Q92 / 180 · COMEDK 2026 (9 May Shift 1) · Mathematics
If A and B are two square matrices of the same order such that $AB = A$ and $BA = B$, then $(A + B)^2$ is equal to:
- A$A^2 + B^2 + 2A$
- B$A^2 + B^2$
- C$A + B$
- Correct answerD$2(A + B)$
Solution
Given $AB = A$ and $BA = B$.
We can find $A^2$ and $B^2$ as follows:
$A^2 = A \cdot A = (AB)A = A(BA)$
Substituting $BA = B$, we get:
$A^2 = AB = A$
Similarly, for $B^2$:
$B^2 = B \cdot B = (BA)B = B(AB)$
Substituting $AB = A$, we get:
$B^2 = BA = B$
Now, expanding $(A + B)^2$:
$(A + B)^2 = (A + B)(A + B)$
$(A + B)^2 = A^2 + AB + BA + B^2$
Substituting $A^2 = A$, $B^2 = B$, $AB = A$, and $BA = B$ into the expansion:
$(A + B)^2 = A + A + B + B$
$(A + B)^2 = 2A + 2B = 2(A + B)$
Answer: $2(A + B)$
We can find $A^2$ and $B^2$ as follows:
$A^2 = A \cdot A = (AB)A = A(BA)$
Substituting $BA = B$, we get:
$A^2 = AB = A$
Similarly, for $B^2$:
$B^2 = B \cdot B = (BA)B = B(AB)$
Substituting $AB = A$, we get:
$B^2 = BA = B$
Now, expanding $(A + B)^2$:
$(A + B)^2 = (A + B)(A + B)$
$(A + B)^2 = A^2 + AB + BA + B^2$
Substituting $A^2 = A$, $B^2 = B$, $AB = A$, and $BA = B$ into the expansion:
$(A + B)^2 = A + A + B + B$
$(A + B)^2 = 2A + 2B = 2(A + B)$
Answer: $2(A + B)$
Hence, the correct answer is option: 'D'.