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Elementary Transformations

Let A and B b...

Question

Let A and B be two non-singular matrices which commute. The $A^{- 1}$ , $B^{- 1}$

A

do not commute

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B

commute

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C

A B = A^{- 1} B^{- 1}

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D

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

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Solution

The correct option is B commute
$A^{- 1} B^{- 1} = (B A)^{- 1} = (A B)^{- 1}$ since $A, B$ commute
$\Rightarrow A^{- 1} B^{- 1} = B^{- 1} A^{- 1}$
Hence $A^{- 1}, B^{- 1}$ also commute

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Similar questions

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

A

B

are non singular matrices which commute then

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

=

Q. Assertion : If

A

is a non-singular matrix, then

| A^{-} 1 | = \frac{1}{| A |}

A

B

are inverse matrices of each other , then

A B = B A = I

| A B | = | A | | B |

A

B

are non-singular, symmetric and commutable matrices, then

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

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

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

[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}

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