Finding the entries in a matrix given that the rank has to be $2$

Question

Find all $y<0$ such that the matrix defined for $y<0$ $$\begin{pmatrix}-3&2&y\\ 0&1&-\frac{1}{y}\\ y&0&y\end{pmatrix}$$ has rank $2$.

It seems ok, we just need a $0$ row, which could be the last row for example, if $y = 0$. However, this would mess up the condition in the second row, where $y≠0$. Im not sure what to do, any suggestions?

Could the determinant help in this case?

Answer 1

Edit: the post has been edited, so the answer is no longer correct, but the method is and it does not require any knowledge of eigenvalues, so I will leave it up as it generally applicable to all of these types of problems. See the other answer for the correct solution to the now edited question.

Original:

What others wrote in the comments will work, but this is the easiest way in my opinion. We can solve this problem using Gaussian elimination to bring the Matrix into upper triangular form, as we know that this preserves Rank (here it also important that $y \neq 0 $, which we already know): $$\begin{pmatrix}-3&2&y\\ \:0&1&\frac{1}{y}\\ \:y&0&y\end{pmatrix} \rightarrow \begin{pmatrix}-y&\frac{2}{3}y&\frac{1}{3}y^2\\ \:\:0&1&\frac{1}{y}\\ \:\:0&\frac{2}{3}y&y+\frac{2}{3}y^2\end{pmatrix}\rightarrow \begin{pmatrix}-y&\frac{2}{3}y&\frac{2}{3}y^2\\ \:\:0&\frac{2}{3}y&\frac{2}{3}\\ \:\:0&0&y+\frac{1}{3}y^2-\frac{2}{3}\end{pmatrix}$$ From this we can see that when $y=\frac{-3+\sqrt{17}}{2}$ or $y=\frac{-3-\sqrt{17}}{2}$ we get a matrix of rank 2.

Answer 2

It should be clear that the second row is linearly independent from the first row, for any valid $y$ (it cannot be zero since we have a term that is $\frac{1}{y}$). So the rank must be at least 2. In that case, we just have to obtain $y$ such that the determinant is zero.

$$\det\left(\begin{bmatrix} -3 & 2 & y\\ 0 & 1 & -\frac{1}{y}\\ y & 0 & y \end{bmatrix}\right) = -3\,y -2 - y^2 = 0 \Rightarrow y = \left\{\begin{align}-2\\-1\end{align}\right.$$

Answer 3

Call the first, the second and the third columns of the matrix $u,v$ and $w$ respectively. Clearly $u$ and $v$ are linearly independent. So, in order that the matrix has rank two, we must have $w=au+bv$ for some scalars $a$ and $b$. In particular, by looking at the bottom elements of the three vectors, we have $3=w_3=au_3+bv_3=a(3)+b(0)=3a$. Hence $a=1$. Consequently, $w-u=bv$, i.e., $$ \pmatrix{y+3\\ -\frac{1}{y}\\ 0}=b\pmatrix{2\\ 1\\ 0}.\tag{$\#$} $$ Since $-\frac{1}{y}$ is nonzero, so must be $b$. Therefore $(\#)$ holds if and only if $\frac{y+3}{-1/y}=2$, i.e., iff $y^2+3y+2=0$. In other words, the matrix has rank two iff $y=-1$ or $y=-2$.