Problem source

\textit{In this question, }\textbf{\textit{$\mathbf{A},\mathbf{B}$
}}\textit{and $\mathbf{X}$ are non-zero $2\times2$ real matrices.}
Are the following assertions true or false? You must provide a proof or a counterexample in each case. 
\begin{questionparts}
\item If $\mathbf{AB=0}$ then $\mathbf{BA=0}.$
\item $(\mathbf{A-B)(A+B)=}\mathbf{A}^{2}-\mathbf{B}^{2}.$
\item The equation $\mathbf{AX=0}$ has a non-zero solution $\mathbf{X}$
if and only if $\det\mathbf{A}=0.$
\item For any $\mathbf{A}$ and $\mathbf{B}$ there are at most two matrices
$\mathbf{X}$ such that $\mathbf{X}^{2}+\mathbf{AX}+\mathbf{B}=\mathbf{0}.$ 
\end{questionparts}

Solution source

\begin{questionparts}
\item This is false, for example let $\mathbf{A} = \begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix}$ and $\mathbf{B} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, then
\begin{align*}
\mathbf{AB} &= \begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \\
&= \begin{pmatrix}0  & 0 \\ 0 & 0\end{pmatrix} \\
\mathbf{BA} &= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix} \\
&= \begin{pmatrix}0  & 1 \\ 0 & 0\end{pmatrix} \\
\end{align*}

\item This is also false, using the same matrices from part (i), we find:

\begin{align*}
(\mathbf{A - B})(\mathbf{A + B}) &= \mathbf{A}^2-\mathbf{BA}+\mathbf{AB}-\mathbf{B}^2 \\
&= \mathbf{A}^2-\mathbf{B}^2+\begin{pmatrix}0  & 1 \\ 0 & 0\end{pmatrix} \\
&\neq \mathbf{A}^2-\mathbf{B}^2
\end{align*}

\item This is true. Claim: The equation $\mathbf{AX=0}$ has a non-zero solution $\mathbf{X}$ if and only if $\det\mathbf{A}=0.$

Proof: $(\Rightarrow)$ Suppose $\det\mathbf{A} \neq 0$ then $\mathbf{A}$ has an inverse, and so we must have $\mathbf{A}^{-1}\mathbf{AX} = \mathbf{0} \Rightarrow \mathbf{X} = \mathbf{0}$. 

$(\Leftarrow)$ Suppose $\det \mathbf{A} = 0$ then $ad-bc=0$, so consider the matrix $\mathbf{X} = \begin{pmatrix} d & d\\ -c & -c\end{pmatrix}$ (or if this is zero, $\mathbf{X} = \begin{pmatrix} a & a\\ -b & -b\end{pmatrix}$)

\item  This is false. Consider $\mathbf{A} = \mathbf{B} = \mathbf{0}$, then $\mathbf{X} = \begin{pmatrix} 0 & x \\ 0 & 0\end{pmatrix}$ has the property that $\mathbf{X}^2 = \mathbf{0}$ for all $x$, so at least more than 2 values

\end{questionparts}