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Question: A: \(AB=A\) and \(BA=B\Rightarrow {{A}^{n}}+{{B}^{n}}=A+B\). R: \(AB=A\) and \...

A: AB=AAB=A and BA=BAn+Bn=A+BBA=B\Rightarrow {{A}^{n}}+{{B}^{n}}=A+B.
R: AB=AAB=A and BA=BABA=B\Rightarrow A and BB are idempotent.
A.Both A and R are true and R is the correct explanation to A.
B.Both A and R are true but R is not the correct explanation for A.
C.A is true R is false.
D.A is false R is true

Explanation

Solution

Hint: Here, to prove A we have to apply the principle of mathematical induction. That is first prove that the statement is true for n = 1 and then assume that the statement is true for n = k and then prove that it is true for n = k+1. From the induction part itself you can prove that R is true.

Complete step by step answer:
Here, we are given that AB=AAB=A and BA=BBA=B.
Now, we have to prove that An+Bn=A+B{{A}^{n}}+{{B}^{n}}=A+B.
Here, we can apply the principle of mathematical induction.
We know that for the principle of mathematical induction first we have to prove that the statement is true for n =1, then we have to assume that it is also true for n = k, k is any integer and prove that it is true for n = k+1.
For n = 1 we have A+B=A+BA+B=A+B
Therefore, we can say that BA=BAn+Bn=A+BBA=B\Rightarrow {{A}^{n}}+{{B}^{n}}=A+B is true for n=1.
Now consider n = 2, i.e. A2+B2{{A}^{2}}+{{B}^{2}}.
We have, AB=AAB=A and BA=BBA=B. Hence, we can write:
A(BA)=A ABA=A (AB)A=A \begin{aligned} & A(BA)=A \\\ & ABA=A \\\ & (AB)A=A \\\ \end{aligned}
Now, by substituting AB=AAB=A we get:
AA=AAA=A
A2=A{{A}^{2}}=A ….. (1)
Similarly we can write:
BA=B B(AB)=B BAB=B (BA)B=B \begin{aligned} & BA=B \\\ & B(AB)=B \\\ & BAB=B \\\ & (BA)B=B \\\ \end{aligned}
Now, by substituting BA=BBA=B, we will get:
BB=BBB=B
B2=B{{B}^{2}}=B ….. (2)
Now, from equation (1) and equation (2), we can write:
A2+B2=A+B{{A}^{2}}+{{B}^{2}}=A+B
Hence, we can say that BA=BAn+Bn=A+BBA=B\Rightarrow {{A}^{n}}+{{B}^{n}}=A+B is true for n = 2.
Now, we can assume that it is true for n = k. Next, we have to prove that it is true for k+1.
For n = k, we have:
Ak+Bk=A+B{{A}^{k}}+{{B}^{k}}=A+B
Now, we can write:
Ak+1+Bk+1=AkA+BkB{{A}^{k+1}}+{{B}^{k+1}}={{A}^{k}}A+{{B}^{k}}B ….. (3)
Here, since A2=A{{A}^{2}}=A and B2=B{{B}^{2}}=B, we can write:
Ak=A{{A}^{k}}=A and Bk=B{{B}^{k}}=B
Hence, our equation (3) becomes:
Ak+1+Bk+1=AA+BB Ak+1+Bk+1=A2+B2 \begin{aligned} & {{A}^{k+1}}+{{B}^{k+1}}=AA+BB \\\ & {{A}^{k+1}}+{{B}^{k+1}}={{A}^{2}}+{{B}^{2}} \\\ \end{aligned}
We already proved that A2+B2=A+B{{A}^{2}}+{{B}^{2}}=A+B. Therefore, we can say that:
Ak+1+Bk+1=A+B{{A}^{k+1}}+{{B}^{k+1}}=A+B
Hence by mathematical induction we can write:
An+Bn=A+B{{A}^{n}}+{{B}^{n}}=A+B
Therefore, we can say that A:AB=AA:AB=A and BA=BAn+Bn=A+BBA=B\Rightarrow {{A}^{n}}+{{B}^{n}}=A+B.
Next, we can consider R:AB=AR:AB=A and BA=BBA=B.
Now, by equation (1) and equation (2) we have:
A2=A{{A}^{2}}=A and B2=B{{B}^{2}}=B
We know that an idempotent matrix is a square matrix which when multiplied by itself yields itself. That is, if A is a matrix then A2=A{{A}^{2}}=A
Hence, by the definition we can say that A and B are idempotent.
Thus we can write, R: AB=AAB=A and BA=BABA=B\Rightarrow A and BB are idempotent.
Therefore, we can say that both A and R are true and R is the correct explanation to A.
Hence, the correct answer for this question is option (a).

Note: Here, you have to use the principle of mathematical induction. Mathematical induction is a technique of proving a statement, theorem or formula which is thought to be true; for each and every natural number n. By generalising this in the form of principle which we would use to prove any mathematical statement is the principle of mathematical induction.