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Question 4 [10 points]
Give a basis for span $(S)$, where $S$ is the set given below.
\[
\left\{\left[\begin{array}{l}
0 \\
0 \\
1
\end{array}\right],\left[\begin{array}{c}
10 \\
3 \\
1
\end{array}\right],\left[\begin{array}{c}
-10 \\
-3 \\
-1
\end{array}\right],\left[\begin{array}{c}
-10 \\
-3 \\
-3
\end{array}\right]\right\}
\]

Number of Vectors: 1
\[
\left\{\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right]\right\}
\]
SUBMT AND MARK

Step 1 ：Given the set of vectors S = \(\left\{\left[\begin{array}{l} 0 \ 0 \ 1 \end{array}\right],\left[\begin{array}{c} 10 \ 3 \ 1 \end{array}\right],\left[\begin{array}{c} -10 \ -3 \ -1 \end{array}\right],\left[\begin{array}{c} -10 \ -3 \ -3 \end{array}\right]\right\}\)

Step 2 ：We need to find a basis for the span of S. A basis for a vector space is a set of vectors that are linearly independent and that span the vector space. In other words, every vector in the vector space can be written as a linear combination of the vectors in the basis.

Step 3 ：To find a basis for the span of S, we need to find a set of vectors from S that are linearly independent. We can do this by putting the vectors into a matrix and performing Gaussian elimination to find the pivot columns. The pivot columns correspond to the vectors in the basis.

Step 4 ：Let's denote the vectors as v1 = \(\left[\begin{array}{l} 0 \ 0 \ 1 \end{array}\right]\), v2 = \(\left[\begin{array}{c} 10 \ 3 \ 1 \end{array}\right]\), v3 = \(\left[\begin{array}{c} -10 \ -3 \ -1 \end{array}\right]\), and v4 = \(\left[\begin{array}{c} -10 \ -3 \ -3 \end{array}\right]\).

Step 5 ：We form a matrix A with these vectors: A = \(\left[\begin{array}{cccc} 0 & 10 & -10 & -10 \ 0 & 3 & -3 & -3 \ 1 & 1 & -1 & -3 \end{array}\right]\).

Step 6 ：After performing Gaussian elimination, we find that the pivot columns are all the columns, which means that all the vectors are linearly independent and therefore form a basis for the span of S.

Step 7 ：Final Answer: The basis for the span of S is \(\boxed{\left\{\left[\begin{array}{l} 0 \ 0 \ 1 \end{array}\right],\left[\begin{array}{c} 10 \ 3 \ 1 \end{array}\right],\left[\begin{array}{c} -10 \ -3 \ -1 \end{array}\right],\left[\begin{array}{c} -10 \ -3 \ -3 \end{array}\right]\right\}}\)