Question

If $A=\left[\begin{array}{cc}ab& b^2\\ -a^2& -ab\end{array}\right]$ and $A^n=0$ then the minimum value of $n$ is

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (A)

$2$

$A=\left[\begin{array}{cc}ab& b^2\\ -a^2& -ab\end{array}\right]$

$A^2=\left[\begin{array}{cc}ab& b^2\\ -a^2& -ab\end{array}\right]\left[\begin{array}{cc}ab& b^2\\ -a^z& -ab\end{array}\right]=\left[\begin{array}{cc}{a}^2{b}^2-a^2{b}^2& a{b}^3-a{b}^3\\ -a^{3}b+a^{3}b& -a^2{b}^2+a^2{b}^2\end{array}\right]$

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

$A^2=0$

$\therefore n=2$.