# 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.

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:

• $$c-b-2d=0$$.
• $$c-b-2d \ne 0$$.
• Sorry I didn't write my whole answer I edited it Mar 9 at 3:04
• looks fine if there is no computational careless mistake. Mar 9 at 3:05
• Thank you very much Mar 9 at 3:07