Arrangement of Matrices using number $-1,0,1$ such that the solution is non-trivial

Question

Let set $A$ contain all the $(3 × 3)$ matrices whose entries are either $0, 1$ or $–1$. Two of these entries are $1$, two are $–1$ and five are $0$. Find the number of matrices $B$ that belong to set $A$ for which the equation, $B\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0\\ 0\\ 0 \end{array}} \right]$, has non-trivial solution.

My approach is as follows. Total number of matrix is $\frac{9!}{51\cdot2!\cdot2!}=756$.

Let us calculate the determinant as

$$D=\left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}}\\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right| ={a_1}{b_2}{c_3}-{a_1}{b_3}{c_2}-{a_2}{b_1}{c_3}+{a_2}{b_3}{c_1}+ {a _3}{b_1}{c_2} - {a_3}{b_2}{c_1}$$

For trivial solution $D\ne0$ , for non trivial solution $D=0$, I am not able to perform the operation such that among $a_1$ to $a_9$, we have five zeroes, two $-1$ and two $1$ such that $D=0$

Answer 1

Hint: There are non-trivial solutions iff the columns are linearly dependent.

Consider the following scenarios:

1. A column of all 0's. Henceforth, every column has at least a non-zero entry.
2. $c_1 = c_2$ (and similar equations).
3. $c_1 = c_2 + c_3$ (and similar equations).

Show that there are no other scenarios that lead to linearly dependent columns.

Thus, what are the total number of cases?