Given $A\in\mathbb{R}^{m\times n}$, $m\geq n$, compute the (economy) QR factorisation. This gives
$$
A = QR, \quad R\in\mathbb{R}^{n\times n}.
$$
Now if $\mathrm{rank}(A)<n$, the upper triangular matrix $R$ has a staircase profile with some of the "steps" of the staircase over more than one column. Select column indices $j_1,\ldots,j_k$ such that if you remove these columns from $R$, you obtain a nonsingular upper triangular matrix (you can consider it as making each step of the staircase of length 1). The columns $j_1,\ldots,j_k$ can be expressed as linear combination of the remaining columns.
Example: The red columns indicate the columns which are linear combinations of the others.
$$
\begin{bmatrix}
\times & \times & \color{red}\times & \times & \color{red}\times & \color{red}\times \\
0 & \times & \color{red}\times & \times & \color{red}\times & \color{red}\times \\
0 & 0 & \color{red}0 & \times & \color{red}\times & \color{red}\times
\end{bmatrix}
$$
Example: For the given matrix from the question, the QR factorisation is:
Q =
0 -0.4472 -0.8944
0 -0.8944 0.4472
-1.0000 0 0
R =
-1.0000 2.0000 -1.0000
0 4.4721 -2.2361
0 0 0
So one can pick the column 2 or 3 to make the matrix $R$ nonsingular and upper triangular (hence either the column 2 or 3 is a linear combination of the others).
You should know that elementary column operations preserve a zero/nonzero determinant. You should also know that if a column is all zeros, then the determinant is zero. All that remains is to convince yourself that if the columns are linearly dependent, then you can make one column all zeros from a sequence of elementary column operators.
Best Answer
What can you say about the following matrix? $$\begin{pmatrix}1&0&1&2\\0&1&1&0\end{pmatrix}$$