If $A = \left[ {\begin{array}{*{20}{c}} {\cos \theta }&{\sin \theta }\\ { - \sin \theta }&{\cos \theta } \end{array}} \right],$ then $\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\;{A^n}$ is

$\left[ {\begin{array}{*{20}{c}} 0&1\\ { - 1}&0 \end{array}} \right]$

JEE MATHEMATICS MCQ: MATRICES

1 Easy

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Application +4 | -1

A = \left[ {\begin{array}{cccccccccccccccccccc} {\cos \theta }&{\sin \theta }\\ { - \sin \theta }&{\cos \theta } \end{array}} \right],

then

\lim\limits_{n \to \infty } \frac{1}{n}\;{A^n}

View answer and explanation

Correct option: A

A null matrix

Option A (A null matrix) is correct because as n approaches infinity, the terms cos(nθ)/n and sin(nθ)/n approach 0.

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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|