Question

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

विकल्प

• det

• `1/(det(A))`

• 1

• 0

Answer 1

`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))`