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.
 A: The mistake is that you assume that $b-a=0, b-2a=0$ and $1-2a=0$ at the same time. This is simply impossible because the first two equations imply $a=b=0$. It is true that you need to get $2$ non zero rows in the Row Echelon form but you can't just assume at some point that all elements in the last row should be equal to $0$ when it is an impossible case. You need to keep doing row operations until you get a sure answer. Maybe you will have to split into cases for that. Anyway, your solution with determinants was probably the best way to solve this exercise anyway. 
A: If $b=a$, then the last row is $\begin{bmatrix}0&0&-a&1-2a\end{bmatrix}$. In which case you can use the second row to kill entries in this row and then get echelon form $$ \begin{bmatrix} 1 &1 &2 &2 \\ 0&0 &-1 &1 \\ 0&0 &0 &(1-3a) \end{bmatrix}$$
Now for rank to be $2$, you need $a=1/3$.
A: 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.
A: Consider the submatrix $M$ obtained from the first two rows 
$$
M = \left[\begin{array}{rrrr}
1 & 1 & 2 & 2 \\
1 & 1 & 1 & 3
\end{array}\right]
$$
The two vectors
\begin{align*}
\vec{n}_1 &= \left\langle1,\,3,\,-1,\,-1\right\rangle & \vec{n}_2 &= \left\langle0,\,4,\,-1,\,-1\right\rangle
\end{align*}
form a basis of $\operatorname{Null}(M)$. Since we want our $3\times 4$ matrix $A$ to have the same rank as $M$, and hence the same nullity as $M$, we must have $A\vec{n}_1=A\vec{n}_2=\vec{O}$. This gives
\begin{align*}
\overset{A\vec{n}_1}{\left\langle0,\,0,\,a + 2 \, b - 1\right\rangle} &= \left\langle0,\,0,\,0\right\rangle & \overset{A\vec{n}_2}{\left\langle0,\,0,\,3 \, b - 1\right\rangle} &= \left\langle0,\,0,\,0\right\rangle
\end{align*}
These two equations give $a=b=\frac{1}{3}$.
