Question

If $A$ and $B$ are two non-zero $n \times n$ matrics such that $A ^2+ B = A ^2 B$, then

• A. $A ^2= I$ or $B = I$
• B. $A ^2 B = BA ^2$
• C. $AB = I$
• D. $A ^2 B = I$

Answer 1

The Correct Option is B

To solve the given problem, we have the equation involving two non-zero \( n \times n \) matrices \( A \) and \( B \):

\(A^2 + B = A^2 B\)

We need to determine which of the given options is correct based on this equation. Let's explore each option systematically.

1. Given equation: \(A^2 + B = A^2 B\).
2. Rearrange the equation: \(A^2 + B - A^2 B = 0\).
3. Factor the equation: \(A^2 (I - B) + B = 0\), where \(I\) is the identity matrix.
4. We can rewrite the equation assuming potential commutativity: \(A^2 B = B A^2\).

Let us verify this by considering each option:

• Option 1: \(A^2 = I\) or \(B = I\)
  Assuming \(A^2 = I\) does not simplify the original equation to an identity unless specific conditions on \(B\) hold. The same applies if \(B = I\). Hence, this is not a general conclusion.
• Option 2: \(A^2 B = B A^2\)
  This option directly stems from observing the factorization \(A^2 (I - B) + B = 0\) which holds if \(A^2 B = B A^2\) since matrix multiplication is generally non-commutative, but here it needs to satisfy this condition for the equation to be valid. Therefore, this is the correct answer.
• Option 3: \(AB = I\)
  This condition would imply a very strong restriction on matrices \(A\) and \(B\) not generally applicable, particularly considering \(A\) and \(B\) would be inverses. Hence, it does not solve the provided equation.
• Option 4: \(A^2 B = I\)
  Similar to the prior options, this would apply too strict a requirement on \(A^2\) and \(B\) to hold true for a matrix identity in general form.

Therefore, the correct answer is Option 2: \(A^2 B = B A^2\).