Rank in row echelon form $$A=
\begin{bmatrix}
  a & 1 & a & 0 & 0 & 0 \\
  0 & b & 1 & b & 0 & 0 \\
  0 & 0 & c & 1 & c & 0 \\
  0 & 0 & 0 & d & 1 & d \\
\end{bmatrix}
$$
Let $A$ be the matrix above and $r$ be the number of non-zero rows in row echelon form.
Show that
$$r>2 \quad\text{always}$$
$$r=3 \quad\text{iff }a=d=0 \text{ and } bc = 1$$
$$r=4 \quad\text{otherwise}$$
Is there a way to prove this question without listing out all the possible combinations of a,b,c,d being zero or non-zero. Can anyone explain please?
 A: we can show that $rank(A) > 2$ by leaving out the first column. second row is replaced by second row minus $b$ times the first row. we have two cases: $ab = 1$ and $ab \neq 1.$ let us take the easier case $ab \neq 1.$ now you have pivots $1, 1-ab, 1$ in rows $1, 2, $ and $3.$ so $rank(A) \ge 3$ 
the second case $ab = 1,$ now we have pivots $1$ on rows $1$ and $3.$ getting rid od $d$ on the last row leads to consider two cases $cd = 1$ which makes a pivot $d$ on the last row and $cd \neq $ amke a pivot $1-cd$ on the last row. so in both case we end up with rank 3.
A: You're given the matrix in row echelon form!
$$A=
\begin{pmatrix}
  a & 1 & a & 0 & 0 & 0 \\
  0 & b & 1 & b & 0 & 0 \\
  0 & 0 & c & 1 & c & 0 \\
  0 & 0 & 0 & d & 1 & d \\
\end{pmatrix}
$$
Now, there's no way row 1 and row $\;2\;$ are linearly dependent , as it'd have to be that
$$(0\;b\;1\;b\;0\;0)=k\cdot(a\;1\;a\;0\;0\;0)\;,\;\;k\in\Bbb F=\text{ some field}$$
and this is impossible (why?), so already $\;r\ge 2\;$ . Now, if row $\;3\;$ is lin. dependent in the first two (as it is easy to see it is not dependent only in the second one, just as before), then it must be $\;a=0\;$, and then looking at the third and fourth coordinates we get again an impossibility, and thus we already have $\;r\ge 3\;$ .
Now, if $\;r=3\;$ then the fourth row must be a lin. combination of the first three, so
$$(0\;0\;0\;d\;1\;d)=\alpha(a\;1\;a\;0\;0\;0)+\beta(0\;b\;1\;b\;0\;0)+\gamma(0\;0\;c\;1\;c\;0)\implies$$
$$\begin{cases}&I\;\;&\alpha a=0\\{}\\
&II\;\;&\alpha+\beta b=0\\{}\\
&III\;\;&\alpha a+\beta+\gamma c=0\\{}\\
&IV\;\;&\beta b+\gamma=d\\{}\\
&V\;\;&\gamma c=1\\{}\\
&VI\;\;&d=0\end{cases}$$
Now, if $\;a\neq 0\;$ then $\;I\implies\alpha=0\;$, but then $\;II\implies\beta b=0\;$ and then $\;IV\implies \gamma=d=0\;$ , and this contradicts $\;V\;$ .
Thus, it must be also $\;a=0\;$ . Take it now from here
A: After two well chosen row operations you can get
$$ \left( \begin {array}{cccccc} a&1&a&0&0&0\\ -ab&0&1-
ab&b&0&0\\ 0&0&c&1-cd&0&-cd\\ 0
&0&0&d&1&d\end {array} \right)
$$
Since we cannot have both $-ab$ and $1-ab$ as zero, the rank is at least 3. The rank can only be 3 if rows 2 and 3 can be combined to make a row of all zeros. 
Now you can suppose $b=0$ and get a contradiction to a row being all zeroes, and then suppose $b \not=0$ and get the following equations if the rank is to be 3:
$$
(1-cd)a=0 ~,~~ -cd=0 ~,~~ bc- (1-ab)(1-cd) = 0
$$
Carefully solving these gives the specified conditions as the only solution.
