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Question: Find characteristic equation of matrix A and hence find matrix represented by
0

$A^8-5A^7+7A^6-3A^5+A^4-5A^3+8A^2-2A+I$ where $A=$ $ \left[ {\begin{array}{cc} -1 & 2 & 3\ 0 & 3 & 5 \ 0 & 0 & -2\ \end{array} } \right] $

Subject: Applied Mathematics 4

Topic: Matrices

Difficulty: Medium

m4m(64) • 79 views
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modified 11 weeks ago  • written 11 weeks ago by gravatar for manasahegde234 manasahegde23420
0

Characteristic equation of A is $\mid A-\lambda I \mid=0$

$\lambda^3-5\lambda^2+[(2+2+4)-(0+1+0)]\lambda-\mid A \mid=0$

$\lambda^3-5\lambda^2+7\lambda-3=0$

Therefore, Cayley-Hamilton theorem,

$A^3-5A^2+7A-3I=0$

from given polynomial in A,

i.e $A^8-5\lambda^7+7A^6-3A^5+A^4-5A^3+8A^2-2A+I$,

we get polynomial in$\lambda$ as,

$\lambda^8-5\lambda^7+7\lambda^6-3\lambda^5+\lambda^4-5\lambda^3+8\lambda^2-2\lambda+1$

Let,

enter image description here

By division algorithm,

$(\lambda^8-5\lambda^7+7\lambda^6-3\lambda^5+\lambda^4-5\lambda^3+8\lambda^2-2\lambda+1)$

$=(\lambda^3-5\lambda^2+7\lambda-3)(\lambda^5+\lambda)+\lambda^2+lambda+1$

By converting this in A we get,

$A^8-5A^7+7A^6-3A^5+A^4-5A^3-2A+I$

$=(A^3-5A^2+7A-3I)(A^5+A)+A^2+A+I$

$=0+A^2+A+I$ from (1)

$ \left[ {\begin{array}{cc} 5 & 4 & 4\\ 0 & 1 & 0 \\ 4 & 4 & 5\\ \end{array} } \right] $+$ \left[ {\begin{array}{cc} 2 & 1 & 1\\ 0 & 1 & 0 \\ 1 & 1 & 2\\ \end{array} } \right] $+$ \left[ {\begin{array}{cc} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1\\ \end{array} } \right] $

=$ \left[ {\begin{array}{cc} 8 & 5 & 5\\ 0 & 3 & 0 \\ 5 & 5 & 8\\ \end{array} } \right] $

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modified 11 weeks ago  • written 11 weeks ago by gravatar for manasahegde234 manasahegde23420
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