Question

Let $A=\left[\begin{array}{cc}0& a\\ 0& 0\end{array}\right]$ and ${(A+I)}^{50}-50A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$, then the value of $a+b+c+d$ is

• A. $2$
• B. $1$
• C. $4$
• D. None of these

Answer 1

The correct option is A $2$

Given,

$A=\left[\begin{array}{cc}0& a\\ 0& 0\end{array}\right]A+I=\left[\begin{array}{cc}0& a\\ 0& 0\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& a\\ 0& 1\end{array}\right]{(A+I)}^2=\left[\begin{array}{cc}1& a\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& a\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 2a\\ 0& 1\end{array}\right]{(A+I)}^3={(A+I)}^2(A+I)=\left[\begin{array}{cc}1& 2a\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& a\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 3a\\ 0& 1\end{array}\right]{(A+I)}^4={(A+I)}^3(A+I)=\left[\begin{array}{cc}1& 3a\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& a\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 4a\\ 0& 1\end{array}\right]\Rightarrow {(A+I)}^{50}=\left[\begin{array}{cc}1& 50a\\ 0& 1\end{array}\right]{(A+I)}^{50}-50A=\left[\begin{array}{cc}1& 50a\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}0& 50a\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\Rightarrow \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\Rightarrow a=1,b=0,c=0,d=1\Rightarrow a+b+c+d=2$

So, option A is correct.