If $A = \begin{bmatrix}ab & b^2 \\ -a^2 & -ab\end{bmatrix}$ and $A^n = 0$, then the minimum value of n is
A. 2
B. 3
C. 4
D. 5

Hint: We need to find the smallest n such that $A^n = 0$. $A^n = 0$ means we are multiplying the matrix A, n times, so n will be the number of times we multiply A to yield 0.

Complete step by step Solution:
To find the smallest n we need to perform matrix multiplication on A itself.
We have,
$A = \begin{bmatrix}ab & b^2 \\ -a^2 & -ab\end{bmatrix}$
Then,
$A^2 = \begin{bmatrix}ab & b^2 \\ -a^2 & -ab\end{bmatrix}\begin{bmatrix}ab & b^2 \\ -a^2 & -ab\end{bmatrix}$
$A^2 = \begin{bmatrix}ab \times ab + b^2\times(-a^2) & ab \times b^2 + b^2 \times (-ab) \\ (-a^2)\times ab+(-ab)\times (-a^2) & (-a^2)\times b^2+(-ab)\times(-ab)\end{bmatrix}$
$A^2 = \begin{bmatrix}a^2b^2-a^2b^2 & ab^3-ab^3\\ -ab^3+ab^3 & -a^2b^2+a^2b^2\end{bmatrix}$
$\implies A^2=0$
$\implies A^3 = A\cdot A^2$
$\implies A^3 = A\cdot 0 = 0$
and $A^n = 0$
Therefore, the smallest n such that $A^n = 0$is n = 2. In other words, $A^n = 0$ for all $n\geqslant2$.

Therefore, the correct option is (A).

Note: Finding the smallest n does not mean that it will be the smallest number among the options. There are cases in which we multiply more than 3 times to get 0. So, do not arrive at the answer simply by looking at the options, also be careful while performing matrix multiplication as we go on doing multiplication may lead to mistakes in computing.