written 5.7 years ago by
teamques10
★ 64k
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modified 5.7 years ago
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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,
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)
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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] $