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

(a) det (A) 

(b) 1/det(A) 

(c) 1 

(d) 0


We know AA-1 = I

Take determinant on both sides

|AA-1| = |I|

|A| |A-1| = 1

=> |A-1| = 1/|A|

Hence option b is the answer.

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