JEE MAIN - Mathematics (2019 - 9th January Morning Slot - No. 9)
If $$A = \left[ {\matrix{
{\cos \theta } & { - \sin \theta } \cr
{\sin \theta } & {\cos \theta } \cr
} } \right]$$, then the matrix A–50 when $$\theta $$ = $$\pi \over 12$$, is equal to :
$$\left[ {\matrix{
{ {{\sqrt 3 } \over 2}} & { - {1 \over 2}} \cr
{{{ 1} \over 2}} & {{{\sqrt 3 } \over 2}} \cr
} } \right]$$
$$\left[ {\matrix{
{{1 \over 2}} & -{{{\sqrt 3 } \over 2}} \cr
{{{\sqrt 3 } \over 2}} & {{{ - 1} \over 2}} \cr
} } \right]$$
$$\left[ {\matrix{
{{{\sqrt 3 } \over 2}} & {{1 \over 2}} \cr
-{{1 \over 2}} & {{{\sqrt 3 } \over 2}} \cr
} } \right]$$
$$\left[ {\matrix{
{{1 \over 2}} & {{{\sqrt 3 } \over 2}} \cr
{-{{\sqrt 3 } \over 2}} & {{{ 1} \over 2}} \cr
} } \right]$$
Explanation
(A$$-$$50) = (A$$-$$1)50
We know,
A$$-$$1 = $${{adjA} \over {\left| A \right|}}$$
$$\left| A \right|$$ = cos2$$\theta $$ + sin2$$\theta $$ = 1
cofactor of A = $$\left[ {\matrix{ {\cos \theta } & { - \sin \theta } \cr {\sin \theta } & {\cos \theta } \cr } } \right]$$
Adjoint of A = Transpose of cofactor matrix
$$ \therefore $$ Adj A = $$\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$$
$$ \therefore $$ A$$-$$1 = $$\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$$
$$ \therefore $$ (A$$-$$1)2 = $$\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$$
= $$\left[ {\matrix{ {\cos 2\theta } & {\sin 2\theta } \cr { - \sin 2\theta } & {\cos 2\theta } \cr } } \right]$$
Similarly,
(A$$-$$1)3 = $$\left[ {\matrix{ {\cos 3\theta } & {\sin 3\theta } \cr { - \sin 3\theta } & {\cos 3\theta } \cr } } \right]$$
:
:
:
(A$$-$$1)50 = $$\left[ {\matrix{ {\cos 50\theta } & {\sin 50\theta } \cr { - \sin 50\theta } & {\cos 50\theta } \cr } } \right]$$
when $$\theta $$ = $${\pi \over {12}}$$ then
$${A^{ - 50}}$$ = $$\left[ {\matrix{ {\cos {{25\pi } \over 6}} & {\sin {{25\pi } \over 6}} \cr { - \sin {{25\pi } \over 6}} & {\cos {{25\pi } \over 6}} \cr } } \right]$$
= $$\left[ {\matrix{ {{{\sqrt 3 } \over 2}} & {{1 \over 2}} \cr { - {1 \over 2}} & {{{\sqrt 3 } \over 2}} \cr } } \right]$$
Note:
$$\cos {{25\pi } \over 6} = \cos \left( {4\pi + {\pi \over 6}} \right) = \cos {\pi \over 6} = {{\sqrt 3 } \over 2}$$
We know,
A$$-$$1 = $${{adjA} \over {\left| A \right|}}$$
$$\left| A \right|$$ = cos2$$\theta $$ + sin2$$\theta $$ = 1
cofactor of A = $$\left[ {\matrix{ {\cos \theta } & { - \sin \theta } \cr {\sin \theta } & {\cos \theta } \cr } } \right]$$
Adjoint of A = Transpose of cofactor matrix
$$ \therefore $$ Adj A = $$\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$$
$$ \therefore $$ A$$-$$1 = $$\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$$
$$ \therefore $$ (A$$-$$1)2 = $$\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$$
= $$\left[ {\matrix{ {\cos 2\theta } & {\sin 2\theta } \cr { - \sin 2\theta } & {\cos 2\theta } \cr } } \right]$$
Similarly,
(A$$-$$1)3 = $$\left[ {\matrix{ {\cos 3\theta } & {\sin 3\theta } \cr { - \sin 3\theta } & {\cos 3\theta } \cr } } \right]$$
:
:
:
(A$$-$$1)50 = $$\left[ {\matrix{ {\cos 50\theta } & {\sin 50\theta } \cr { - \sin 50\theta } & {\cos 50\theta } \cr } } \right]$$
when $$\theta $$ = $${\pi \over {12}}$$ then
$${A^{ - 50}}$$ = $$\left[ {\matrix{ {\cos {{25\pi } \over 6}} & {\sin {{25\pi } \over 6}} \cr { - \sin {{25\pi } \over 6}} & {\cos {{25\pi } \over 6}} \cr } } \right]$$
= $$\left[ {\matrix{ {{{\sqrt 3 } \over 2}} & {{1 \over 2}} \cr { - {1 \over 2}} & {{{\sqrt 3 } \over 2}} \cr } } \right]$$
Note:
$$\cos {{25\pi } \over 6} = \cos \left( {4\pi + {\pi \over 6}} \right) = \cos {\pi \over 6} = {{\sqrt 3 } \over 2}$$
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