Answer:
One of the obvious non-trivial solutions is [tex](x_1, x_2, x_3)=(-2, 3, -5)[/tex].
Step-by-step explanation:
Suppose the matrix A is as follows:
[tex]A=\left[\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&3_{23}\\a_{31}&a_{32}&a_{33}\end{array}\right][/tex]
The observed system [tex]Ax=0[/tex] after multiplying looks like this
[tex]Ax=0 \iff \left[\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{array}\right] \cdot \left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right] =0 \iff \\ \\a_{11}x_1+a_{12}x_2+a_{13}x_3=0\\a_{21}x_1+a_{22}x_2+a_{23}x_3=0\\a_{31}x_1+a_{32}x_2+a_{33}x_3=0\\\\[/tex]
Since we now that [tex]-2A_1+3A_2-5A_3=0[/tex], where [tex]A_i\ ,\ i=1, 2, 3[/tex] are the columns of the matrix A, we actually know this:
[tex]-2\cdot \left[\begin{array}{ccc}a_{11}\\a_{21}\\a_{31}\end{array}\right] +3\cdot \left[\begin{array}{ccc}a_{12}\\a_{22}\\a_{32}\end{array}\right] -5\cdot \left[\begin{array}{ccc}a_{13}\\a_{23}\\a_{33}\end{array}\right] =\left[\begin{array}{ccc}0\\0\\0\end{array}\right][/tex]
Once we multiply and sum up these 3 by 1 matrices, we get that these equations hold:
[tex]-2a_{11}+3a_{12}-5a_{13}=0\\-2a_{21}+3a_{22}-5a_{23}=0\\-2a_{31}+3a_{32}-5a_{33}=0[/tex]
This actually means that the solution to the previously observed system of equations (or equivalently, our system [tex]Ax=0[/tex]) has a non-trivial solution [tex](x_1, x_2, x_3)=(-2, 3, -5)[/tex].
