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#### Please Solve R.D.Sharma Class 12 Chapter 4 Exercise MCQs Question 2 Maths Textbook Solution.

Answer: The correct option is $\text { (C), } A^{4}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$

Given:$A=\left[\begin{array}{ll} i & 0 \\ 0 & i \end{array}\right]$

Solution:

Now,$A^{2}=A \times A$

\begin{aligned} &=\left[\begin{array}{ll} i & 0 \\ 0 & i \end{array}\right] \times\left[\begin{array}{cc} i & 0 \\ 0 & i \end{array}\right] \\ &\quad=\left[\begin{array}{cc} i^{2} & 0 \\ 0 & i^{2} \end{array}\right] \\ &\quad=\left[\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right] \\ &A^{3}=A^{2} \times A \\ &=\left[\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right] \times\left[\begin{array}{cc} i & 0 \\ 0 & i \end{array}\right] \\ &=\left[\begin{array}{cc} -i & 0 \\ 0 & -i \end{array}\right] \\ &A^{4}=A^{3} \times A \\ &=\left[\begin{array}{cc} -i & 0 \\ 0 & -i \end{array}\right] \times\left[\begin{array}{cc} i & 0 \\ 0 & i \end{array}\right] \\ &=\left[\begin{array}{cc} -i^{2} & 0 \\ 0 & -i^{2} \end{array}\right] \\ &=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \end{aligned}

So,$A^{4}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$