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(C) Given,$a^2+b^2+c^2+d^2=1$ and $A=\left[\begin{array}{cc}a+ib & c+id \ -c+id & a-ib\end{array}ight]$.First,we calculate the determinant $|A|$:$|

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$\left[\begin{array}{cc}a-ib & -c-id \ c-id & a+ib\end{array}ight]$

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(C) Given,$a^2+b^2+c^2+d^2=1$ and $A=\left[\begin{array}{cc}a+ib & c+id \ -c+id & a-ib\end{array}ight]$.First,we calculate the determinant $|A|$:$|A| = (a+ib)(a-ib) - (c+id)(-c+id)$$|A| = (a^2 - (ib)^2) - ((id)^2 - c^2)$$|A| = (a^2 + b^2) - (-d^2 - c^2) = a^2+b^2+c^2+d^2 = 1$.Since $|A|=1$,the inverse $A^{-1}$ is given by the adjugate matrix $	ext{adj}(A)$:$A^{-1} = \frac{1}{|A|} 	ext{adj}(A) = \frac{1}{1} \left[\begin{array}{cc}a-ib & -(c+id) \ -(-c+id) & a+ib\end{array}ight]$$A^{-1} = \left[\begin{array}{cc}a-ib & -c-id \ c-id & a+ib\end{array}ight]$.

If $a, b, c$ and $d$ are real numbers such that $a^2+b^2+c^2+d^2=1$ and $A=\left[\begin{array}{cc}a+ib & c+id \\ -c+id & a-ib\end{array}\right]$,then $A^{-1}$ is equal to

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