Question

Let A=$\left[\begin{array}{cc}0& \alpha \\ 0& 0\end{array}\right]$ 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

$A=\left[\begin{array}{cc}0& \alpha \\ 0& 0\end{array}\right]$

$A+I=\left[\begin{array}{cc}0& \alpha \\ 0& 0\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& \alpha \\ 0& 1\end{array}\right]$

Let's find $(A+I{)}^{50}$ by simple analysis,

$(A+I{)}^2=\left[\begin{array}{cc}1& \alpha \\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& \alpha \\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 2\alpha \\ 0& 1\end{array}\right]$

$(A+I{)}^3=\left[\begin{array}{cc}1& \alpha \\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& \alpha \\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& \alpha \\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 2\alpha \\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& \alpha \\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 3\alpha \\ 0& 1\end{array}\right]$

So, $(A+I{)}^{50}=\left[\begin{array}{cc}1& 50\alpha \\ 0& 1\end{array}\right]$

Now, $(A+I{)}^{50}-50A=\left[\begin{array}{cc}1& 50\alpha \\ 0& 1\end{array}\right]-\left[\begin{array}{cc}0& 50\alpha \\ 0& 0\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

Thus, $a+b+c+d=2$