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Q12.     If A and B are square matrices of the same order such that AB = BA, then prove by induction that AB^n = B^n A. Further, prove that (AB)^n = A^n B^nfor all n \in N.

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A and B are square matrices of the same order such that AB = BA,

To prove : AB^n = B^n An \in N

For n=1, we have AB^1 = B^1 A

Thus, the result is true for n=1.

Let the result be true for n=k,then we have  AB^k = B^k A

 

Now, taking n=k+1 , we have  AB^{k+1} = AB^k .B

                                                AB^{k+1} = (B^kA) .B

                                              AB^{k+1} = (B^k) .AB

                                            AB^{k+1} = (B^k) .BA

                                            AB^{k+1} = (B^k.B) .A

                                            AB^{k+1} = (B^k^+^1) .A

Thus, the result is true for n=k+1.

Hence, we have AB^n = B^n An \in N.

To prove: (AB)^n = A^n B^n

For n=1, we have (AB)^1 = A^1 B^1

Thus, the result is true for n=1.

Let the result be true for n=k,then we have  (AB)^k = A^k B^k

 

Now, taking n=k+1 , we have  (AB)^{k+1} = (A B)^k.(AB)

                                                (AB)^{k+1} = A^k B^k.(AB)

                                              (AB)^{k+1} = A^{K}( B^kA)B

                                            (AB)^{k+1} = A^{K}( AB^k)B

                                            (AB)^{k+1} = (A^{K}A)(B^kB)

                                            (AB)^{k+1} = (A^{k+1})(B^{k+1})

Thus, the result is true for n=k+1.

Hence, we have  AB^n = B^n A         and    (AB)^n = A^n B^nfor all n \in N.

 

 

 

 

 

 

       

                              

                

Posted by

seema garhwal

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