Question

If A and B are 3×3 matrices such that $AB=A$ and $BA=B$, then
A. ${{A}^ {2}} =A $ and ${{B}^ {2}} \ne B$
B. ${{A}^ {2}} \ne A $ and ${{B}^ {2}} =B$
C. ${{A}^ {2}} =A $ and ${{B}^ {2}} =B$
D. ${{A}^ {2}} \ne A $ and ${{B}^ {2}} \ne B$

Answer 1

Hint: As we have to find the value of ${{A}^ {2}} $and ${{B}^ {2}} $, we have to solve the given equations present in the question by post multiplying both of them which will be $AB=A $ to A and $BA=B $ to B. then we will get our required solution.

Formula Used: We have to do post multiply in this question as in this case AB = A has the post multiply A and BA = B has the post multiply B

Complete step by step solution:
First of all solve equation 1 which is $AB=A$
Since $AB=A$, we will post multiply by A
Which will give $ABA=AA$. Now as given in the question $BA=B$, substitute $BA=B $ in this $ABA=AA$.
Which will give $AB={{A}^ {2}} $ and also given in the question $AB=A$, substituting it will get
$A={{A}^ {2}} $→ (1)
Similarly for equation 2 which is $BA=B$
Since $BA=B$, we will post multiply by B
Which will give $BAB={{B}^ {2}} $. Now as given in the question $AB=A$, substitute $AB=A$ in this
$BAB={{B}^ {2}} $.
Which will give $BA={{B}^ {2}} $and also given in the question $BA=B$, substituting it will get
$B={{B}^ {2}} $→ (2)
 from equation (1) and (2) we can clearly see that $A={{A}^ {2}} $and $B={{B}^ {2}} $

Option ‘C’ is correct

Note: Always remember that whenever we post multiply anything like for $AB=A $ we post multiplied by A so always multiply by last alphabet which is $ABA=AA$ don’t multiply before first alphabet
And as in this equation given in the question don’t take AB as it will give ${{A}^ {2}} ={{A}^ {2}} $which is not possible similarly to the other equation always take second last and last alphabet while post multiplying.