Advertisements
Advertisements
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))`
shaalaa.com
Adjoint of a Matrix
Is there an error in this question or solution?