Activity 2.4.3 Linear independence and homogeneous equations
Explain why the homogenous equation \(A\xvec = \zerovec\) is consistent no matter the matrix \(A\text{.}\)

Consider the matrix
\begin{equation*} A = \left[\begin{array}{rrr} 3 \amp 2 \amp 0 \\ 1 \amp 0 \amp 2 \\ 2 \amp 1 \amp 1 \end{array}\right] \end{equation*}whose columns we denote by \(\vvec_1\text{,}\) \(\vvec_2\text{,}\) and \(\vvec_3\text{.}\) Are the vectors \(\vvec_1\text{,}\) \(\vvec_2\text{,}\) and \(\vvec_3\) linearly dependent or independent?
Give a description of the solution space of the homogeneous equation \(A\xvec = \zerovec\text{.}\)

We know that \(\zerovec\) is a solution to the homogeneous equation. Find another solution that is different from \(\zerovec\text{.}\) Use your solution to find weights \(c_1\text{,}\) \(c_2\text{,}\) and \(c_3\) such that
\begin{equation*} c_1\vvec_1 + c_2\vvec_2 + c_3\vvec_3 = \zerovec\text{.} \end{equation*} Use the expression you found in the previous part to write one of the vectors as a linear combination of the others.