Question

If $A$ and $B$ are two matrices such that $AB=BA$, then $\forall$ $n\in N$

• A. $A^{n}B=B{A}^n$
• B. ${\left(AB\right)}^n=A^n{B}^n$
• C. ${(A+B)}^n{=}^n{C}_0{A}^n{+}^n{C}_1{A}^{n-1}B{+}^n{C}_2{A}^{n-2}{B}^2+...{+}^n{C}_n{B}^n$
• D. $A^{2n}-B^{2n}=(A^n-B^n)(A^n+B^n)$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $A^{n}B=B{A}^n$
B ${\left(AB\right)}^n=A^n{B}^n$
C ${(A+B)}^n{=}^n{C}_0{A}^n{+}^n{C}_1{A}^{n-1}B{+}^n{C}_2{A}^{n-2}{B}^2+...{+}^n{C}_n{B}^n$
D $A^{2n}-B^{2n}=(A^n-B^n)(A^n+B^n)$

Given $AB=BA$

We check all the options

A. The result $A^{n}B=B{A}^n$ is true for $n=2$ since

$A^{2}B=A\left(AB\right)=A\left(BA\right)=\left(AB\right)A=\left(BA\right)A=B{A}^2$

Let the result be true for $n=k$

$\Rightarrow A^{k}B=B{A}^k$

For $n=k+1$, we have

$A^{k+1}B=A\left(A^{k}B\right)=A\left(B{A}^k\right)=\left(AB\right)A^k=\left(BA\right)A^k=B{A}^{k+1}$

Thus the result is true for $n=k+1$

Hence by principle of Mathematical induction, the result is true for all $n\in N$

Options B and C : since A and B commute, both these option hold

Option D: $(A^n-B^n)(A^n+B^n)=A^{2n}+A^n{B}^n-B^n{A}^n-B^{2n}=A^{2n}-B^{2n}$, A and B commute.