I was presented with the following Matrix in an assignment and asked to find the value for B that would make the rank of the matrix 2.
\begin{pmatrix}3&-3&0\\ 3&-1&2\\ b&0&2\end{pmatrix}
I solved this by using row operations to make the last row all zeros using the following steps.
Step 1: Row 2 = Row 2 - row 1. \begin{pmatrix}3&-3&0\\ 0&2&2\\ b&0&2\end{pmatrix}
Step 2: Row 3 = Row 3 - Row2.
\begin{pmatrix}3&-3&0\\ 0&2&2\\ b&-2&0\end{pmatrix}
Step 3: Row 3 = Row 3 - 2/3*row1.
\begin{pmatrix}3&-3&0\\ 0&2&2\\ b-2&0&0\end{pmatrix}
From this b must be 2 to make the rank of the matrix 2, no? I was told this was wrong as I had used Gauss-Jordan elimination incorrectly. The answer to the question is two but I was awarded no marks for it. Why is this so wrong? Thank you!
Your steps to arrive at the conclusion that the rank is at least $2$ are correct. But you fail to show that the rank is $2$ if and only if $b=2$.
You're only showing that if $b=2$ then the rank is $2$.
Well, you were not far from finishing correctly the argument, but as it stands it's incomplete: the form you get is not triangular. Some row swaps would suffice.
Where's the real problem? You took advantage from the particular matrix, but, as I use to say, you should think to the given matrix as a $1000\times1000$ matrix, maybe with several parameters: you don't want to start looking for promising simplifications, because it would take longer than directly performing Gauss-Jordan.
\begin{align} \begin{pmatrix} 3 & -3 & 0 \\ 3 & -1 & 2 \\ b & 0 & 2 \end{pmatrix} &\to \begin{pmatrix} 1 & -1 & 0 \\ 3 & -1 & 2 \\ b & 0 & 2 \end{pmatrix} && R_1\gets \tfrac{1}{3}R_1 \\[6px]&\to \begin{pmatrix} 1 & -1 & 0 \\ 0 & 2 & 2 \\ 0 & b & 2 \end{pmatrix} && \begin{aligned} R_2 &\gets R_2-3R_1 \\ R_3 &\gets R_3-bR_1 \end{aligned} \\[6px]&\to \begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & 1 \\ 0 & b & 2 \end{pmatrix} && R_2 \gets \tfrac{1}{2}R_2 \\[6px]&\to \begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 2-b \end{pmatrix} && R_3 \gets R_3-bR_2 \end{align} At this point you have two possibilities: if $b=2$, the elimination stops and the rank is $2$; if $b\ne2$, you have another step and the rank is $3$.