There are exactly three $2\times 2$ row reduced matrices $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ such that $a+b+c+d=0$ Let $A$ be $2\times 2$ matrix with complex entries,
$$A=\begin{bmatrix} a & b\\ c& d\end{bmatrix}$$
Suppose that $A$ is row reduced and also that $a+b+c+d=0$. Prove that there are exactly three such matrices. 
 A: I have seen this question in Section 1.4, Exercise 6 of Linear Algebra by Hoffman & Kunze. I believe both the comment to the original question and the already-existing answer are assuming row-reduced is the same as row-reduced echelon. 
In this text, a matrix is row-reduced if: 


*

*the first non-zero entry in each non-zero row is equal to 1

*each column of the matrix which contains the leading nonzero entry of some row has all its other entries equal to zero. 


With this definition, the three matrices are: 
$ \begin{bmatrix} 0 & 0\\ 0& 0\end{bmatrix}, \begin{bmatrix} 1 & -1\\ 0& 0\end{bmatrix}, \begin{bmatrix} 0 & 0\\ 1& -1\end{bmatrix} $
A: According to http://en.wikipedia.org/wiki/Row_echelon_form. The following criteria must be met: 
A matrix is in reduced row echelon form (also called row canonical form) if it satisfies the following conditions.


*

*All nonzero rows are above any rows of all zeroes.

*The leading coefficient of a nonzero row is always strictly to the right of the leading coefficient of the row above it.

*Every leading coefficient is 1 and is the only nonzero entry in its column.


That means c must be 0. Furthermore a=1 or a=0. 
We can then work on the relation $a+b+c+d=0$. 
So let's work this out:
First c=0. 


*

*a=0 -> Then b = -d. But since nonzero coefficients are strictly to the right of the ones on the previous line (rule 2) this means b=d=0. 

*a=1 -> Then b = -1 - d.
By requirement three b=0, so d must equal -1. Which violates rule 3

*a = 1 and d = 0 -> b = - a


This gives two distinct shapes of matrices: 
$ \begin{bmatrix} 0 & 0\\ 0& 0\end{bmatrix}, \begin{bmatrix} 1 & -1\\ 0& 0\end{bmatrix} $
