Find the values of $a$ and $b$ such that
$$\mbox{rank} \begin{pmatrix}
3 & 2 & 5 \\
1 & a & -1\\
1 & 3 & b\\
\end{pmatrix} = 2$$
I can only seem to get single values for $a$ and $b$. To get a matrix of rank 2 we need a matrix in row echelon form with one row of $0$'s at the bottom of the matrix. I.e. two linearly independent rows.
I am able to reduce the above matrix to:
$$\begin{pmatrix}
1 & 3 & b\\
0 & -7 & 5-b\\
0 & a-3 & -1-b\\
\end{pmatrix}$$
We get a row of $0$'s at the bottom when $b=-1$ and $a = 3$. Since the questions asks for an equation for a and b I feel like my solution is incomplete. Any advice?
Best Answer
Here is why your argument is incomplete. One way for your reduced matrix to have rank 2 is to have row three all zeros, as you said. HOWEVER, there is another possibility: row 2 and 3 being multiple of each other. (With row 1 this is not a question b/c row 1 cannot be multiple of either row 2 or row 3 since we have 1 versus 0s on first column.
So let us look at this second possibility: we will have rank 2 if $$ \frac{-7}{a-3} = \frac{5-b}{-1-b} \, . $$ Which hopefully is the same equation as the one in the answer below reached by computing determinant.