Let $\alpha $ and $\beta $ be the roots of the equation ${x^2} + x + 1 = 0.$ Then, for $y \ne 0$ in $R,\,\,\left| {\begin{array}{*{20}{c}} {y + 1}&\alpha &\beta \\ \alpha &{y + \beta }&1\\ \beta &1&{y + \alpha } \end{array}} \right|$ is equal to

JEE MATHEMATICS MCQ: MATRICES

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Application +4 | -1

Let

\alpha

and

\beta

be the roots of the equation

{x^2} + x + 1 = 0.

Then, for

y \ne 0

R,\,\,\left| {\begin{array}{cccccccccccccccccccc} {y + 1}&\alpha &\beta \\ \alpha &{y + \beta }&1\\ \beta &1&{y + \alpha } \end{array}} \right|

is equal to

View answer and explanation

Correct option: D

{y^3}

Option D (

{y^3}

) is correct because the determinant simplifies to this expression using properties of determinants and the roots of the given equation.

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\left|\begin{array}{ccc}a & b-c & b+c \\ c+a & b & c-a \\ a-b & a+b & c\end{array}\right|=0

a, b, c

a x+b y+c=0

{x_2} - {x_3} = 1,\; - {x_1} + 2{x_3} = - 2,\;{x_1} - 2,\;{x_1} - 2{x_2} = 3

2x + 3y + 1 = 0,3x + y - 2 = 0

ax + 2y - b = 0

\left|\begin{array}{ccc}1 & \left(\alpha^{2 x}-\alpha^{-2 x}\right)^2 & \left(\alpha^{2 x}+\alpha^{-2 x}\right)^2 \\ 1 & \left(\beta^{2 x}-\beta^{-2 x}\right)^2 & \left(\beta^{2 x}+\beta^{-2 x}\right)^2 \\ 1 & \left(\gamma^{2 x}-\gamma^{-2 x}\right)^2 & \left(\gamma^{2 x}+\gamma^{-2 x}\right)^2\end{array}\right|

\omega

\left| {\begin{array}{cccccccccccccccccccc} 1&\omega &{{\omega ^2}}\\ \omega &{{\omega ^2}}&1\\ {{\omega ^2}}&1&\omega \end{array}} \right|

A

2 \times 2

A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]

|A|