Question

If $A=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$ then $(aI+bA{)}^n$ is (where $I$ is the identity matrix of order $2$ )

• A. $a^{2}I+a^{n-1}b\cdot A$
• B. $a^{n}I+n{a}^{n}bA$
• C. $a^{n}I+n\cdot a^{n-1}b\cdot A$
• D. $a^{n}I+b^{n}A$

Answer 1

Answer: (C)

$A=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$
$[aI+bA{]}^1=\left[\begin{array}{cc}a& 0\\ 0& a\end{array}\right]+\left[\begin{array}{cc}0& b\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}a& b\\ 0& a\end{array}\right]$
$[aI+IA{]}^2=\left[\begin{array}{cc}a& b\\ 0& a\end{array}\right]\left[\begin{array}{cc}a& b\\ 0& a\end{array}\right]=\left[\begin{array}{cc}{a}^2& 2ab\\ 0& a^2\end{array}\right]$
$[aI+bA{]}^3=\left[\begin{array}{cc}{a}^2& 2ab\\ 0& a^2\end{array}\right]\left[\begin{array}{cc}a& b\\ 0& a\end{array}\right]=\left[\begin{array}{cc}{a}^3& 3^{2}b\\ 0& a^3\end{array}\right]$
$\therefore [aI+bA{]}^n=\left[\begin{array}{cc}{a}^n& na{a}^{n-1}b\\ 0& a^n\end{array}\right]$
$=a^{n}I+n\cdot a^{n-1}bA$