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If B is a non-singular matrix and A is a square matrix, then det (B-1AB) =
- a)det(A-1)
- b)det(B-1)
- c)det(A)
- d)det(B)
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
If B is a non-singular matrix and A is a square matrix, then det (B-1A...
Solution:
Given, B is a non-singular matrix and A is a square matrix, then we need to find det(B-1AB).
Using the property of determinant, we can write det(B-1AB) as
det(B-1) × det(A) × det(B)
Now, as B is a non-singular matrix, det(B) ≠ 0. Therefore, we can write det(B-1) as det(B)-1.
Substituting the values, we get
det(B-1AB) = det(B)-1 × det(A) × det(B)
= det(B) × det(A) × det(B)-1
As det(B) × det(B)-1 = 1, we get
det(B-1AB) = det(A)
Hence, the correct option is (c) det(A).
Given, B is a non-singular matrix and A is a square matrix, then we need to find det(B-1AB).
Using the property of determinant, we can write det(B-1AB) as
det(B-1) × det(A) × det(B)
Now, as B is a non-singular matrix, det(B) ≠ 0. Therefore, we can write det(B-1) as det(B)-1.
Substituting the values, we get
det(B-1AB) = det(B)-1 × det(A) × det(B)
= det(B) × det(A) × det(B)-1
As det(B) × det(B)-1 = 1, we get
det(B-1AB) = det(A)
Hence, the correct option is (c) det(A).
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