$\left| {\begin{array}{*{20}{c}} {\rm{\alpha }}&{ - {\rm{\beta }}}&0\\ 0&{\rm{\alpha }}&{\rm{\beta }}\\ {\rm{\beta }}&0&{\rm{\alpha }} \end{array}} \right| = 0$, then

${\rm{\alpha }}$ is one of the cube roots of unity

${\rm{\alpha \beta }}$ is one of the cube roots of unity

JEE MATHEMATICS MCQ: MATRICES

1 Easy

1 of 1

Application +4 | -1

\left| {\begin{array}{cccccccccccccccccccc} \mathrm{\alpha }&{ - \mathrm{\beta }}&0\\ 0&\mathrm{\alpha }&\mathrm{\beta }\\ \mathrm{\beta }&0&\mathrm{\alpha } \end{array}} \right| = 0

, then

View answer and explanation

Correct option: A

\frac{\mathrm{\alpha }}{\mathrm{\beta }}

is one of the cube roots of unity

Option A (

\frac{{\rm{{\alpha }}}}{{\rm{{\beta }}}}}

is one of the cube roots of unity) is correct because the determinant equation simplifies to

(\frac{{\rm{{\alpha }}}}{{\rm{{\beta }}}})^3 = 1

, indicating that

\frac{{\rm{{\alpha }}}}{{\rm{{\beta }}}}

is a cube root of unity.

Practice All Chapter MCQs →

\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|