Question

If $A,B$ are two square matrices such that $AB=B;BA=A$ and $nϵN$ then $(A+B{)}^n=$

• A. $2^n(A+B)$
• B. $2^{n-1}(A+B)$
• C. $2^{n+1}(A+B)$
• D. $2^{n/2}(A+B)$

Answer 1

The correct option is B $2^{n-1}(A+B)$

Given: $AB=B;BA=A$

In first, multiply $A$ on both sides- $ABA=BA$

Put $BA=A$ on both sides$\Rightarrow A.A=A\Rightarrow A^2=A$

$A^3=A^2.A=A$

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$A^n=A$

Similarly in second multiply both sides by $B$

$BAB=AB$

Put $AB=B$ on both sides $\Rightarrow B.B=B\Rightarrow B^2=B$

$B^3=B^2.B=B$

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$B^n=B$

Expansion of $(A+B{)}^n=(\frac{n}{0})A^n+(\frac{n}{1})A^{n-1}B+........+(\frac{n}{n-1})A{B}^{n-1}+(\frac{n}{n})B^n$

Putting $A^n=A$ and $B^n=B$

we get $(A+B{)}^n=(\frac{n}{0})A+(\frac{n}{1})AB+........+(\frac{n}{n-1})AB+(\frac{n}{n})B$

Put $AB=B$

$=(\frac{n}{0})A+(\frac{n}{1})B+........+(\frac{n}{n-1})B+(\frac{n}{n})B$.......$\left(1\right)$

Also $(A+B{)}^n=(B+A{)}^n$

So similarly expanding $(B+A{)}^n$

$(A+B{)}^n=(\frac{n}{0})A^n+(\frac{n}{1})A^{n-1}B+........+(\frac{n}{n-1})A{B}^{n-1}+(\frac{n}{n})B^n$

Putting $A^n=A$ and $B^n=B$

we get $(B+A{)}^n=(\frac{n}{0})B+(\frac{n}{1})BA+........+(\frac{n}{n-1})BA+(\frac{n}{n})A$

Put $BA=A$

$=(\frac{n}{0})B+(\frac{n}{1})A+........+(\frac{n}{n-1})A+(\frac{n}{n})A$........$\left(2\right)$

Adding equation $\left(1\right)+\left(2\right)$

$(A+B{)}^n+(B+A{)}^n=(\frac{n}{0})A+(\frac{n}{1})B+........+(\frac{n}{n-1})B+(\frac{n}{n})B+(\frac{n}{0})B+(\frac{n}{1})A+........+(\frac{n}{n-1})A+(\frac{n}{n})A$

$2(A+B{)}^n=((\frac{n}{0})+(\frac{n}{1})+........+(\frac{n}{n-1})+(\frac{n}{n})\left)\right(A+B)$

We know that, $((\frac{n}{0})+(\frac{n}{1})+........+(\frac{n}{n-1})+(\frac{n}{n}))=2^n$

$\therefore 2(A+B{)}^n=2^n(A+B)$

$(A+B{)}^n=2^{n-1}(A+B)$

Hence, $\left(B\right)$