Q69 / 170 · Manipal (MET) 2015 · Mathematics
If $A$ is a square matrix such that $A^2=A$ and $(1+A)^n=I+\lambda A$, then $\lambda$ is equal to
- A$2 n-1$
- Correct answerB$2^n-1$
- C$2 n+1$
- DNone of these
Solution
We have,
$A^2=A$
$\therefore(I+A)^2=(I+A)(I+A)=I+2 A+A^2=I+3 A$
and
$(I+A)^3=(I+A)^2(I+A)$
$\begin{array}{lr}=(I+3 A)(I+A) & {\left[\because(I+A)^2=I+3 A\right]} \\ =I+4 A+3 A^2 & \\ =I+7 A & {\left[\because A^2=A\right]}\end{array}$
Thus, we have
$(I+A)^2=I+3 A$ and $(I+A)^3=I+7 A$
$\Rightarrow \quad(I+A)^2=I+\left(2^2-1\right) A$
and $\quad(I+A)^3=I+\left(2^3-1\right) A$
Hence, $(I+A)^n=I+\left(2^n-1\right) A$
$\begin{aligned} \therefore & & (I+A)^n & =I+\lambda A \\ \Rightarrow & & \lambda & =2^n-1\end{aligned}$
Hence, the correct answer is option: 'B'.