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Question

(a) verify that A is diagonalizable by finding P1AP,P^{ - 1}AP, and (b) use the result of part (a) and Theorem to find the eigenvalues of A.

A=[110030425],P=[013040122]A=\left[\begin{array}{rrr} {-1} & {1} & {0} \\ {0} & {3} & {0} \\ {4} & {-2} & {5} \end{array}\right], P=\left[\begin{array}{rrr} {0} & {1} & {-3} \\ {0} & {4} & {0} \\ {1} & {2} & {2} \end{array}\right]

Solution

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Answered 1 year ago
Answered 1 year ago
Step 1
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Begin by determining P1P^{-1}. Since PP1=IPP^{-1}=I,

[013040122][x11x12x13x21x22x23x31x32x33]=[100010001]\begin{bmatrix} 0&1&-3\\ 0&4&0\\ 1&2&2 \end{bmatrix} \begin{bmatrix} x_{11}&x_{12}&x_{13}\\ x_{21}&x_{22}&x_{23}\\ x_{31}&x_{32}&x_{33} \end{bmatrix}= \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix}

from which

{x213x31=1x223x32=0x233x33=04x21=04x22=14x23=0x11+2x21+2x31=0x12+2x22+2x32=0x13+2x23+2x33=1\begin{cases} x_{21}-3x_{31}=1\\ x_{22}-3x_{32}=0\\ x_{23}-3x_{33}=0\\ 4x_{21}=0\\ 4x_{22}=1\\ 4x_{23}=0\\ x_{11}+2x_{21}+2x_{31}=0\\ x_{12}+2x_{22}+2x_{32}=0\\ x_{13}+2x_{23}+2x_{33}=1 \end{cases}

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