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(C) Given $A + B = I$ and $A^{-1} + B^{-1} = 2I$.Since $A^{-1} + B^{-1} = \frac{A+B}{AB} = 2I$,we have $I = 2AB$,which implies $AB = \frac{1}{2}I$.Tak

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$64$

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(C) Given $A + B = I$ and $A^{-1} + B^{-1} = 2I$.Since $A^{-1} + B^{-1} = \frac{A+B}{AB} = 2I$,we have $I = 2AB$,which implies $AB = \frac{1}{2}I$.Taking the determinant on both sides,$|AB| = |\frac{1}{2}I| = (\frac{1}{2})^3 |I| = \frac{1}{8}$.We need to find $|adj(4AB)|$.Since $4AB = 4(\frac{1}{2}I) = 2I$,we have $|4AB| = |2I| = 2^3 |I| = 8$.The property of the adjoint matrix is $|adj(M)| = |M|^{n-1}$,where $n$ is the order of the matrix.Here $n = 3$,so $|adj(4AB)| = |4AB|^{3-1} = |4AB|^2$.Substituting $|4AB| = 8$,we get $|adj(4AB)| = 8^2 = 64$.

Let $A$ and $B$ be two non-singular matrices of order $3$ such that $A + B = I$ and $A^{-1} + B^{-1} = 2I$. Then $|adj(4AB)|$ is equal to (where $adj(A)$ is the adjoint of matrix $A$):

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