<s>[INST] Question of rank of a matrix $4\times 6$ Let $a,b,c,d$ be complex numbers. Find the rank of the following matrix
\begin{align*}
    \begin{pmatrix}
    2& -3& 6& 0& -6& a\\
    -1& 2& -4& 1& 8& b\\
    1& 0& 0& 1& 6& c\\
    1& -1& 2& 0& -1& d\\
    \end{pmatrix}
\end{align*}
Idea: Proceed the elementary row operations and reduce to the raw Echelon form then the number of nonzero rows are the rank of the matrix.
Solution
\begin{align*}
&\begin{pmatrix}
    2& -3& 6& 0& -6& a\\
    -1& 2& -4& 1& 8& b\\
    1& 0& 0& 1& 6& c\\
    1& -1& 2& 0& -1& d\\
    \end{pmatrix}\overset{r_1  \xrightarrow{} r_4} {}
    \begin{pmatrix}
    1& -1& 2& 0& -1& d\\
    -1& 2& -4& 1& 8& b\\
    1& 0& 0& 1& 6& c\\
    2& -3& 6& 0& -6& a\\
    \end{pmatrix} \underset{\overset{r_1(-2) + r_4 \xrightarrow{}r_4{}}{}}{\overset{r_1 + r_2 \xrightarrow{}r_2,\, (-1)r_1 + r_3 \xrightarrow{}r_3}{}}\\
    &\begin{pmatrix}
    1& -1& 2& 0& -1& d\\
    0& 1& -2& 1& 7& b+d\\
    0& 1& -2& 1& 7& c-d\\
    0& -1& 2& 0& -4& a-2d\\
    \end{pmatrix}  \underset{\overset{r_2 + r_4 \xrightarrow{}r_4{}}{}}{\overset{r_2(-1) + r_3 \xrightarrow{}r_3}{}}\begin{pmatrix}
    1& -1& 2& 0& -1& d\\
    0& 1& -2& 1& 7& b+d\\
    0& 0& 0& 0& 0& c-b-2d\\
    0& 0& 0& 1& 3& a+b-d\\
    \end{pmatrix}\overset{r_3  \xrightarrow{}\, r_4} {}\\
    & \begin{pmatrix}
    1& -1& 2& 0& -1& d\\
    0& 1& -2& 1& 7& b+d\\
    0& 0& 0& 1& 3& a+b-d\\
    0& 0& 0& 0& 0& c-b-2d\\
    \end{pmatrix}
\end{align*}
Answer: If $c-b-2d = 0$ the rank of the matrix is 3, If $c-b-2d \neq 0$ the rank of the matrix is 4
This is my answer, am I correct here? I feel it is not enough for an answer. Is there anything I have to do? Any advice would be appreciated.
 [/INST] Guide:
Your idea is fine.
The question is what is the rank of the matrix, the answer should be a number, of which you have not provided.
Consider two cases:

*

*$c-b-2d=0$.

*$c-b-2d \ne 0$.

</s>