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If A=\frac{1}{2}\left[\begin{array}{cc} 1 & \sqrt{3} \\ -\sqrt{3} & 1 \end{array}\right] , then :

Option: 1

\mathrm{A}^{30}+\mathrm{A}^{25}+\mathrm{A}=\mathrm{I}


Option: 2

A^{30}=A^{25}


Option: 3

\mathrm{A}^{30}+\mathrm{A}^{25}-\mathrm{A}=\mathrm{I}


Option: 4

\mathrm{A}^{30}-\mathrm{A}^{25}=2 \mathrm{I}


Answers (1)

best_answer

4=\frac{1}{2}\left[\begin{array}{cc} 1 & f_3 \\ -f^3 & 1 \end{array}\right]=\left[\begin{array}{cc} \frac{1}{2} & \frac{f_3}{2} \\ -\frac{f_3}{2} & \frac{1}{2} \end{array}\right]

|4-\lambda| \mid=0

\left|\begin{array}{cc} \frac{1}{2}-\lambda & \frac{f_3}{2} \\ -\frac{f_3}{2} & \frac{1}{2}-\lambda \end{array}\right|=0 \quad \Rightarrow \lambda^2+\frac{1}{4}-\lambda+\frac{3}{4}=0

\begin{aligned} \lambda^2-\lambda+1=0 & \Rightarrow A^2-A+1=0 \\ & \Rightarrow A^3-A^2+A=0 \end{aligned}

\begin{aligned} A^4=(A-I)^2 & \Rightarrow A^4=A^2+I-2 A \\ & \Rightarrow A^4=A-I+I-2 A=-A \end{aligned}

  

\mathrm{A}^4=-A

\begin{aligned} & \Rightarrow A^{30}=\left(A^4\right)^7 A^2=-A^4=-\left(A^4\right) A=-A^3=A-A^2 \\ & A^{25}=\left(A^4\right)^6 A=A^6 A=A^7=A^4 A^3=-A A^3=-A^4=A \end{aligned}

Put these values on all options, we, get,

\Rightarrow A^{30}+A^{25}-A=I

So, option (3) is correct.

 

Posted by

Rishabh

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