Your question only makes sense to me if the matrix given is the coefficient matrix for the constraints. Let's denote that matrix $A$. Each column in $A$ corresponds to a variable. Let's suppose these are $x_1, x_2, x_3, x_4,$ and $x_5$.
There is a one-to-one correspondence between (1) bases of the column space of $A$ that consist of columns of $A$ and (2) basic solutions to the linear program. (That's why they're called basic solutions.) Since columns 1 and 2 are linearly dependent, they can't both be in a basis of the column space of $A$. Similarly for columns 3 and 4. Thus there are four bases for the column space of $A$ that can be formed by choosing columns from $A$:
- Column vectors 1, 3, and 5
- Column vectors 1, 4, and 5
- Column vectors 2, 3, and 5
- Column vectors 2, 4, and 5
This means that there are four basic solutions to the linear program; i.e., those in which the following are the basic variables:
- $\{x_1, x_3, x_5\}$
- $\{x_1, x_4, x_5\}$
- $\{x_2, x_3, x_5\}$
- $\{x_2, x_4, x_5\}$
Depending on the right-hand sides of the constraints, not all of the basic solutions may be feasible, but your question does not ask about feasibility.