To determine the row rank of the matrix we reduce the matrix into reduced echelon form.

First we add  times the 1st row to the 2nd row

add  times the 1st row to the 3rd row

Switch the 2nd row and the 3rd row

multiply the 2nd row by

add  times the 2nd row to the 1st row

And we find that the row rank is 

Yes, the set is linearly independent. There are multiple ways to see this

Way 1) Put the vectors into matrix form,

The matrix is already in reduced echelon form. Notice there are three rows that have a nonzero number in them and we started with 3 vectors. Thus the set is linearly independent.

Way 2) Consider the equation

If when we solve the equation, we get  then it is linearly independent. Let's solve the equation and see what we get.

Distribute the scalar constants to get

Thus we get a system of 3 equations

Since the vectors are linearly independent.

The vectors have dimension 3. Therefore the largest possible size for a linearly independent set is 3. But there are 4 vectors given. Thus, the set cannot be linearly independent and must be linearly dependent

Another way to see this is by noticing that  can be written as a linear combination of the other vectors:

The dimension of a vector space is the maximum number of vectors possible in a linearly independent set. (notice you can have linearly independent sets with 5 or less, but never more than 5)

Since there are three linearly independent vectors, they span a 3 dimensional space. 

Notice that the vectors each have 5 coordinates to them. Therefore they actually span a 3 dimensional subspace of a 5 dimensional space.

Since there are three linearly independent vectors, they span a 3 dimensional space.

Notice that the vectors each have 5 coordinates to them. Therefore they actually span a 3 dimensional subspace of a 5 dimensional space.

Since  is a  matrix, . Since  has three linearly independent columns, it must have a column space (and hence row space) of dimension , causing  by the definition of rank. Hence.

 is equal to the number of linearly independent columns of . The first and third columns are the same, so one of these columns is redundant in the column space of . The second column evidently cannot be a multiple of the first, since the second has two 's, and the first has none. Hence .

A test to determine whether these matrices form a basis is to set up a matrix with each row comprising the coefficients of one polynomial, and performing row reductions until the matrix is in row-echelon form.   is a vector space of dimension 4, so these four polynomials will form a basis if and only if the resulting   matrix has rank 4. The initial matrix is:

Perform the following row operations:

The matrix is now in row-echelon form. Each row has a leading 1, so the matrix has rank 4, the dimension of . It follows that the given polynomials comprise a basis of .

Elements of a vector space form a basis if the elements are linearly independent and if they span the space - that is, every element in that space can be uniquely expressed as the sum of the elements. 

A test to determine whether these matrices form a basis is to set up a matrix with each row comprising the coefficients of one polynomial, and performing row reductions until the matrix is in row-echelon form.   is a vector space of dimension 4, so these four polynomials will form a basis if and only if the resulting   matrix has rank 4. The initial matrix is:

Perform the following row operations:

The matrix is now in row-echelon form. There is a row of zeroes at the bottom, so the rank of the matrix is 3. Therefore, the four polynomials are not linearly independent, and they do not form a basis for .