If the matrices $A, B, (A + B)$ are non singular then ${[A {(A + B)}^{- 1} B]}^{- 1}$ is equal to-

A + B

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

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A {(A + B)}^{- 1}

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None

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Solution

The correct option is C $A^{- 1} + B^{- 1}$
$(A (A + B)^{- 1} B)^{- 1}$

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

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

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

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

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Similar questions

Q. Assertion :If the matrices A, B, (A + B) are nonsingular, then

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

Reason:

[A (A + B)^{- 1} B]^{- 1} = [A (A^{- 1} + B^{- 1}) B]^{- 1} = [(I + A B^{- 1}) B]^{- 1} = [(B + A B^{- 1} B)]^{- 1} = [(B + A I)]^{- 1} = [(B + 1)]^{- 1} = B^{- 1} + A^{- 1}

Q. If

A

and

B

are non singular matrices which commute then

A (A (A + B)^{- 1} B)^{- 1} B

=

Q. If A and B are invertible matrices, then which one of the following is not correct?
(a) adj A =

|A|

A^-1 (b) det (A^-1) = [det(A)]^-1
(c) (AB)^-1 = B^-1A^-1 (d) (A+B)^-1 = B^-1 + A^-1

Q. If A, B are two n × n non-singular matrices, then
(a) AB is non-singular
(b) AB is singular
(c)

{(A B)}^{- 1} A^{- 1} B^{- 1}

(d) (AB)⁻¹ does not exist

If A, B are square matrices of order 3, A is non-singular and AB = O, then B is a
(a) null matrix
(b) singular matrix
(c) unit-matrix
(d) non-singular matrix

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If the matrices A,B,(A+B) are non singular then [A(A+B)−1B]−1 is equal to-

The correct option is C A−1+B−1(A(A+B)−1B)−1=[(A(A−1+B−1))B]−1=[(AA−1+AB−1)B]−1=((I+AB−1)B)−1=(B+AB−1B)−1=(B+A)−1=A−1+B−1

If the matrices $A, B, (A + B)$ are non singular then ${[A {(A + B)}^{- 1} B]}^{- 1}$ is equal to-

The correct option is C $A^{- 1} + B^{- 1}$
$(A (A + B)^{- 1} B)^{- 1}$

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

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

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

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