Question

Let $A$ and $B$ are two matrices such that $AB=BA$, then for every $n\in N$,

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

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

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

Similarly, $A^{3}B=B{A}^3$

In general $A^{n}B=B{A}^n\forall n\ge 1$

(b) and (c) hold as $AB=BA$.

Also, $(A^n-B^n)(A^n+B^n)$

$=A^n{A}^n-B^n{A}^n+A^n{B}^n-B^n{B}^n$

$=A^{2n}-B^{2n}$

Hence, options A,B,C and D.