I have a $3 \times4$ matrix $$ \begin{pmatrix} i & 4c & -1 & 0 \\ 2ic & 4 & -2c & 0 \\ -i & -2 & 1 & 6c-3 \\ \end{pmatrix} $$
to figure out the complex numbers $c$ such that the row space has dimension 2, how would I approach this problem? Also, for each $c$ I would like to try to find the subset of the columns of A that makes it a $\Bbb C$-basis of the column-space A.
So far, i've taken the determinate's of the 3 sub $3 \times3$ matrices. In particular: $$(A_1) = \begin{pmatrix} 4c & -1 & 0 \\ 4 & -2c & 0 \\ -2 & 1 & 6c-3 \\ \end{pmatrix} = det(A_1) -48c^2+48c-12$$ $$(A_2) = \begin{pmatrix} i & -1 & 0 \\ 2ic & -2 & 0 \\ -2 & 1 & 6c-3 \\ \end{pmatrix} = det(A_2) 12ic^2-18ic-12i$$ $$(A_2) = \begin{pmatrix} i & 4c & -1 \\ 2ic & 4 & -2 \\ -i & -2 & 1 \\ \end{pmatrix} = det(A_3) -8c^2i+12ic-4i$$ From here, I'm not sure what to do.I set the determinate equal to zero and got the following (below).
$$c(A_1)=\frac{1}{2}$$ $$c(A_2)= \frac{3i \pm 5i}{4i}$$ $$c(A_3)= \frac{-3i\pm i}{4}$$ Does anyone have any clue how or where I have messed up because I feel as though I have made a mistake somewhere.
I think reducing by rows can help:$${}$$
$$ \begin{pmatrix} i & 4c & -1 & 0 \\ 2ic & 4 & -2c & 0 \\ -i & -2 & 1 & 6c-3 \\ \end{pmatrix}\longrightarrow\begin{pmatrix} i & 4c &\!\!-1 & 0 \\ 0 & 4-8c^2 & 0 & 0 \\ 0 & 4c-2 & 0 & 6c-3 \\ \end{pmatrix}$$
Observe that if $\;6c-3=0\iff c=\frac12\;$ , then it follows that the last row becomes all zeros and none of the two first one is all zeros, and also if $\;8c^2=4\iff c=\pm\frac1{\sqrt2}\;$, and with this value the second row is all zeros.
Thus, only for $\;c=\frac12,\,\pm\frac1{\sqrt2}\;$ the matrix has row dimension equal to two