If A=⎣⎡ 2i−1+i32i1+i3 2i−1−i32i1−i3⎦⎤,i=−i and f(x)=x2+2, then f(A) is equal to

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If A = {bmatrix} {-1 + i {3}}{2i}& {-1 - i {3}}{2i} {1 + i

Question

If $A = ⎣ ⎡ \frac{- 1 + i 3}{2 i} \frac{1 + i 3}{2 i} \frac{- 1 - i 3}{2 i} \frac{1 - i 3}{2 i} ⎦ ⎤, i = - i$ and $f (x) = x^{2} + 2$ , then $f (A)$ is equal to

(\frac{5 - i 3}{2}) [1001]

(\frac{3 - i 3}{2}) [1001]

[1001]

(2 + i 3) [1001]

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Noah Discussed

A = ⎣ ⎡ \frac{- 1 + i 3}{2 i} \frac{1 + i 3}{2 i} \frac{- 1 - i 3}{2 i} \frac{1 - i 3}{2 i} ⎦ ⎤, i = - i

and

f (x) = x^{2} + 2

, then

f (A)

is equal to

14 mins ago

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14 mins ago

From the given matrix, we have

A = ⎣ ⎡ \frac{ω}{i} - \frac{ω ^{2}}{i} \frac{ω ^{2}}{i} - \frac{ω}{i} ⎦ ⎤ = \frac{ω}{i} [1 - ω ω - 1]

Thus

A^{2} = ω^{2} [1 - ω^{2} 0 0 1 - ω^{2}]

= - [- ω^{2} + ω^{4} 0 0 - ω^{2} + ω]

= [- ω^{2} + ω 0 0 - ω^{2} + ω]

....

Since f (x) = x^{2} + 2

Therefore,

f (A) = A^{2} + 2 = [- ω^{2} + ω 0 0 - ω^{2} + ω] + [2002]

- [ω^{2} + ω + 2] [1001]

= (3 + 2 ω) [1001]

= (2 + i 3) [1001]

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Question Text	If $A = ⎣ ⎡ \frac{- 1 + i 3}{2 i} \frac{1 + i 3}{2 i} \frac{- 1 - i 3}{2 i} \frac{1 - i 3}{2 i} ⎦ ⎤, i = - i$ and $f (x) = x^{2} + 2$ , then $f (A)$ is equal to
Answer Type	Text solution:1
Upvotes	150