For non-singular square matrix A, B and C of the same order $(A B^{- 1} C) =$
(a) $A^{- 1} B C^{- 1}$
(b) $C^{- 1} B^{- 1} A^{- 1}$
(c) $C B A^{- 1}$
(d) $C^{- 1} B A^{- 1}$

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Solution

Disclaimer: In Quesion, We are to find the inverse of $(A B^{- 1} C)$ . The inverse is missing in the question.

(d) $C^{- 1} B A^{- 1}$

We have,
${(A B^{- 1} C)}^{- 1} = C^{- 1} {(B^{- 1})}^{- 1} A^{- 1} = C^{- 1} B A^{- 1}$

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Similar questions

a

\frac{1}{b}

c

\frac{1}{d}

Prove that:

(i) $\frac{a + b + c}{a^{- 1} b^{- 1} + b^{- 1} c^{- 1} + c^{- 1} a^{- 1}} = a b c$

(ii) $(a^{- 1} + b^{- 1})^{- 1} = \frac{a b}{a + b}$

\frac{a + b + c}{a^{- 1} b^{- 1} + b^{- 1} c^{- 1} + a^{- 1} c^{- 1}} = a b c

A, B

C

B A C = I,

A

A

B

| A^{- 1} B A |

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