Find a $3\times 3$ matrix whose minimal polynomial is $x^2$. Find  a $3\times 3$ matrix whose minimal polynomial is $x^2$.
My try:
Since  a characteristic polynomial and a minimal polynomial have the same roots ,so the characteristic polynomial must be $x^3$ since $0$ is the only characteristic value of multiplicity $2$.
So the matrix $A$ must be of the form  \begin{bmatrix} 0 & 0 & 0\\ b & 0 & 0 \\ c & a & 0 \end{bmatrix}
Since the minimal polynomial is $x^2$ so rank $A=2$,so we must have a non-zero minor of order $2$ .Hence we should have $a\neq 0,b\neq 0;a,b,c\in \mathbb R$ .
Is the solution correct?Please suggest edits if required.

I think this would work:
Let A be a $3\times3$ matrix whose minimal polynomial is $x^{2}$ (the only root of which is x=0).
Since the minimal polynomial and the characteristic polynomial have same roots, the characteristic polynomial of A must then be $x^{3}$.
So we start by considering A as a triangular matrix (say lower triangular) with all diagonal entries as zero.
(Why? Since then the characteristic polynomial = determinant of the triangular matrix (xI-A) = product of diagonal entries (x-0),(x-0),(x-0) = $x^{3}$)
Thus, A as of now, is of the form
$$ A = \left[\begin{array}\\0&0&0\\a&0&0\\b&c&0\end{array}\right]$$
where a,b,c$\in \Bbb R$
Now we want $A\neq 0$ and $A^2=0$  so that the polynomial $x^2$ is the minimum degree polynomial that annihilates A, i.e.we need,
$$ A = \left[\begin{array}\\0&0&0\\a&0&0\\b&c&0\end{array}\right]\neq 0,\space \space \space and \space A^2 = \left[\begin{array}\\0&0&0\\0&0&0\\ac&0&0\end{array}\right] = 0$$,
$\Rightarrow$a,b,c cannot all be zero, and both a,c cannot be non-zero.
Hence, $$ \left[\begin{array}\\0&0&0\\0&0&0\\9&0&0\end{array}\right] ,  \left[\begin{array}\\0&0&0\\0&0&0\\-1&1&0\end{array}\right],  \left[\begin{array}\\0&0&4\\0&0&13\\0&0&0\end{array}\right],  \left[\begin{array}\\0&-7&4\\0&0&0\\0&0&0\end{array}\right] $$ are all examples of $3\times3$ matrices whose minimal polynomial is $x^2$.
Thank You!
