If $A$ and $B$ are invertible matrices such that $A B = C$ then $d e t (B)$ equals

\frac{d e t (C)}{| C^{- 1} |}

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\frac{d e t (A)}{A^{- 1}}

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\frac{d e t (C)}{d e t (A)}

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None of these

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Solution

The correct option is C $\frac{d e t (C)}{d e t (A)}$
Given, $A B = C$
$| A B | = | C |$
$| A | | B | = | C |$
$\Rightarrow | B | = \frac{| C |}{| A |}$

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Similar questions

|A|

If A and B are invertible matrices, then which of the following is not correct?

(a) $a d j A = | A | . A^{- 1}$
(b) $d e t (A)^{- 1} = [d e t (A)]^{- 1}$
(c) $(A B)^{- 1} = B^{- 1} A^{- 1}$
(d) $(A + B)^{- 1} = B^{- 1} + A^{- 1}$

If A is an invertible matrix of order 2, then det (A⁻¹) is equal to

A. det (A) B. C. 1 D. 0

adj A = |A| A^{- 1}

\det (A^{- 1}) = {(\det A)}^{- 1}

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

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

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