7- I want solution of this question with all steps, please. Thanks 7.21]Given that matrix P =0 -12. Find the matrix A such that A = PBP-1, where11]-2 0 0]3 00 0 6]B is diagonal matrix given by 0

Given:The matrix P=2310-12101 , B=-200030006To find:we have to find the matrix A such…

Answered: 2 3. 1] Given that matrix P =0 2. Find…

Linear Algebra: A Modern Introduction

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter4: Eigenvalues And Eigenvectors

Section4.4: Similarity And Diagonalization

Problem 50EQ

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7- I want solution of this question with all steps, please. Thanks

7.
2
1]
Given that matrix P =
0 -1
2. Find the matrix A such that A = PBP-1, where
1
1]
-2 0 0]
3 0
0 0 6]
B is diagonal matrix given by 0

Expert Solution

Step 1

$G i v e n : T h e m a t r i x P = [\begin{matrix} 2 & 3 & 1 \\ 0 & - 1 & 2 \\ 1 & 0 & 1 \end{matrix}], B = [\begin{matrix} - 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 6 \end{matrix}] T o f i n d : w e h a v e t o f i n d t h e m a t r i x A s u c h t h a t A = P B P^{- 1}$

Step 2

$H e r e P = [\begin{matrix} 2 & 3 & 1 \\ 0 & - 1 & 2 \\ 1 & 0 & 1 \end{matrix}], B = [\begin{matrix} - 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 6 \end{matrix}] F i r s t w e f i n d t h e i n v e r s e o f m a t r i x P T o f i n d t h e i n v e r s e m a t r i x, a u g m e n t i t w i t h t h e i d e n t i t y m a t r i x a n d p e r f o r m r o w o p e r a t i o n s t o m a k e t h e i d e n t i t y m a t r i x t o l e f t a n d r i g h t w i l l b e t h r i n v e r s e m a t r i x . T h e a u g m e n t m a t r i x w i t h i d e n t i t y m a t r i x i s [\begin{matrix} 2 & 3 & 1 & 1 & 0 & 0 \\ 0 & - 1 & 2 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 & 0 & 1 \end{matrix}] R_{1} = \frac{R_{1}}{2} [\begin{matrix} 1 & \frac{3}{2} & \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ 0 & - 1 & 2 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 & 0 & 1 \end{matrix}]$

Step 3

$R_{3} = R_{3} - R_{1} [\begin{matrix} 1 & \frac{3}{2} & \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ 0 & - 1 & 2 & 0 & 1 & 0 \\ 0 & - \frac{3}{2} & \frac{1}{2} & - \frac{1}{2} & 0 & 1 \end{matrix}] R_{2} = - R_{2} [\begin{matrix} 1 & \frac{3}{2} & \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 1 & - 2 & 0 & - 1 & 0 \\ 0 & - \frac{3}{2} & \frac{1}{2} & - \frac{1}{2} & 0 & 1 \end{matrix}] R_{1} = R_{1} - \frac{3}{2} R_{2} [\begin{matrix} 1 & 0 & \frac{7}{2} & \frac{1}{2} & \frac{3}{2} & 0 \\ 0 & 1 & - 2 & 0 & - 1 & 0 \\ 0 & - \frac{3}{2} & \frac{1}{2} & - \frac{1}{2} & 0 & 1 \end{matrix}]$

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