If a, b, c and d are real numbers such that a | vedclass

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

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$\left[\begin{array}{cc}a-i b & -c-i d \ c-i d & a+i b\end{array}ight]$

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(C) Given that $a^2+b^2+c^2+d^2=1$ and $A=\left[\begin{array}{cc}a+i b & c+i d \ -c+i d & a-i b\end{array}ight]$.First,we calculate the determinant of $A$,denoted as $|A|$.$|A| = (a+i b)(a-i b) - (c+i d)(-c+i d)$$|A| = (a^2 - (i b)^2) - ((i d)^2 - c^2)$$|A| = (a^2 + b^2) - (-d^2 - c^2)$$|A| = a^2 + b^2 + c^2 + d^2 = 1$.The inverse of a $2 	imes 2$ matrix $A = \left[\begin{array}{cc}x & y \ z & w\end{array}ight]$ is given by $A^{-1} = \frac{1}{|A|} \left[\begin{array}{cc}w & -y \ -z & x\end{array}ight]$.Substituting the values of $A$ and $|A|=1$,we get:$A^{-1} = \frac{1}{1} \left[\begin{array}{cc}a-i b & -(c+i d) \ -(-c+i d) & a+i b\end{array}ight]$$A^{-1} = \left[\begin{array}{cc}a-i b & -c-i d \ c-i d & a+i b\end{array}ight]$.

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

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