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Which of the following is not a property of invertible matrices if A and B are matrices of the same order?
- a)(AB)-1 = A-1 B-1
- b)(AA-1) = (A-1 A) = I
- c)(AB)-1 = B-1 A-1
- d)AB = BA = I
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
Which of the following is not a property of invertible matrices if A a...
Properties of Invertible Matrices:
1. (AA-1) = (A-1 A) = I
2. AB = BA = I
3. (A-1)-1 = A
4. (kA)-1 = (1/k)A-1, where k is a non-zero scalar
5. (AB)-1 = B-1 A-1
Explanation:
Option A states that (AB)-1 = A-1 B-1, which is not a property of invertible matrices. This statement is false because in general, (AB)-1 ≠ A-1 B-1.
To see why this is true, consider the case where A and B are both 2x2 matrices:
A = [a b]
[c d]
B = [e f]
[g h]
Then AB is given by:
AB = [ae+bg af+bh]
[ce+dg cf+dh]
The inverse of AB, assuming it exists, is given by:
(AB)-1 = 1/det(AB) [dh -bh -cf af]
[-dg ae ce -af]
where det(AB) = (ae+bg)(cf+dh)-(af+bh)(ce+dg)
On the other hand, the inverse of A and B are given respectively by:
A-1 = 1/det(A) [d -b]
[-c a]
B-1 = 1/det(B) [h -f]
[-g e]
where det(A) = ad-bc and det(B) = eh-fg
Now, if we compute A-1 B-1, we get:
A-1 B-1 = 1/det(A)det(B) [(dh-bgf-eh+fg) -(bh-ae-hf+cf)]
[(-dg+ce+bg-af) (ag-ce-df+ae)]
Notice that A-1 B-1 is not equal to (AB)-1 in general, unless A and B commute (i.e. AB = BA).
Therefore, option A is not a property of invertible matrices.
1. (AA-1) = (A-1 A) = I
2. AB = BA = I
3. (A-1)-1 = A
4. (kA)-1 = (1/k)A-1, where k is a non-zero scalar
5. (AB)-1 = B-1 A-1
Explanation:
Option A states that (AB)-1 = A-1 B-1, which is not a property of invertible matrices. This statement is false because in general, (AB)-1 ≠ A-1 B-1.
To see why this is true, consider the case where A and B are both 2x2 matrices:
A = [a b]
[c d]
B = [e f]
[g h]
Then AB is given by:
AB = [ae+bg af+bh]
[ce+dg cf+dh]
The inverse of AB, assuming it exists, is given by:
(AB)-1 = 1/det(AB) [dh -bh -cf af]
[-dg ae ce -af]
where det(AB) = (ae+bg)(cf+dh)-(af+bh)(ce+dg)
On the other hand, the inverse of A and B are given respectively by:
A-1 = 1/det(A) [d -b]
[-c a]
B-1 = 1/det(B) [h -f]
[-g e]
where det(A) = ad-bc and det(B) = eh-fg
Now, if we compute A-1 B-1, we get:
A-1 B-1 = 1/det(A)det(B) [(dh-bgf-eh+fg) -(bh-ae-hf+cf)]
[(-dg+ce+bg-af) (ag-ce-df+ae)]
Notice that A-1 B-1 is not equal to (AB)-1 in general, unless A and B commute (i.e. AB = BA).
Therefore, option A is not a property of invertible matrices.
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Which of the following is not a property of invertible matrices if A a...
(AB)-1 = A-1 B-1 is incorrect. The correct formula is (AB)-1 = B-1 A-1. B-1 A-1 ≠ A-1 B-1 as matrix multiplication is not commutative.
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Which of the following is not a property of invertible matrices if A and B are matrices of the same order?a)(AB)-1 = A-1 B-1b)(AA-1) = (A-1 A) = Ic)(AB)-1 = B-1 A-1d)AB = BA = ICorrect answer is option 'A'. Can you explain this answer?
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