Let A, B, A+B are non singular matrices of order 2×2 and satisfying A−1+B−1=(A+B)−1 then value of AB= ?

A−1(A+B)+B−1(A+B)=I⇒A−1B+B−1A≠I=0Parallelly BA−1+AB−1+I=0B+AB−1A+A=0and A+BA−1B+B=0⇒AB−1A=BA−1B

Let A, B, A+B are non singular matrices of order 2×2 and satisfying A−1+B−1=(A+B)−1 then value of AB= ?

A−1(A+B)+B−1(A+B)=I⇒A−1B+B−1A≠I=0Parallelly BA−1+AB−1+I=0B+AB−1A+A=0and A+BA−1B+B=0⇒AB−1A=BA−1B

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Let A, B, A+B are non singular matrices of order $2 \times 2$ and satisfying $A^{- 1} + B^{- 1} = (A + B)^{- 1}$ then value of $|\frac{A}{B}| = ?$

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answer is C.

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Detailed Solution

$\begin{array}{l} A^{- 1} (A + B) + B^{- 1} (A + B) = I \\ \Rightarrow A^{- 1} B + B^{- 1} A \neq I = 0 \end{array}$
Parallelly ${BA}^{- 1} + {AB}^{- 1} + I = 0$
$B + {AB}^{- 1} A + A = 0$
and $A + {BA}^{- 1} B + B = 0$
$\Rightarrow {AB}^{- 1} A = {BA}^{- 1} B$