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If the matrix $$A = \left( {\begin{matrix} 0 & 2 \\ K & { - 1} \\ \end{matrix} } \right)$$ satisfies $$A({A^3} + 3I) = 2I$$, then the value of K is :

JEE · Math · previous-year question

  1. A.$${1 \over 2}$$correct
  2. B.$$-$$$${1 \over 2}$$
  3. C.$$-$$1
  4. D.1

Answer

A. $${1 \over 2}$$

Explanation

Given matrix $$A = \left[ {\begin{matrix} 0 & 2 \\ k & { - 1} \\ \end{matrix} } \right]$$ $${A^4} + 3IA = 2I$$ $$ \Rightarrow {A^4} = 2I - 3A$$ Also characteristic equation of A is $$|A - \lambda I|\, = 0$$ $$ \Rightarrow \left| {\begin{matrix} {0 - \lambda } & 2 \\ k & { - 1 - \lambda } \\ \end{matrix} } \right| = 0$$ $$ \Rightarrow \lambda + {\lambda ^2} - 2k = 0$$ $$ \Rightarrow A + {A^2} = 2K.I$$ $$ \Rightarrow {A^2} = 2KI - A$$ $$ \Rightarrow {A^4} = 4{K^2}I + {A^2} - 4AK$$ Put $${A^2} = 2KI - A$$ and $${A^4} = 2I - 3A$$ $$2I - 3A = 4{K^2}I + 2KI - A - 4AK$$ $$ \Rightarrow I(2 - 2K - 4{K^2}) = A(2 - 4K)$$ $$ \Rightarrow - 2I(2{K^2} + K - 1) = 2A(1 - 2K)$$ $$ \Rightarrow - 2I(2K - 1)(K + 1) = 2A(1 - 2K)$$ $$ \Rightarrow (2K - 1)(2A) - 2I(2K - 1)(K + 1) = 0$$ $$ \Rightarrow (2K - 1)[2A - 2I(K + 1)] = 0$$ $$ \Rightarrow K = {1 \over 2}$$

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