જો $A = \left( {\begin{array}{*{20}{c}}
{\cos \,\alpha }&{ - \sin \,\alpha }\\
{\sin \,\alpha }&{\cos \,\alpha }
\end{array}} \right)$, $\left( {\alpha \in R} \right)$ આપલે છે કે જેથી ${A^{32}} = \left( {\begin{array}{*{20}{c}}
0&{ - 1}\\
1&0
\end{array}} \right)$ તો $\alpha $ ની કિમંત મેળવો.
{\cos \,\alpha }&{ - \sin \,\alpha }\\
{\sin \,\alpha }&{\cos \,\alpha }
\end{array}} \right)$, $\left( {\alpha \in R} \right)$ આપલે છે કે જેથી ${A^{32}} = \left( {\begin{array}{*{20}{c}}
0&{ - 1}\\
1&0
\end{array}} \right)$ તો $\alpha $ ની કિમંત મેળવો.
JEE MAIN 2019, Difficult
Download our app for free and get started
$A = \left[ {\begin{array}{*{20}{c}}
{\cos \alpha }&{ - \sin \alpha }\\
{\sin \alpha }&{\cos \alpha }
\end{array}} \right]$
{\cos \alpha }&{ - \sin \alpha }\\
{\sin \alpha }&{\cos \alpha }
\end{array}} \right]$
${A^2} = \left[ {\begin{array}{*{20}{c}}
{\cos \alpha }&{ - \sin \alpha }\\
{\sin \alpha }&{\cos \alpha }
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{\cos \alpha }&{ - \sin \alpha }\\
{\sin \alpha }&{\cos \alpha }
\end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}}
{\cos 2\alpha }&{ - \sin 2\alpha }\\
{\sin 2\alpha }&{\cos 2\alpha }
\end{array}} \right]$
${A^3} = \left[ {\begin{array}{*{20}{c}}
{\cos 2\alpha }&{ - \sin 2\alpha }\\
{\sin 2\alpha }&{\cos 2\alpha }
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{\cos \alpha }&{ - \sin \alpha }\\
{\sin \alpha }&{\cos \alpha }
\end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}}
{\cos 3\alpha }&{ - \sin 3\alpha }\\
{\sin 3\alpha }&{\cos 3\alpha }
\end{array}} \right]$
Simiarly ${A^{32}} = \left[ {\begin{array}{*{20}{c}}
{\cos 32\alpha }&{ - \sin 32\alpha }\\
{\sin 32\alpha }&{\cos 32\alpha }
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0&{ - 1}\\
1&0
\end{array}} \right]$
$ \Rightarrow \cos 32\alpha = 0$ and $\sin 32\alpha = 0$
$ \Rightarrow 32\alpha = \left( {4n + 1} \right)\frac{\pi }{2},n \in 1$
$\alpha = \left( {4n + 1} \right)\frac{\pi }{{64}},n \in 1$
$\alpha = \frac{\pi }{{64}}$ for $n = 0$
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1જો $A = \left[ {\begin{array}{*{20}{c}}1&0\\2&0\end{array}} \right],B = \left[ {\begin{array}{*{20}{c}}0&0\\1&{12}\end{array}} \right]$, તોView Solution
- 2$\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&{1 - x}&1\\1&1&{1 + y}\end{array}\,} \right|$ = . . .View Solution
- 3જો $A = \left[ {\begin{array}{*{20}{c}}{ab}&{{b^2}}\\{ - {a^2}}&{ - ab}\end{array}} \right]$ અને ${A^n} = O$, તો $ n$ ની ન્યૂનતમ કિમત મેળવો.View Solution
- 4$\left| {\,\begin{array}{*{20}{c}}{{{({a^x} + {a^{ - x}})}^2}}&{{{({a^x} - {a^{ - x}})}^2}}&1\\{{{({b^x} + {b^{ - x}})}^2}}&{{{({b^x} - {b^{ - x}})}^2}}&1\\{{{({c^x} + {c^{ - x}})}^2}}&{{{({c^x} - {c^{ - x}})}^2}}&1\end{array}\,} \right| = $View Solution
- 5જો ${a_1},{a_2},{a_3}.....{a_n}....$ એ સમગુણોતર શ્રેણીમાં હોય તો $\left| {\,\begin{array}{*{20}{c}}{\log {a_n}}&{\log {a_{n + 1}}}&{\log {a_{n + 2}}}\\{\log {a_{n + 3}}}&{\log {a_{n + 4}}}&{\log {a_{n + 5}}}\\{\log {a_{n + 6}}}&{\log {a_{n + 7}}}&{\log {a_{n + 8}}}\end{array}\,} \right|$ ની કિમંત મેળવો.View Solution
- 6જો $A$ એ $ 3$ કક્ષાવાળો ચોરચ શ્રેણિક હોય , તો $. . . . .. \ ($કે જ્યાં $I$ એ એકમ શ્રેણિક છે.$)$View Solution
- 7$A$ અને $B$ એ $3 \times 3$ કક્ષાનો શ્રેણિક છે કે જેથી $AB + A + B = 0$ હોય તો . . . .View Solution
- 8જો $\left| {\,\begin{array}{*{20}{c}}{ - {a^2}}&{ab}&{ac}\\{ab}&{ - {b^2}}&{bc}\\{ac}&{bc}&{ - {c^2}}\end{array}\,} \right| = K{a^2}{b^2}{c^2} $ તો $K = $View Solution
- 9જો $A = \left[ {\begin{array}{*{20}{c}}{ab}&{{b^2}}\\{ - {a^2}}&{ - ab}\end{array}} \right]$ અને ${A^n} = O$, તો $ n$ ની ન્યૂનતમ કિમત મેળવો.View Solution
- 10ધારો કે $x , y , z > 1$ અને $A=\left[\begin{array}{lll}1 & \log _x y & \log _x z \\ \log _y x & 2 & \log _y z \\ \log _z x & \log _z y & 3\end{array}\right]$ તો $\left|\operatorname{adj}\left(\operatorname{adj} A^2\right)\right| =.........$View Solution