Hi, I can not get rank/spark relation correctly, I know that rank is a number of linearly independent columns of a matrix and spark is linearly dependent ones. In this question I understand option a), but I can not realize option b) and c). Could anyone explain that 2 option?
The answer to (b) is the same as for (a). The only difference is that instead of adding three base vectors together, you only add 2: $$e_{12} = e_1 + e_2 = \begin{bmatrix}1\\1\\0\\0\\0\end{bmatrix}$$
As far as I can tell, you only need to add one column to the base vectors to get a matrix of rank 5, spark 3. I don't see anything that demands a higher column count: $$\begin{bmatrix}1&0&0&0&0&1\\0&1&0&0&0&1\\0&0&1&0&0&0\\0&0&0&1&0&0\\0&0&0&0&1&0\end{bmatrix}$$
The answer to (c) is just to throw out everything involving $e_5$ from the answer to (a) (including the combo-vectors built from it). Since the bottom row is always $0$, the rank can be no higher than $4$. Again, I don't see the need to add more than one combo vector: $$\begin{bmatrix}1&0&0&0&1\\0&1&0&0&1\\0&0&1&0&1\\0&0&0&1&0\\0&0&0&0&0\end{bmatrix}$$