Suppose $ \begin{vmatrix} 1 & 1 & 0 & 1 & 0 \\ 2 & 1 & 0 & 1 & 0 \\ 1 & 2 & 0 & 1 & 1 \\ 1 & 1 & 0 & 2 & 1 \\ 0 & 0 & 2 & 0 & 2 \\ \end{vmatrix} $ = 4.
(1) $ \{(2, 1,0,1,0), (1,1,0,1,0), (1,1,0,2,1)\}$ is linearly independent.
(2) the rows form a basis for $r^5$
(3) $ \left\{ \begin{pmatrix} 1 \\ 2 \\ 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 2 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 1 \\ 2 \\ 0 \end{pmatrix}\right\} $ is linearly independent.
I think (1), (2), (3) are correct, but it is not possible for the sets in (1) and (3) to span $R^5$ since a minimum of 5 vectors is required, which is also the reason why (2) is correct? Please correct me if I am wrong thank you.
To set things as clear as possible: don't forget that a basis is characterised as
When they ask whether a set of vectors is linearly independent, they do not mean it is a maximal linearly independent set. So there's no incompatibility.
As to question c), you just have to compute the determinant of the matrix, either by row reduction, or a combination of row reduction and expansion along a row or a column, when it has enough $0$s. I'll show a way to start: $$\begin{vmatrix}1&1&0&1&0 \\ 2&1&0&1&0 \\1&2&0&1&1 \\1&1&0&2&1 \\0&0&2&0&2 \end{vmatrix}= 2\begin{vmatrix}\color{red}1&1&0&1&0 \\ 0&-1&0&-1&0 \\0&1&0&0&1 \\0&0&0&1&1 \\0&0&1&0&1 \end{vmatrix}=2\begin{vmatrix}-1&0&-1&0 \\1&0&0&1 \\0&0&1&1 \\0&\color{red}1&0&1 \end{vmatrix}=2\begin{vmatrix}-1&-1&0 \\1 &0&1 \\0 &1&1 \end{vmatrix}\\[10ex]=\cdots$$