JEE MAIN - Mathematics (2019 - 9th April Morning Slot - No. 15)
Let $$\alpha $$ and $$\beta $$ be the roots of the equation
x2 + x + 1 = 0. Then for y $$ \ne $$ 0 in R,
$$$\left| {\matrix{ {y + 1} & \alpha & \beta \cr \alpha & {y + \beta } & 1 \cr \beta & 1 & {y + \alpha } \cr } } \right|$$$ is equal to
$$$\left| {\matrix{ {y + 1} & \alpha & \beta \cr \alpha & {y + \beta } & 1 \cr \beta & 1 & {y + \alpha } \cr } } \right|$$$ is equal to
y(y2 – 1)
y(y2 – 3)
y3
y3 – 1
Explanation
$$\alpha $$ and $$\beta $$ are the roots of the equation
x2 + x + 1 = 0.
$$ \therefore $$ $$\alpha $$ = $$\omega $$ and $$\beta $$ = $${\omega ^2}$$
$$\left| {\matrix{ {y + 1} & \alpha & \beta \cr \alpha & {y + \beta } & 1 \cr \beta & 1 & {y + \alpha } \cr } } \right|$$
= $$\left| {\matrix{ {y + 1} & \omega & {{\omega ^2}} \cr \omega & {y + {\omega ^2}} & 1 \cr {{\omega ^2}} & 1 & {y + \omega } \cr } } \right|$$
C1 $$ \to $$ C1 + C2 + C3
= $$\left| {\matrix{ {y + 1 + \omega + {\omega ^2}} & \omega & {{\omega ^2}} \cr {y + 1 + \omega + {\omega ^2}} & {y + {\omega ^2}} & 1 \cr {y + 1 + \omega + {\omega ^2}} & 1 & {y + \omega } \cr } } \right|$$
= $$\left| {\matrix{ y & \omega & {{\omega ^2}} \cr y & {y + {\omega ^2}} & 1 \cr y & 1 & {y + \omega } \cr } } \right|$$
As $$1 + \omega + {\omega ^2}$$ = 0
= $$y\left| {\matrix{ 1 & \omega & {{\omega ^2}} \cr 1 & {y + {\omega ^2}} & 1 \cr 1 & 1 & {y + \omega } \cr } } \right|$$
R2 $$ \to $$ R2 - R1
R3 $$ \to $$ R3 - R1
= $$y\left| {\matrix{ 1 & \omega & {{\omega ^2}} \cr 0 & {y + {\omega ^2} - \omega } & {1 - {\omega ^2}} \cr 0 & {1 - \omega } & {y + \omega - {\omega ^2}} \cr } } \right|$$
= y$$\left[ {\left( {y + {\omega ^2} - \omega } \right)\left( {y + \omega - {\omega ^2}} \right) - \left( {1 - {\omega ^2}} \right)\left( {1 - \omega } \right)} \right]$$
= y(y2) = y3
$$ \therefore $$ $$\alpha $$ = $$\omega $$ and $$\beta $$ = $${\omega ^2}$$
$$\left| {\matrix{ {y + 1} & \alpha & \beta \cr \alpha & {y + \beta } & 1 \cr \beta & 1 & {y + \alpha } \cr } } \right|$$
= $$\left| {\matrix{ {y + 1} & \omega & {{\omega ^2}} \cr \omega & {y + {\omega ^2}} & 1 \cr {{\omega ^2}} & 1 & {y + \omega } \cr } } \right|$$
C1 $$ \to $$ C1 + C2 + C3
= $$\left| {\matrix{ {y + 1 + \omega + {\omega ^2}} & \omega & {{\omega ^2}} \cr {y + 1 + \omega + {\omega ^2}} & {y + {\omega ^2}} & 1 \cr {y + 1 + \omega + {\omega ^2}} & 1 & {y + \omega } \cr } } \right|$$
= $$\left| {\matrix{ y & \omega & {{\omega ^2}} \cr y & {y + {\omega ^2}} & 1 \cr y & 1 & {y + \omega } \cr } } \right|$$
As $$1 + \omega + {\omega ^2}$$ = 0
= $$y\left| {\matrix{ 1 & \omega & {{\omega ^2}} \cr 1 & {y + {\omega ^2}} & 1 \cr 1 & 1 & {y + \omega } \cr } } \right|$$
R2 $$ \to $$ R2 - R1
R3 $$ \to $$ R3 - R1
= $$y\left| {\matrix{ 1 & \omega & {{\omega ^2}} \cr 0 & {y + {\omega ^2} - \omega } & {1 - {\omega ^2}} \cr 0 & {1 - \omega } & {y + \omega - {\omega ^2}} \cr } } \right|$$
= y$$\left[ {\left( {y + {\omega ^2} - \omega } \right)\left( {y + \omega - {\omega ^2}} \right) - \left( {1 - {\omega ^2}} \right)\left( {1 - \omega } \right)} \right]$$
= y(y2) = y3
Comments (0)
