AP EAMCET · Maths · Matrices
If \(a, b, c\) and \(d\) are real numbers such that \(a^2+b^2+c^2+d^2=1 \quad\) and \(A=\left[\begin{array}{c}a+i b c+i d \\ -c+i d a-i b\end{array}\right]\), then \(A^{-1}\) equals to
- A \(\left[\begin{array}{cc}a+i b & -c-i d \\ c-i d & a-i b\end{array}\right]\)
- B \(\left[\begin{array}{cc}a-i b & c+i d \\ -c+i d & a+i b\end{array}\right]\)
- C \(\left[\begin{array}{cc}a-i b & -c-i d \\ c-i d & a+i b\end{array}\right]\)
- D \(\left[\begin{array}{ll}a+i b & c+i d \\ c-i d & a-i b\end{array}\right]\)
Answer & Solution
Correct Answer
(C) \(\left[\begin{array}{cc}a-i b & -c-i d \\ c-i d & a+i b\end{array}\right]\)
Step-by-step Solution
Detailed explanation
Given, \[ a^2+b^2+c^2+d^2=1 \] \[ \text { and } \quad A=\left[\begin{array}{cc} a+i b & c+i d \\ -c+i d & a-i b \end{array}\right] \]…
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