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Let A, B be n x n matrices such that BA+ B2 = I - BA2, where I is the nxn identity matrix, which of the following is always true ?
- a)A is non - singular
- b)B is non - singular
- c)A + B is non - singular
- d)AB is non - singular
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Let A, B be n x n matrices such that BA+ B2 = I - BA2, where I is the ...
Given
⇒ A + B + AB be inverse of B
⇒ B is invertible => B is non - singular.
Most Upvoted Answer
Let A, B be n x n matrices such that BA+ B2 = I - BA2, where I is the ...
Explanation:
We are given that BA B2 = I - BA2, where A and B are n x n matrices and I is the nxn identity matrix.
To find which of the given options is always true, let's analyze each option one by one.
a) A is non-singular:
If A is non-singular, it means that A has an inverse, denoted as A^(-1). Multiplying both sides of the given equation by A^(-1) gives us:
BA B2 = I - BA2
A^(-1)(BA B2) = A^(-1)(I - BA2)
(A^(-1)BA) (A^(-1)B2) = A^(-1)I - A^(-1)(BA2)
A^(-1)B(AB) - A^(-1)(BA2) = A^(-1) - A^(-1)(BA2)
A^(-1)B(AB - A2) = A^(-1) - A^(-1)(BA2)
Now, if A is non-singular, then A^(-1) exists. However, from the equation above, we can see that A^(-1)B(AB - A2) = A^(-1) - A^(-1)(BA2) depends on the values of A and B. Therefore, we cannot conclude that A is always non-singular.
b) B is non-singular:
To prove that B is always non-singular, we need to show that B has an inverse, denoted as B^(-1). Multiplying both sides of the given equation by B^(-1) gives us:
BA B2 = I - BA2
BA(B^(-1)B) B2(B^(-1)B) = (I - BA2)(B^(-1)B)
(BA)I B2(B^(-1)B) = B^(-1)B - BA2(B^(-1)B)
BA B2(B^(-1)B) = B^(-1)B - BA2(B^(-1)B)
Now, if B is non-singular, then B^(-1) exists. We can rewrite the equation as:
BA B2(B^(-1)B) = B^(-1)B - BA2(B^(-1)B)
BA B^(-1)(BB^(-1))B = B^(-1)B - BA2(B^(-1)B)
BA B^(-1)I B = B^(-1)B - BA2(B^(-1)B)
BA^2 B = B^(-1)B - BA2(B^(-1)B)
From the equation above, we can conclude that BA^2 B = B^(-1)B - BA2(B^(-1)B) only if B is non-singular.
Therefore, the correct answer is option 'B': B is non-singular.
We are given that BA B2 = I - BA2, where A and B are n x n matrices and I is the nxn identity matrix.
To find which of the given options is always true, let's analyze each option one by one.
a) A is non-singular:
If A is non-singular, it means that A has an inverse, denoted as A^(-1). Multiplying both sides of the given equation by A^(-1) gives us:
BA B2 = I - BA2
A^(-1)(BA B2) = A^(-1)(I - BA2)
(A^(-1)BA) (A^(-1)B2) = A^(-1)I - A^(-1)(BA2)
A^(-1)B(AB) - A^(-1)(BA2) = A^(-1) - A^(-1)(BA2)
A^(-1)B(AB - A2) = A^(-1) - A^(-1)(BA2)
Now, if A is non-singular, then A^(-1) exists. However, from the equation above, we can see that A^(-1)B(AB - A2) = A^(-1) - A^(-1)(BA2) depends on the values of A and B. Therefore, we cannot conclude that A is always non-singular.
b) B is non-singular:
To prove that B is always non-singular, we need to show that B has an inverse, denoted as B^(-1). Multiplying both sides of the given equation by B^(-1) gives us:
BA B2 = I - BA2
BA(B^(-1)B) B2(B^(-1)B) = (I - BA2)(B^(-1)B)
(BA)I B2(B^(-1)B) = B^(-1)B - BA2(B^(-1)B)
BA B2(B^(-1)B) = B^(-1)B - BA2(B^(-1)B)
Now, if B is non-singular, then B^(-1) exists. We can rewrite the equation as:
BA B2(B^(-1)B) = B^(-1)B - BA2(B^(-1)B)
BA B^(-1)(BB^(-1))B = B^(-1)B - BA2(B^(-1)B)
BA B^(-1)I B = B^(-1)B - BA2(B^(-1)B)
BA^2 B = B^(-1)B - BA2(B^(-1)B)
From the equation above, we can conclude that BA^2 B = B^(-1)B - BA2(B^(-1)B) only if B is non-singular.
Therefore, the correct answer is option 'B': B is non-singular.
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