If A, B and are non-singular matrices of same order, then the inverse of , is equal to [2025]

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Q.
If A, B and $(adj (A^{- 1}) + adj (B^{- 1}))$ are non-singular matrices of same order, then the inverse of $A (adj (A^{- 1}) + adj {(B^{- 1}))}^{- 1} B$ , is equal to [2025]

1 $A B^{- 1} + A^{- 1} B$

2 $\frac{A B^{- 1}}{| A |} + \frac{B A^{- 1}}{| B |}$

3 $adj (B^{- 1}) + adj (A^{- 1})$

4 $\frac{1}{| A B |} (adj (B) + adj (A))$

Ans.
(4)

${[A (adj (A^{- 1}) + adj {(B^{- 1}))}^{- 1} B]}^{- 1}$

$= B^{- 1} (adj (A^{- 1}) + adj (B^{- 1})) A^{- 1}$

$= B^{- 1} (adj (A^{- 1})) A^{- 1} + B^{- 1} (adj (B^{- 1})) A^{- 1}$

$= B^{- 1}$ $| A^{- 1} |$ $+$ $| B^{- 1} |$ $A^{- 1}$

$= \frac{B^{- 1}}{| A |} + \frac{A^{- 1}}{| B |} = \frac{adj (B)}{| B | | A |} + \frac{adj (A)}{| A | | B |}$

$= \frac{1}{| A | | B |} (adj (B) + adj (A))$

$= \frac{1}{| A B |} (adj (B) + adj (A))$ .

Solution

Q.
If A, B and $(adj (A^{- 1}) + adj (B^{- 1}))$ are non-singular matrices of same order, then the inverse of $A (adj (A^{- 1}) + adj {(B^{- 1}))}^{- 1} B$ , is equal to [2025]

1 $A B^{- 1} + A^{- 1} B$

2 $\frac{A B^{- 1}}{| A |} + \frac{B A^{- 1}}{| B |}$

3 $adj (B^{- 1}) + adj (A^{- 1})$

4 $\frac{1}{| A B |} (adj (B) + adj (A))$

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Solution

Q. If A, B and (adj(A–1)+adj(B–1)) are non-singular matrices of same order, then the inverse of A(adj(A–1)+adj(B–1))–1B, is equal to [2025]

1 AB–1+A–1B

2 AB–1|A|+BA–1|B|

3 adj(B–1)+adj(A–1)

4 1|AB|(adj(B)+adj(A))

Ans. (4) [A(adj(A–1)+adj(B–1))–1B]–1 =B–1(adj(A–1)+adj(B–1))A–1 =B–1(adj(A–1))A–1+B–1(adj(B–1))A–1 =B–1|A–1|+|B–1|A–1 =B–1|A|+A–1|B|=adj(B)|B||A|+adj(A)|A||B| =1|A||B|(adj(B)+adj(A)) =1|AB|(adj(B)+adj(A)).

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Q.
If A, B and $(adj (A^{- 1}) + adj (B^{- 1}))$ are non-singular matrices of same order, then the inverse of $A (adj (A^{- 1}) + adj {(B^{- 1}))}^{- 1} B$ , is equal to [2025]

1 $A B^{- 1} + A^{- 1} B$

2 $\frac{A B^{- 1}}{| A |} + \frac{B A^{- 1}}{| B |}$

3 $adj (B^{- 1}) + adj (A^{- 1})$

4 $\frac{1}{| A B |} (adj (B) + adj (A))$