**Solution to Linear Algebra Hoffman & Kunze Chapter 1.3 Exercise 1.3.6**

Solution:

Case $a\not=0$: Then to be in row-reduced form it must be that $a=1$ and $A=\left[\begin{array}{cc}1&b\\c&d\end{array}\right]$ which implies $c=0$, so $A=\left[\begin{array}{cc}1&b\\0&d\end{array}\right]$. Suppose $d\not=0$. Then to be in row-reduced form it must be that $d=1$ and $b=0$, so $A=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$ which implies $a+b+c+d\not=0$. So it must be that $d=0$, and then it follows that $b=-1$. So $a\not=0$ $\Rightarrow$ $A=\left[\begin{array}{cc}1&-1\\0&0\end{array}\right]$.

Case $a=0$: Then $A=\left[\begin{array}{cc}0&b\\c&d\end{array}\right]$. If $b\not=0$ then $b$ must equal $1$ and $A=\left[\begin{array}{cc}0&1\\c&d\end{array}\right]$ which forces $d=0$. So $A=\left[\begin{array}{cc}0&1\\c&0\end{array}\right]$ which implies (since $a+b+c+d=0$) that $c=-1$. But $c$ cannot be $-1$ in row-reduced form. So it must be that $b=0$. So $A=\left[\begin{array}{cc}0&0\\c&d\end{array}\right]$. If $c\not=0$ then $c=1$, $d=-1$ and $A=\left[\begin{array}{cc}0&0\\1&-1\end{array}\right]$. Otherwise $c=0$ and $A=\left[\begin{array}{cc}0&0\\0&0\end{array}\right]$.

Thus the three possibilities are:

$$\left[\begin{array}{cc}0&0\\0&0\end{array}\right],\quad\left[\begin{array}{cc}1&-1\\0&0\end{array}\right],\quad\left[\begin{array}{cc}0&0\\1&-1\end{array}\right].$$