JEE MAIN - Mathematics (2021 - 27th August Morning Shift - No. 3)
If the matrix $$A = \left( {\matrix{
0 & 2 \cr
K & { - 1} \cr
} } \right)$$ satisfies $$A({A^3} + 3I) = 2I$$, then the value of K is :
$${1 \over 2}$$
$$-$$$${1 \over 2}$$
$$-$$1
1
Explanation
Given matrix $$A = \left[ {\matrix{
0 & 2 \cr
k & { - 1} \cr
} } \right]$$
$${A^4} + 3IA = 2I$$
$$ \Rightarrow {A^4} = 2I - 3A$$
Also characteristic equation of A is $$|A - \lambda I|\, = 0$$
$$ \Rightarrow \left| {\matrix{ {0 - \lambda } & 2 \cr k & { - 1 - \lambda } \cr } } \right| = 0$$
$$ \Rightarrow \lambda + {\lambda ^2} - 2k = 0$$
$$ \Rightarrow A + {A^2} = 2K.I$$
$$ \Rightarrow {A^2} = 2KI - A$$
$$ \Rightarrow {A^4} = 4{K^2}I + {A^2} - 4AK$$
Put $${A^2} = 2KI - A$$
and $${A^4} = 2I - 3A$$
$$2I - 3A = 4{K^2}I + 2KI - A - 4AK$$
$$ \Rightarrow I(2 - 2K - 4{K^2}) = A(2 - 4K)$$
$$ \Rightarrow - 2I(2{K^2} + K - 1) = 2A(1 - 2K)$$
$$ \Rightarrow - 2I(2K - 1)(K + 1) = 2A(1 - 2K)$$
$$ \Rightarrow (2K - 1)(2A) - 2I(2K - 1)(K + 1) = 0$$
$$ \Rightarrow (2K - 1)[2A - 2I(K + 1)] = 0$$
$$ \Rightarrow K = {1 \over 2}$$
$${A^4} + 3IA = 2I$$
$$ \Rightarrow {A^4} = 2I - 3A$$
Also characteristic equation of A is $$|A - \lambda I|\, = 0$$
$$ \Rightarrow \left| {\matrix{ {0 - \lambda } & 2 \cr k & { - 1 - \lambda } \cr } } \right| = 0$$
$$ \Rightarrow \lambda + {\lambda ^2} - 2k = 0$$
$$ \Rightarrow A + {A^2} = 2K.I$$
$$ \Rightarrow {A^2} = 2KI - A$$
$$ \Rightarrow {A^4} = 4{K^2}I + {A^2} - 4AK$$
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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