Replace $R_2$ with $R_2 - R_1$:
\[ \begin{pmatrix} 1 & 1 & 2 & 2 \\ 0 & 0 & -1 & 1 \\ a & b & b & 1 \end{pmatrix} \]
For the rank to be 2, the third row ($R_3$) must be a linear combination of the first row ($R_1$) and the second row ($R_2'$). There must exist constants $c_1$ and $c_2$
\(R_{3}=c_{1}(R_{1})+c_{2}(R_{2}^{\prime })\)
\((a,b,b,1)=c_{1}(1,1,2,2)+c_{2}(0,0,-1,1)\)
By comparing the components, we get four equations:
- \(c_1 = a\)
- \(c_1 = b\) (Therefore, \(a = b\))
- \(2c_1 - c_2 = b\)
- \(2c_1 + c_2 = 1\)
Using equation (2), substitute $c_1 = b$ into equation (3):
\[ 2b - c_2 = b \implies c_2 = b \]
Now, substitute both $c_1 = b$ and $c_2 = b$ into equation (4):
\[ 2(b) + (b) = 1 \]\[ 3b = 1 \]\[ b = 1/3 \]
The correct answer is D
