Using row reduction to solve for a variable in a matrix

Question

With given column vectors x = [1,3,2], y = [3,1,2], and z = [7,a,6], I need to use row reduction to find the value of a such that the vectors {x, y, z} are linearly dependent.

I am seeking to understand how to find this variable via row reduction. I reduced the matrix entirely to RREF when I remembered that I was supposed to be solving for a. Could I get some assistance?

Answer 1

Hint: A set of vectors $v_1, ..., v_n$ is linearly independent if the $c_1v_1 + ... + c_nv_n = 0 \implies c_i = 0$ for $i = 1, ..., n$. When row-reducing a matrix, you are essentially solving a system of equations like this and trying to show that there is only one solution s.t. all the unknowns of the system are $0$. If you pay close attention, you'll notice that a system has only one solution when the coefficient matrix is the identity i.e. if the system of equations being solved has one solution that is $0$ in every entry. Thus, a matrix $A \in R^{nxn}$ has linearly independent columns if RREF($A$) = $I_{nxn}$, where $I_{nxn}$ is the identity matrix.

Answer 2

$\begin{pmatrix} 1 & 3& 2\\3 & 1& 2\\7 & a& 6 \end{pmatrix}\rightarrow \begin{pmatrix} 1 & 3& 2\\0 & -8& -4\\0 & a-21& -8 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 3& 2\\0 & -8& -4\\0 & a-21+16& -8+4*2 \end{pmatrix} $, so $a=5$