\[6{{A}^{-1}}={{A}^{2}}+cA+dI,\] then (c, d) is
A) (-6, 11)
B) (-11,6)
C) (11,6)
D) (6,11)
Correct Answer: A
Solution :
Using ??cayley Hamilton theorem??. ie,\[|A-\lambda I|=0\] \[\Rightarrow \]\[\left| \begin{matrix} 1-\lambda & 0 & 0 \\ 0 & 1-\lambda & 1 \\ 0 & -2 & 4-\lambda \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[(1-\lambda )\{(1-\lambda )(4-\lambda )+2\}=0\] \[\Rightarrow \]\[{{\lambda }^{3}}-6{{\lambda }^{2}}+11\lambda -6=0\] or\[{{A}^{3}}-6{{A}^{2}}+11A-6I=0\] ?(i) Given \[6{{A}^{-1}}={{A}^{2}}+cA+dI,\]multiplying both sides by A, we get\[6I={{A}^{3}}+c{{A}^{2}}+dA\] \[\Rightarrow \]\[{{A}^{3}}+c{{A}^{2}}+dA-6I=0\] ?(ii) \[c=-6\]and\[d=11\] On comparing Eqs. (i) and (ii), we get C = - 6 and d = ll Solutions for Q. No. 4 to 5. Given, \[\cos \frac{\pi }{7},\cos \frac{3\pi }{7},\cos \frac{5\pi }{7}\]are the roots of the equation \[8{{x}^{3}}-4{{x}^{2}}-4x+1=0\] ..(i) Replacing x by \[\frac{1}{x}\]in Eq. (i), then, we ge, \[{{x}^{3}}-4{{x}^{2}}-4x+8=0\] ?(ii)
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