The value of the determinant $\left| {\begin{array}{*{20}{c}} 1&{\cos \left( {{\rm{\alpha }} - {\rm{\beta }}} \right)}&{{\rm{cos\alpha }}}\\ {\cos \left( {{\rm{\alpha }} - {\rm{\beta }}} \right)}&1&{{\rm{cos\beta }}}\\ {\cos {\rm{\alpha }}}&{{\rm{cos\beta }}}&1 \end{array}} \right|$ is

${{\rm{\alpha }}^2} - {{\rm{\beta }}^2}$

${{\rm{\alpha }}^2} + {{\rm{\beta }}^2}$

JEE MATHEMATICS MCQ: MATRICES

1 Hard

1 of 1

Analyzing +4 | -1

The value of the determinant

\left| {\begin{array}{cccccccccccccccccccc} 1&{\cos \left( {\mathrm{\alpha } - \mathrm{\beta }} \right)}&{\mathrm{cos\alpha }}\\ {\cos \left( {\mathrm{\alpha } - \mathrm{\beta }} \right)}&1&{\mathrm{cos\beta }}\\ {\cos \mathrm{\alpha }}&{\mathrm{cos\beta }}&1 \end{array}} \right|

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Correct option: A

Option A (0) is correct because the given determinant can be simplified to 0 using trigonometric identities.

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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|