Question

Show that if A and B are square matrices such that AB = BA then, $ {{\left( A+B \right)}^{2}}={{A}^{2}}+2AB+{{B}^{2}}$?

Answer 1

Hint: In order to do this question, you need to first write $ {{\left( A+B \right)}^{2}}=\left( A+B \right)\left( A+B \right)$ in this form. After writing it like this, you need to multiply all the terms using the distribution law. Then you get two terms AB and BA but since AB = BA you can write it as 2AB or 2BA. But in the question, in the right hand side, there is 2AB. Therefore, we write the 2AB term instead of 2BA.

Complete step by step solution:
The first step to do this question is to write the square terms as a product of two terms. Therefore, by writing it in that form, we get
$ \Rightarrow {{\left( A+B \right)}^{2}}=\left( A+B \right)\left( A+B \right)$
Now , we have to use the distribution law to find the product of all the terms. According to the distribution law, we have $ \left( a+b \right)\left( c+d \right)=ac+ad+bc+bd$. Therefore, by using this law, we get
$ \Rightarrow \left( A+B \right)\left( A+B \right)={{A}^{2}}+AB+BA+{{B}^{2}}$
Here, we got two terms as AB and BA. But according to the question, AB = BA. Hence we can write the AB + BA as 2AB. Hence, by using this, we get
$ \Rightarrow \left( A+B \right)\left( A+B \right)={{A}^{2}}+2AB+{{B}^{2}}$
From here, we can see that we get the answer as $ {{\left( A+B \right)}^{2}}={{A}^{2}}+2AB+{{B}^{2}}$
Hence, we proved that $ {{\left( A+B \right)}^{2}}={{A}^{2}}+2AB+{{B}^{2}}$ when A and B are square matrices and AB = BA.

Note: For solving this question, you need to know the distribution law, otherwise you cannot solve this problem. Also, the equation in the question only applies when A and B are square matrices. In other cases, it does not apply.