Question

If $A$ is a square matrix such that $A^2=A$ nd $(1+A{)}^n=I+\lambda A$, then $\lambda$ is equal to

• A. $2n-1$
• B. $2^n-1$
• C. $2n+1$
• D. None of these

Answer 1

The correct option is B $2^n-1$

We have,

$A^2=A$

$\therefore (I+A{)}^2=(I+A\left)\right(I+A)=I+2A+A^2=I+3A$

and $(I+A{)}^3=(I+A{)}^2(I+A)$

$=(I+3A)(I+A)$ ....... $[\because (I+A{)}^2=I+3A]$

$=I+4A+3A^2$

$=I+7A$ $[\because A^2=A]$

Thus, we have

$(I+A{)}^2=I+3A$ and $(I+A{)}^3=I+7A$

$\Rightarrow (I+A{)}^2=I+(2^2-1)A$

and $(I+A{)}^3=I+(2^3-1)A$

Hence, $(I+A{)}^n=I+(2^n-1)A$

$\therefore (I+A{)}^n=I+\lambda A$

$\Rightarrow \lambda =2^n-1$