Suppose that for a 2 × 2 matrix $ A $, $ A\vec{v} = 3\vec{v} $ for $ \vec{v} = \begin{bmatrix} -1 \\ -1 \end{bmatrix} $.- How is the pair $ (3, \vec{v}) $ called for a matrix $ A $?Suppose further that the only non-zero vectors $ x $, for which $ Ax = rx $ for some $ r $, must be multiples of $ \vec{v} $ above.- What more can you now say about the number $ r = 3 $?Suppose further that $ A\vec{w} - 3\vec{w} = \vec{v} $ for $ \vec{w} = \begin{bmatrix} -3 \\ -1 \end{bmatrix} $. Write down the general solution of the differential equation $ x'(t) = Ax $.

Given: Av→=3v→ for v→=-1-1. Consider the matrix A as abcd Av→=abcd-1-1=-a-b-c-d If a+b=c+d, then…

Answered: Suppose that for a 2 x 2 matrix A, Au =…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Suppose that for a 2 × 2 matrix $ A $, $ A\vec{v} = 3\vec{v} $ for $ \vec{v} = \begin{bmatrix} -1 \\ -1 \end{bmatrix} $.

- How is the pair $ (3, \vec{v}) $ called for a matrix $ A $?

Suppose further that the only non-zero vectors $ x $, for which $ Ax = rx $ for some $ r $, must be multiples of $ \vec{v} $ above.

- What more can you now say about the number $ r = 3 $?

Suppose further that $ A\vec{w} - 3\vec{w} = \vec{v} $ for $ \vec{w} = \begin{bmatrix} -3 \\ -1 \end{bmatrix} $. Write down the general solution of the differential equation $ x'(t) = Ax $.

Expert Solution

Solution:

Given: $A \vec{v} = 3 \vec{v} for \vec{v} = [\begin{matrix} - 1 \\ - 1 \end{matrix}]$ .

Consider the matrix A as $[\begin{matrix} a & b \\ c & d \end{matrix}]$

$\begin{array}{rcl} A \vec{v} & = & [\begin{matrix} a & b \\ c & d \end{matrix}] [\begin{matrix} - 1 \\ - 1 \end{matrix}] \\ = & [\begin{matrix} - a - b \\ - c - d \end{matrix}] \end{array}$

$\begin{array}{rcl} If \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} (a + b = c + d) \end{array} \begin{array}{rcl} , \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} then \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} (a + b) \end{array} \begin{array}{rcl} [\begin{matrix} - 1 \\ - 1 \end{matrix}] \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} or \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} (c + d) \end{array} \begin{array}{rcl} [\begin{array}{l} - 1 \\ - 1 \end{array}] \end{array}$

Keep a+b = c+d =3, that gives $3 \begin{array}{rcl} [\begin{array}{l} - 1 \\ - 1 \end{array}] \end{array}$ .

Hence, the pair $(3, \vec{v})$ is called for a matrix A.

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