Question

If \(AB = A\) and \(BA = B\), where \(A\) and \(B\) are square matrices of same order, then

• A. \(B^2 = B, A^2 = A\)
• B. \(B^2 \neq B\) and \(A^2 = A\)
• C. \(A^2 \neq A, B^2 = B\)
• D. \(A^2 \neq A, B^2 \neq B\)

Hint:

Write \(A^2 = (AB)A = A(BA)\) and use \(BA = B\). Do the same for \(B^2\).

Answer 1

The Correct Option is A

Step 1: Find \(A^2\):
Start with \(A^2 = A \cdot A\). Replace the first \(A\) by \(AB\), since \(AB = A\).
\[ A^2 = (AB)A = A(BA) \] Now use \(BA = B\):
\[ A^2 = AB = A \] So \(A^2 = A\).

Step 2: Find \(B^2\):
Start with \(B^2 = B \cdot B\). Replace the first \(B\) by \(BA\), since \(BA = B\).
\[ B^2 = (BA)B = B(AB) \] Now use \(AB = A\):
\[ B^2 = BA = B \] So \(B^2 = B\).

Step 3: Match with the options:
Both \(A^2 = A\) and \(B^2 = B\) hold. Option 2 says \(B^2 \neq B\), option 3 says \(A^2 \neq A\) and option 4 denies both. All three contradict what we proved, so they are wrong. Only option 1 is correct.

Final Answer:
Both matrices are idempotent. This is option 1. \[ \boxed{B^2 = B,\ A^2 = A} \]