If rank of a given matrix of order $3 \times 4$ is $2$ then the value of $b$ is Q) Suppose the rank of the matrix
$\begin{pmatrix} 1 &1 &2 &2 \\ 1&1 &1 &3 \\ a&b &b &1 \end{pmatrix}$
is $2$ for some real numbers $a$ and $b$. Then $b$ equals
$(A)$ $1\;\;\;$ $(B)$ $3\;\;\;$ $(C)$ $1/2\;\;\;$ $(D)$ $1/3\;\;\;$
My Approach :- Since rank is $2$ , So, the determinant of all the submatrices of order $3 \times 3$ must be zero.
So, $\begin{vmatrix} 1&1 &2 \\ 1&1 &1 \\ a&b &b \end{vmatrix}$ = $0$. After solving it, I am getting $a=b$ 
and $\begin{vmatrix} 1&2 &2 \\ 1&1 &3 \\ b&b &1 \end{vmatrix}$ = $0$. After Solving it , $b=\frac{1}{3}$ 
Now, Rank of a matrix is also defined as no. of non-zero rows in row echelon form of that matrix. So, If I convert it into Row Echelon form then
$\begin{bmatrix} 1 &1 &2 &2 \\ 1&1 &1 &3 \\ a&b &b &1 \end{bmatrix}$ 
Applying $R_{2} \leftarrow R_{2}-R_{1}$ 
$\begin{bmatrix} 1 &1 &2 &2 \\ 0&0 &-1 &1 \\ a& b &b &1 \end{bmatrix}$
Now, Applying $R_{3}\leftarrow R_{3}-aR_{1}$ 
$\begin{bmatrix} 1 &1 &2 &2 \\ 0&0 &-1 &1 \\ 0&(b-a) &(b-2a) &(1-2a) \end{bmatrix}$ 
Now, to make rank of this matrix = $2$ means I should have 2 non-zero rows or all the elements of the last row must be zero simultaneously.
So, $(b-a) =0$ and $(b-2a) =0$ and $(1-2a) =0$ 
So, $b=a$ and $b=2a$ and $a=\frac{1}{2}$ 
Now, My doubt is how $b=a$ and $b=2a$ is possible here simultaneously and why this method is giving wrong result. I must be doing some mistake here but I am not getting what mistake I am doing. Please help.

After getting$$\begin{bmatrix}1&1&2&1\\0&0&-1&1\\0&b-a&b-2a&1-2a\end{bmatrix},$$you can't just say (as you did) that the all the elements of the last row must be zero simultaneously. At this point, you should consider the cases $b=a$ and $b\neq a$. In the first case, you now have the matrix$$\begin{bmatrix}1&1&2&1\\0&0&-1&1\\0&0&-a&1-2a\end{bmatrix}$$and you carry on. If $b\neq a$, you divide the third row by $b-a$. You'll get$$\begin{bmatrix}1&1&2&1\\0&0&-1&1\\0&1&\frac{b-2a}{b-a}&\frac{1-2a}{b-a}\end{bmatrix}$$and, again, you carry on.
