If A is an invertible matrix of order 2, then det $({A^{ - 1}})$ is equal to

(A) det (A)

(B) $\cfrac{1}{{\det (A)}}$

(C) 1

(D) 0

Opton b is corect

When A is an invertible matrix of order 2, $A{A^{ - 1}} = {I_2} = {A^{ - 1}}A,$

where ${I_2}$ is identity matrix of order 2.
$\Rightarrow$ $\det (A{A^{ - 1}}) = \det I \Rightarrow \det A.\det ({A^{ - 1}}) = 1$

$\Rightarrow$ $\det ({A^{ - 1}}) = \cfrac{1}{{\det \;A}},\det A \ne 0$

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