If A=−1+i32i −1−i32i1+i32i 1−i32i,i=−1 and fx=x2+2 then fA=

A=ωiω2i−ω2i−ωi⇒A2=−ω2+ω00ω2+ωfx=x2+2⇒fA=A2+2I=−ω2+ω00−ω2+ω+2002=−ω2+ω+21001=2+i31001

If A=−1+i32i −1−i32i1+i32i 1−i32i,i=−1 and fx=x2+2 then fA=

A=ωiω2i−ω2i−ωi⇒A2=−ω2+ω00ω2+ωfx=x2+2⇒fA=A2+2I=−ω2+ω00−ω2+ω+2002=−ω2+ω+21001=2+i31001

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If $A = [\begin{array}{l} \frac{- 1 + i \sqrt{3}}{2 i} \frac{- 1 - i \sqrt{3}}{2 i} \\ \frac{1 + i \sqrt{3}}{2 i} \frac{1 - i \sqrt{3}}{2 i} \end{array}], i = \sqrt{- 1}$ and $f (x) = x^{2} + 2$ then $f (A) =$

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$[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

$(\frac{3 - i \sqrt{3}}{2}) [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

$(\frac{5 - i \sqrt{3}}{2}) [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

$(2 + i \sqrt{3}) [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

answer is D.

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$A = [\begin{matrix} \frac{ω}{i} & \frac{ω^{2}}{i} \\ \frac{- ω^{2}}{i} & \frac{- ω}{i} \end{matrix}]$ $\Rightarrow$ $A^{2} = (\begin{matrix} - ω^{2} + ω & 0 \\ 0 & ω^{2} + ω \end{matrix})$

$f (x) = x^{2} + 2$ $\Rightarrow$ $f (A) = A^{2} + 2 I$

$= (\begin{matrix} - ω^{2} + ω & 0 \\ 0 & - ω^{2} + ω \end{matrix}) + (\begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix})$

$= (- ω^{2} + ω + 2) [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = (2 + i \sqrt{3}) [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

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If A=−1+i32i −1−i32i1+i32i 1−i32i,i=−1 and fx=x2+2 then fA=

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If $A = [\begin{array}{l} \frac{- 1 + i \sqrt{3}}{2 i} \frac{- 1 - i \sqrt{3}}{2 i} \\ \frac{1 + i \sqrt{3}}{2 i} \frac{1 - i \sqrt{3}}{2 i} \end{array}], i = \sqrt{- 1}$ and $f (x) = x^{2} + 2$ then $f (A) =$