JEE MAIN - Mathematics (2021 - 16th March Morning Shift - No. 6)
Let $$A = \left[ {\matrix{
i & { - i} \cr
{ - i} & i \cr
} } \right],i = \sqrt { - 1} $$. Then, the system of linear equations $${A^8}\left[ {\matrix{
x \cr
y \cr
} } \right] = \left[ {\matrix{
8 \cr
{64} \cr
} } \right]$$ has :
Exactly two solutions
Infinitely many solutions
A unique solution
No solution
Explanation
$$A = \left[ {\matrix{
i & { - i} \cr
{ - i} & i \cr
} } \right]$$
$${A^2} = \left[ {\matrix{ i & { - i} \cr { - i} & i \cr } } \right]\left[ {\matrix{ i & { - i} \cr { - i} & i \cr } } \right] = \left[ {\matrix{ { - 2} & 2 \cr 2 & { - 2} \cr } } \right] = 2\left[ {\matrix{ { - 1} & 1 \cr 1 & { - 1} \cr } } \right]$$
$${A^4} = 4\left[ {\matrix{ { - 1} & 1 \cr 1 & { - 1} \cr } } \right]\left[ {\matrix{ { - 1} & 1 \cr 1 & { - 1} \cr } } \right] = 4\left[ {\matrix{ 2 & { - 2} \cr { - 2} & 2 \cr } } \right] = 8\left[ {\matrix{ 1 & { - 1} \cr { - 1} & 1 \cr } } \right]$$
$${A^8} = 64\left[ {\matrix{ 1 & { - 1} \cr { - 1} & 1 \cr } } \right]\left[ {\matrix{ 1 & { - 1} \cr { - 1} & 1 \cr } } \right] = 64\left[ {\matrix{ 2 & { - 2} \cr { - 2} & 2 \cr } } \right] = 128\left[ {\matrix{ 1 & { - 1} \cr { - 1} & 1 \cr } } \right]$$
$$128\left[ {\matrix{ 1 & { - 1} \cr { - 1} & 1 \cr } } \right]\left[ {\matrix{ x \cr y \cr } } \right] = \left[ {\matrix{ 8 \cr {64} \cr } } \right]$$
$$128\left[ {\matrix{ {x - y} \cr { - x + y} \cr } } \right] = \left[ {\matrix{ 8 \cr {64} \cr } } \right] $$
$$\Rightarrow 128(x - y) = 8$$
$$ \Rightarrow x - y = {1 \over {16}}$$ .... (1)
and $$128( - x + y) = 64 \Rightarrow x - y = {{ - 1} \over 2}$$ .... (2)
$$ \Rightarrow $$ No solution (from eq. (1) & (2))
$${A^2} = \left[ {\matrix{ i & { - i} \cr { - i} & i \cr } } \right]\left[ {\matrix{ i & { - i} \cr { - i} & i \cr } } \right] = \left[ {\matrix{ { - 2} & 2 \cr 2 & { - 2} \cr } } \right] = 2\left[ {\matrix{ { - 1} & 1 \cr 1 & { - 1} \cr } } \right]$$
$${A^4} = 4\left[ {\matrix{ { - 1} & 1 \cr 1 & { - 1} \cr } } \right]\left[ {\matrix{ { - 1} & 1 \cr 1 & { - 1} \cr } } \right] = 4\left[ {\matrix{ 2 & { - 2} \cr { - 2} & 2 \cr } } \right] = 8\left[ {\matrix{ 1 & { - 1} \cr { - 1} & 1 \cr } } \right]$$
$${A^8} = 64\left[ {\matrix{ 1 & { - 1} \cr { - 1} & 1 \cr } } \right]\left[ {\matrix{ 1 & { - 1} \cr { - 1} & 1 \cr } } \right] = 64\left[ {\matrix{ 2 & { - 2} \cr { - 2} & 2 \cr } } \right] = 128\left[ {\matrix{ 1 & { - 1} \cr { - 1} & 1 \cr } } \right]$$
$$128\left[ {\matrix{ 1 & { - 1} \cr { - 1} & 1 \cr } } \right]\left[ {\matrix{ x \cr y \cr } } \right] = \left[ {\matrix{ 8 \cr {64} \cr } } \right]$$
$$128\left[ {\matrix{ {x - y} \cr { - x + y} \cr } } \right] = \left[ {\matrix{ 8 \cr {64} \cr } } \right] $$
$$\Rightarrow 128(x - y) = 8$$
$$ \Rightarrow x - y = {1 \over {16}}$$ .... (1)
and $$128( - x + y) = 64 \Rightarrow x - y = {{ - 1} \over 2}$$ .... (2)
$$ \Rightarrow $$ No solution (from eq. (1) & (2))
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