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Question: If \(A = \begin{bmatrix} i & 0 \\ 0 & i \end{bmatrix},n \in N\), then \(A^{4n}\)equals...

If A=[i00i],n∈NA = \begin{bmatrix} i & 0 \\ 0 & i \end{bmatrix},n \in N, then A4nA^{4n}equals

A

[1001]\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

B

[i00i]\begin{bmatrix} i & 0 \\ 0 & i \end{bmatrix}

C

[0ii0]\begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}

D

[0000]\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}

Answer

[1001]\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

Explanation

Solution

$A^{2} = \begin{bmatrix} i & 0 \ 0 & i \end{bmatrix}\begin{bmatrix} i & 0 \ 0 & i \end{bmatrix} = \begin{bmatrix}

  • 1 & 0 \ 0 & - 1 \end{bmatrix},, A^{4} = A^{2}.A^{2} = \begin{bmatrix}
  • 1 & 0 \ 0 & - 1 \end{bmatrix}\begin{bmatrix}
  • 1 & 0 \ 0 & - 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = I;; (A^{4})^{n} = I^{n} = I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$