If A = begin{bmatrix} a+ib & c+id -c+id & a- | vedclass

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(A) Given the matrix $A = \begin{bmatrix} a+ib & c+id \ -c+id & a-ib \end{bmatrix}$.First,we calculate the determinant $|A|$:$|A| = (a+ib)(a-ib) - (c

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(A) Given the matrix $A = \begin{bmatrix} a+ib & c+id \ -c+id & a-ib \end{bmatrix}$.First,we calculate the determinant $|A|$:$|A| = (a+ib)(a-ib) - (c+id)(-c+id) = (a^2+b^2) - (-(c^2+d^2)) = a^2+b^2+c^2+d^2$.We know that $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$.The adjoint of $A$ is obtained by swapping the diagonal elements and changing the signs of the off-diagonal elements:$	ext{adj}(A) = \begin{bmatrix} a-ib & -c-id \ c-id & a+ib \end{bmatrix}$.Thus,$A^{-1} = \frac{1}{a^2+b^2+c^2+d^2} \begin{bmatrix} a-ib & -c-id \ c-id & a+ib \end{bmatrix}$.Comparing this with the given $A^{-1} = \begin{bmatrix} a+ib & -c-id \ -c+id & a-ib \end{bmatrix}$,we observe that for the equality to hold,the determinant $|A|$ must be $1$.Therefore,$a^2+b^2+c^2+d^2 = 1$.

If $A = \begin{bmatrix} a+ib & c+id \\ -c+id & a-ib \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} a+ib & -c-id \\ -c+id & a-ib \end{bmatrix}$,find $(a^2+b^2+c^2+d^2)$.

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