JEE MAIN - Mathematics (2020 - 7th January Morning Slot - No. 5)
Let $$\alpha $$ be a root of the equation x2 + x + 1 = 0 and the
matrix A = $${1 \over {\sqrt 3 }}\left[ {\matrix{ 1 & 1 & 1 \cr 1 & \alpha & {{\alpha ^2}} \cr 1 & {{\alpha ^2}} & {{\alpha ^4}} \cr } } \right]$$
then the matrix A31 is equal to
matrix A = $${1 \over {\sqrt 3 }}\left[ {\matrix{ 1 & 1 & 1 \cr 1 & \alpha & {{\alpha ^2}} \cr 1 & {{\alpha ^2}} & {{\alpha ^4}} \cr } } \right]$$
then the matrix A31 is equal to
A2
A
I3
A3
Explanation
x2 + x + 1 = 0
$$ \Rightarrow $$ x = $${{ - 1 + i\sqrt 3 } \over 2}$$ = $$\omega $$ or $${{ - 1 - i\sqrt 3 } \over 2}$$ = $${\omega ^2}$$
Let $$\alpha $$ = $$\omega $$
$$ \therefore $$ A = $${1 \over {\sqrt 3 }}\left[ {\matrix{ 1 & 1 & 1 \cr 1 & \omega & {{\omega ^2}} \cr 1 & {{\omega ^2}} & {{\omega ^4}} \cr } } \right]$$
A2 = $${1 \over 3}\left[ {\matrix{ 3 & 0 & 0 \cr 0 & 0 & 3 \cr 0 & 3 & 0 \cr } } \right]$$ = $$\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right]$$
Now A4 = $$\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$$ = I
$$ \therefore $$ A31 = A28A3 = A3
$$ \Rightarrow $$ x = $${{ - 1 + i\sqrt 3 } \over 2}$$ = $$\omega $$ or $${{ - 1 - i\sqrt 3 } \over 2}$$ = $${\omega ^2}$$
Let $$\alpha $$ = $$\omega $$
$$ \therefore $$ A = $${1 \over {\sqrt 3 }}\left[ {\matrix{ 1 & 1 & 1 \cr 1 & \omega & {{\omega ^2}} \cr 1 & {{\omega ^2}} & {{\omega ^4}} \cr } } \right]$$
A2 = $${1 \over 3}\left[ {\matrix{ 3 & 0 & 0 \cr 0 & 0 & 3 \cr 0 & 3 & 0 \cr } } \right]$$ = $$\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right]$$
Now A4 = $$\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$$ = I
$$ \therefore $$ A31 = A28A3 = A3
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