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JEE MATHEMATICS MCQ: MATRICES

1 Easy

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Analyzing +4 | -1

Let

\alpha

be a root of the equation

{x^2} + x + 1 = 0

and the matrix

A = \frac{1}{{\sqrt 3 }}\left[ {\begin{array}{cccccccccccccccccccc} 1&1&1\\ 1&\alpha &{{\alpha ^2}}\\ 1&{{\alpha ^2}}&{{\alpha ^4}} \end{array}} \right]

, then the matrix

{A^{31}}

is equal to

View answer and explanation

Correct option: A

{A^3}

Option A (

{A^3}

) is correct because after calculating

A^2

and using the property

{\alpha^3} = 1

, we find that

{A^4} = I

, so

{A^{31}} = {A^{28}} \cdot {A^3} = {A^3}

.

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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

are real numbers then the line

a x+b y+c=0

passes through the fixed point is The number of solutions of the system of equations

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

is If the equations

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

ax + 2y - b = 0

are consistent, then The value of the determinant

\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

is the cube root of unity, then

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

is equal to For any matrix

A

2 \times 2

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

|A|