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Question: Answered & Verified by Expert
If $\mathrm{A}=\left[\begin{array}{ll}\mathrm{i} & 1 \\ 1 & 0\end{array}\right]$ where $\mathrm{i}=\sqrt{-1}$ and $\mathrm{B}=\mathrm{A}^{2029}$, then $\mathrm{B}^{-1}=$
MathematicsMatricesMHT CETMHT CET 2023 (11 May Shift 2)
Options:
  • A $-\mathrm{A}$
  • B $\operatorname{adj} \mathrm{A}$
  • C $-I$
  • D $-\operatorname{adj} \mathrm{A}$
Solution:
2077 Upvotes Verified Answer
The correct answer is: $-\operatorname{adj} \mathrm{A}$
$\begin{aligned} & A=\left[\begin{array}{ll}i & 1 \\ 1 & 0\end{array}\right] \\ & \therefore \quad A^2=\left[\begin{array}{ll}i & 1 \\ 1 & 0\end{array}\right] \times\left[\begin{array}{ll}i & 1 \\ 1 & 0\end{array}\right]=\left[\begin{array}{ll}0 & i \\ i & 1\end{array}\right] \\ & \therefore \quad A^3=\left[\begin{array}{ll}0 & i \\ i & 1\end{array}\right]\left[\begin{array}{ll}i & 1 \\ 1 & 0\end{array}\right]=\left[\begin{array}{ll}i & 0 \\ 0 & i\end{array}\right] \\ & \therefore \quad \mathrm{A}^6=\mathrm{A}^3 \times \mathrm{A}^3 \\ & =\left[\begin{array}{ll}\mathrm{i} & 0 \\ 0 & \mathrm{i}\end{array}\right] \times\left[\begin{array}{ll}\mathrm{i} & 0 \\ 0 & \mathrm{i}\end{array}\right] \\ & =\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right]=(-1) I_2 \\ & \text { Now, } B=A^{2029}=A^{(6 \times 338+1)} \\ & \therefore \quad \mathrm{B}=\left(\mathrm{A}^6\right)^{338} \times \mathrm{A} \\ & =\left((-1) \mathrm{I}_2\right)^{338} \times \mathrm{A} \\ & =\mathrm{I}_2 \times \mathrm{A} \\ & \therefore \quad \mathrm{B}=\mathrm{A} \\ & \text { Now, }|A|=-1 \\ & \therefore \quad B^{-1}=A^{-1}=\frac{1}{|A|} \operatorname{adj} A \\ & \therefore \quad \mathrm{B}^{-1}=-\operatorname{adj} \mathrm{A} \\ & \end{aligned}$

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