrix} 12 & 14 \\ 14 & 33 \end{matrix}]$

A symmetric 2x2 matrix $B = \frac{1}{5} [\begin{matrix} 6 & 2 \\ 2 & 9 \end{matrix}]$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Symmetric matrix:

In linear algebra, a symmetric matrix is a square matrix that stays unchanged when its transpose is calculated. In that instance, a symmetric matrix is one whose transpose equals the matrix itself.

Step 2: To determine the eigenvalues of the matrix A :

Let $A = B^{3}$ , determine the eigenvalues of the matrix A

$d e t (A - λ l_{n}) = 0 [\begin{matrix} \frac{12}{5} - λ & \frac{14}{5} \\ \frac{14}{5} & \frac{33}{5} - λ \end{matrix}] = 0 (\frac{12}{5} - λ) (\frac{33}{5} - λ) - (\frac{14}{5}) (\frac{14}{5}) = 0 25 λ^{2} - 225 λ + 200 = 0 (λ - 8) (λ - 1) = 0$

The eigenvalues are $λ = 1$ , and $λ = 8$ . Now we obtain the eigenvectors.

CASE 1: When $λ = 1$

$[A - l] \tilde{x} = 0 \Rightarrow [\begin{matrix} \frac{12}{5} - 1 & \frac{14}{5} \\ \frac{14}{5} & \frac{33}{5} - 1 \end{matrix}] [\begin{matrix} a \\ b \end{matrix}] = 0 [\begin{matrix} \frac{7}{5} & \frac{14}{5} \\ \frac{14}{5} & \frac{28}{5} \end{matrix}] [\begin{matrix} a \\ b \end{matrix}] = 0$

apply row operation $R_{2} \to 2 R_{1} - R_{2}$ :

$[\begin{matrix} \frac{7}{5} & \frac{14}{5} \\ 0 & 0 \end{matrix}] [\begin{matrix} a \\ b \end{matrix}] = 0$

apply row operation $R_{1} \to 5 R_{1}$

$[\begin{matrix} 7 & 14 \\ 0 & 0 \end{matrix}] [\begin{matrix} a \\ b \end{matrix}] = 0$

Step 3: Compute eigenvector corresponding:

so here we have $a = - 2 b$ , choosing $b = 1$ yields $a = - 2$ . The eigenvector corresponding to the eigenvalue $λ = 1$ is

$\vec{v} = [\begin{matrix} - 2 \\ 1 \end{matrix}]$

therefore,

$E_{1} = s p a n (\frac{1}{\sqrt{5}} [\begin{matrix} - 2 \\ 1 \end{matrix}])$

CASE 2: When $λ = 8$

role="math" localid="1660104429366" $[A - l] \tilde{x} = 0 \to [\begin{matrix} \frac{12}{5} - 8 & \frac{14}{5} \\ \frac{14}{5} & \frac{33}{5} - 8 \end{matrix}] [\begin{matrix} a \\ b \end{matrix}] = 0 [\begin{matrix} - \frac{28}{5} & \frac{14}{5} \\ \frac{14}{5} & - \frac{7}{5} \end{matrix}] [\begin{matrix} a \\ b \end{matrix}] = 0$

apply row operation $R_{2} \to 2 R_{2} + R_{1}$ and $R_{1} \to 5 R_{1}$ :

$[\begin{matrix} - 28 & 14 \\ 0 & 0 \end{matrix}] [\begin{matrix} a \\ b \end{matrix}] = 0$

so here we have $a = \frac{1}{2} b$ , choosing b = 2 yields a = 1 . The eigenvector corresponding to the eigenvalue $λ = 8$ is

${\vec{v}}_{2} = [\begin{matrix} 1 \\ 2 \end{matrix}]$

therefore,

$E_{8} = s p a n (\frac{1}{\sqrt{5}} [\begin{matrix} 1 \\ 2 \end{matrix}])$

Step 4: Find a symmetric 2x2 matrix B :

In decomposition $A = S D S^{- 1}$ , the orthogonal matrix is given by

$S = \frac{1}{\sqrt{5}} [\begin{matrix} 1 & - 2 \\ 2 & 1 \end{matrix}]$

and the diagonal matrix is given by

$D = [\begin{matrix} 8 & 0 \\ 0 & 1 \end{matrix}]$

Now let

$P = [\begin{matrix} 2 & 0 \\ 0 & 1 \end{matrix}]$

notice that $P^{3} = D$ so ${(S D S^{- 1})}^{3} = S P^{3} S^{- 1} = A$ . So, we compute B such that

$S^{- 1} B S = P \to B = S P S^{- 1} = \frac{1}{\sqrt{5}} [\begin{matrix} 1 & - 2 \\ 2 & 1 \end{matrix}] [\begin{matrix} 2 & 0 \\ 0 & 1 \end{matrix}] {[\begin{matrix} 1 \\ \sqrt{5} \end{matrix} [\begin{matrix} 1 & - 2 \\ 2 & 1 \end{matrix}]]}^{- 1} = \frac{1}{\sqrt{5}} [\begin{matrix} 2 & - 2 \\ 4 & 1 \end{matrix}] \frac{1}{\sqrt{5}} [\begin{matrix} 1 & 2 \\ - 2 & 1 \end{matrix}] = \frac{1}{5} [\begin{matrix} 6 & 2 \\ 2 & 9 \end{matrix}]$

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Linear Algebra With Applications

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Short Answer

Find a symmetric 2x2 matrix B such that $B^{3} = \frac{1}{5} [\begin{matrix} 12 & 14 \\ 14 & 33 \end{matrix}]$

Step by Step Solution

Step 1: Symmetric matrix:

Step 2: To determine the eigenvalues of the matrix A :

Step 3: Compute eigenvector corresponding:

Step 4: Find a symmetric 2x2 matrix B :

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Linear Algebra With Applications

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Find a symmetric 2x2 matrix B such that B3=15[12141433]

Step by Step Solution

Step 1: Symmetric matrix:

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Step 4: Find a symmetric 2x2 matrix B :

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Find a symmetric 2x2 matrix B such that $B^{3} = \frac{1}{5} [\begin{matrix} 12 & 14 \\ 14 & 33 \end{matrix}]$