The given matrix is
$\begin{pmatrix}
1 & 1 & 0 & 0\\
-1 & 0 & 1 & 0\\
1 & 3 & 4 & 1\\
1 & 2 & 1 & 0\end{pmatrix}$
To obtain the determinant, I use Guass method on this 4×4 matrix until I get a matrix of a form where there are 1 zero in front of the second row, 2 zeros in front of the third row and 3 zeros in front of the fourth row.
1) do p1+p2, -p1+p3 and -p1+p4 and you get the matrix:
$\begin{pmatrix}
1 & 1 & 0 & 0\\
0 & 1 & 1 & 0\\
0 & 2 & 4 & 1\\
0 & 1 & 1 & 0\end{pmatrix}$
2) then do -2p2+p3,-p2+p4 and you get the final matrix as:
$\begin{pmatrix}
1 & 1 & 0 & 0\\
0 & 1 & 1 & 0\\
0 & 0 & 2 & 1\\
0 & 0 & 0 & 0\end{pmatrix}$
My answer:
The determinant also known as the product of the diagonal is zero because
$1 * 1 * 2 * 0 = 0$
The column space is $|S| = \{c_1s_1 + \dots + c_ns_n \mid c_1,\dots,c_m \in \mathbb{R} \text{ and } s_1,\dots, s_n \in S\}$
http://i33.photobucket.com/albums/d86/warnexus/basis_zps4b301a1b.png
I do not know what value are the constants of my column space. Should I keep it general like the above?
Best Answer
Look at the "pivots" of your final matrix---that is, in each row, pick out the first non-zero entry, if one exists, and write down what column number it's in (i.e. column 1, column 3, etc.). Since your original matrix has the same number of columns as your final matrix, each column number you wrote down specifies a column of your original matrix. These columns then form a basis for the column space.
Let's go through your specific problem. The pivots in rows 1, 2, and 3 of your final matrix occur in columns 1, 2, and 3, respectively. However, row 4 is entirely 0's, so it doesn't have any pivots. Therefore, we can completely ignore row 4. Thus, the only column numbers we write down are 1, 2, and 3. Picking out columns 1, 2, and 3 of our original matrix, we find that $$ \begin{pmatrix}1\\-1\\1\\1\end{pmatrix}, \begin{pmatrix}1\\0\\3\\2\end{pmatrix},\begin{pmatrix}0\\1\\4\\1\end{pmatrix}$$ forms a basis for the column space.