Answer
The given matrix is singular.
Work Step by Step
1. Put an identity $I$ matrix on the right of the given matrix, to get a $4 \times 8$ one.
\[ \left( \begin{array}{ccc}
1 & 2 & 3 & 4 & 1 & 0 & 0 & 0\\
2 & 3 & 3 & 3 & 0 & 1 & 0 & 0\\
0 & 1 & 2 & 3 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & 2 & 0 & 0 & 0 & 1 \end{array} \right)\]
2. Row-reduce the whole matrix:
$R_2 = R_2 - 2R_1$
\[ \left( \begin{array}{ccc}
1 & 2 & 3 & 4 & 1 & 0 & 0 & 0\\
0 & -1 & -3 & -5 & -2 & 1 & 0 & 0\\
0 & 1 & 2 & 3 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & 2 & 0 & 0 & 0 & 1 \end{array} \right)\]
$R_1 = R_1 + 2R_2$
$R_3 = R_3 + R_2$
\[ \left( \begin{array}{ccc}
1 & 0 & -3 & -6 & -3 & 2 & 0 & 0\\
0 & -1 & -3 & -5 & -2 & 1 & 0 & 0\\
0 & 0 & -1 & -2 & -2 & 1 & 1 & 0 \\
0 & 0 & 1 & 2 & 0 & 0 & 0 & 1 \end{array} \right)\]
$R_1 = R_1 - 3R_3$
$R_2 = R_2 - 3R_3$
$R_4 = R_4 + R_3$
\[ \left( \begin{array}{ccc}
1 & 0 & 0& 0 & 3 & -1 & -3 & 0\\
0 & -1 & 0 & 1 & 4 & -2 & -3 & 0\\
0 & 0 & -1 & -2 & -2 & 1 & 1 & 0 \\
0 & 0 & 0 & 0 & -2 & 1 & 1 & 1 \end{array} \right)\]
- As we can see, the left part of the last row is made of zeros. Thus, the original matrix is singular.
