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Let R denote the set of all real numbers and let i = -1 . Consider the matrices S = bmatrix 0 & -1 1 & 0 bmatrix and T = bmatrix 1 & 1 0 & 1 bmatrix . Let a, b, c, d be real numbers such that ST = bmatrix a & b c & d bmatrix . Let H = x + iy : x, y R and y > 0 . Then which of the following statements is (are) TRUE ?

Options

  1. Ab + ia d + ic = i
  2. BIf = -1 + i 3 2 , then a + b c + d =
  3. CIf m is an integer greater than 2 such that (ST)^2 = (ST)^m , then m is an integer multiple of 8
  4. DIf z H , then az + b cz + d H

Correct answer

B. If = -1 + i 3 2 , then a + b c + d =

Step-by-step solution

We are given the matrices S = bmatrix 0 & -1 1 & 0 bmatrix and T = bmatrix 1 & 1 0 & 1 bmatrix . First, we compute the product ST : ST = bmatrix 0 & -1 1 & 0 bmatrix bmatrix 1 & 1 0 & 1 bmatrix = bmatrix 0 & -1 1 & 1 bmatrix Comparing this with ST = bmatrix a & b c & d bmatrix , we get a = 0 , b = -1 , c = 1 , and d = 1 . Evaluating Option (A): b + ia d + ic = -1 + i(0) 1 + i(1) = -1 1 + i = -1(1 - i) (1 + i)(1 - i) = -1 + i 2 This is not equal to i , so statement (A) is FALSE. Evaluating Option (B): Given = -1 + i

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