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

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Question

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

Options

  • det

  • `1/(det(A))`

  • 1

  • 0

MCQ

Solution

`1/(det(A))`

Explanation:

Since A is an invertible matrix A-1 exists and A-1 = `1/|A|` adj A.

As matrix A is of order 2, let A = `[(a, b),(c, d)]`

Then |A| = ad – bc and adj A = `[(d, -b),(-c, a)]`

Now, A-1 = `1/|A|` adj A = `[(d/|A|, (-b)/|A|),((-c)/|A|, a/|A|)]`

∴ `|A^-1| = [(d/|A|, (-b)/|A|),((-c)/|A|, a/|A|)]`

= `1/|A^2| |(d, -b),(-c, a)|`

= `1/|A^2| (ad - bc)`

= `1/|A^2| |A|`

= `1/A`

∴ det (A-1) = `1/(det(A))`

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