Given Information
We have to construct $3 \times 2$ matrices A and B such that $Ax = 0$ has only the trivial solution and $Bx = 0$ has a nontrivial solution.

Step-1:
Let, \[ A = \left[ \begin{array} { l l } { a } & { d } \\ { b } & { e } \\ { c } & { f } \end{array} \right] \text { and } \mathbf { x } = \left[ \begin{array} { l } { x _ { 1 } } \\ { x _ { 2 } } \end{array} \right] \]

Step-2: Augmented matrix for $Ax=0$
\[\begin{array}{l} \left[ {A\,\,\,{\bf{0}}} \right] = \left[ {\begin{array}{*{20}{c}} a&d&0\\ b&e&0\\ c&f&0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} a&d&0\\ b&{e - \dfrac{b}{a}d}&0\\ c&{f - \dfrac{c}{a}d}&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{R_2} \to {R_2} - \dfrac{b}{a}{R_1}}\\ {{R_3} \to {R_3} - \dfrac{c}{a}{R_1}} \end{array}} \right\} \end{array}\]

Step-4:
The system has a trivial solution if $e - \dfrac { b } { a } d \neq 0$ and $f - \dfrac { c } { a } d \neq 0$.

Hence, \[\begin{array}{l} e - \dfrac{b}{a}d \ne 0 \Rightarrow \dfrac{{ae - bd}}{a} \ne 0 \Rightarrow ae - bd \ne 0 \Rightarrow \dfrac{a}{d} \ne \dfrac{b}{e}\\ \\ f - \dfrac{c}{a}d \ne 0 \Rightarrow \dfrac{{af - cd}}{a} \ne 0 \Rightarrow af - cd \ne 0 \Rightarrow \dfrac{a}{d} \ne \dfrac{c}{f} \end{array}\] Suppose: \[\dfrac{a}{d} \ne \dfrac{b}{e} \ne \dfrac{c}{f} = k \Rightarrow a \ne kd,b \ne ke,c \ne kf\] Then, \[\left[ {\begin{array}{*{20}{l}} a\\ b\\ c \end{array}} \right] \ne \left[ {\begin{array}{*{20}{l}} {kd}\\ {ke}\\ {kf} \end{array}} \right] \ne k\left[ {\begin{array}{*{20}{l}} d\\ e\\ f \end{array}} \right]\] Also, if the one of the column is zero, then it is zero times of the other, hence, this case is also considered to have trivial solution. Therefore,

The columns of $2\times 3$ matrix should be linearly independent then the equation has trivial solution.

Step-5: Augmented matrix for $Bx=0$
\[\begin{array}{l} [B|{\bf{0}}] = \left[ {\begin{array}{*{20}{c}} {{l_1}}&{{l_2}}&0\\ {{m_1}}&{{m_2}}&0\\ {{n_1}}&{{n_2}}&0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {{l_1}}&{{l_2}}&0\\ {{m_1}}&{{m_2} - \dfrac{{{m_1}}}{{{l_1}}}{l_2}}&0\\ {{n_1}}&{{n_2} - \dfrac{{{n_1}}}{{{l_1}}}{l_2}}&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{R_2} \to {R_2} - \dfrac{{{m_1}}}{{{l_1}}}{R_1}}\\ {{R_3} \to {R_3} - \dfrac{{{n_1}}}{{{l_1}}}{R_1}} \end{array}} \right\} \end{array}\]

Step-6:
The system has a nontrivial solution if: \[\begin{array}{l} {m_2} - \dfrac{{{m_1}}}{{{l_1}}}{l_2} = 0 \Rightarrow \dfrac{{{m_2}{l_1} - {m_1}{l_2}}}{{{l_1}}} = 0 \Rightarrow {m_2}{l_1} - {m_1}{l_2} = 0 \Rightarrow \dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{{m_1}}}{{{m_2}}}\\ \\ {n_2} - \dfrac{{{n_1}}}{{{l_1}}}{l_2} = 0 \Rightarrow \dfrac{{{n_2}{l_1} - {n_1}{l_2}}}{{{l_1}}} = 0 \Rightarrow {n_2}{l_1} - {n_1}{l_2} = 0 \Rightarrow \dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{{n_1}}}{{{n_2}}} \end{array}\] The above conditions yield: \[ \dfrac { l _ { 1 } } { l _ { 2 } } = \dfrac { m _ { 1 } } { m _ { 2 } } = \dfrac { n _ { 1 } } { n _ { 2 } } \] Therefore, \[\left[ {\begin{array}{*{20}{l}} {{l_1}}\\ {{m_1}}\\ {{n_1}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} {k{l_2}}\\ {k{m_2}}\\ {k{n_2}} \end{array}} \right] = k\left[ {\begin{array}{*{20}{c}} {{l_2}}\\ {{m_2}}\\ {{n_2}} \end{array}} \right]\] Therefore,

If the columns of matrix are linearly dependent , then the equation $Bx=0$ has a nontrivial solution.