Determine whether the vector is a spanning set, dependent set, and if it has a basis. Determine whether the vector is a spanning set, dependent set, and if it has a basis.
A) The set $\begin{bmatrix}1&0\\0&1\\\end{bmatrix}$,$\begin{bmatrix}0&1\\1&0\\\end{bmatrix}$,$\begin{bmatrix}0&1\\-1&0\\\end{bmatrix}$,$\begin{bmatrix}1&1\\-1&1\\\end{bmatrix}$ of vectors in $M_{22}$.
$\rightarrow$  I believe it is linearly dependent because the last vector has a det. of 2 and all the others have det. = 0 therefore the last vector is a linear combination of the others making the set linearly dependent.
$\rightarrow$ Since it is linear. dependent it does not form a basis.
$\rightarrow$ By Gaussian elimination i got a rank of 2 equaling the dimensions hence it spans.
B) The set {x, 2, $x^2-x^3$, $x^3$} of vectors in $P_3$.
$\rightarrow$ this one has me very confused.
EDIT- This does form a basis since the polynomials will always be in $P_3$
C) The set {(3,17),(51,5),(97,103)} of vectors in $R^2$.
$\rightarrow$ It is spanning since the rank is 2 and the Dimension is 2
$\rightarrow$ Not sure about the rest since the fact that this forms a 2 x 3 matrix which confuses me about the conditions.
 A: Let's see exercise A. The determinant of the four matrices is just irrelevant. What you need is to see whether you can find a zero linear combination with not all coefficients zero. So write
$$
a\begin{bmatrix}1&0\\0&1\\\end{bmatrix}+
b\begin{bmatrix}0&1\\1&0\\\end{bmatrix}+
c\begin{bmatrix}0&1\\-1&0\\\end{bmatrix}+
d\begin{bmatrix}1&1\\-1&1\\\end{bmatrix}=
\begin{bmatrix}0&0\\0&0\\\end{bmatrix}
$$
This becomes
$$
\begin{bmatrix}
a+d & b+c+d \\
b-c-d & a+d
\end{bmatrix}=
\begin{bmatrix}0&0\\0&0\\\end{bmatrix}
$$
This is the linear system
\begin{cases}
a+d=0\\
b+c+d=0\\
b-c-d=0\\
a+d=0
\end{cases}
which has a matrix
$$
\begin{bmatrix}
1 & 0 & 0 & 1 \\
0 & 1 & 1 & 1 \\
0 & 1 & -1 & -1 \\
1 & 0 & 0 & 1
\end{bmatrix}
$$
whose reduced row echelon form is easily seen to be
$$
\begin{bmatrix}
1 & 0 & 0 & 1 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix}
$$
This means that the system has infinitely many solutions, so the four vectors form a linearly independent set. But this also shows that the fourth matrix is a linear combination of the first three, precisely
$$
\begin{bmatrix}1&1\\-1&1\\\end{bmatrix}=
1\begin{bmatrix}1&0\\0&1\\\end{bmatrix}+
0\begin{bmatrix}0&1\\1&0\\\end{bmatrix}+
1\begin{bmatrix}0&1\\-1&0\\\end{bmatrix}
$$
and that the first three matrices form a linearly independent set.
Thus the set is not a spanning set for $M_{22}$ which has dimension $4$. The spanned subspace has indeed dimension $3$.
You should do similarly for the other exercises.
