Ax = b tem somente solução trivial

a) Para mostrar essa afirmação, realizaremos os passos abaixo:

\(\begin{align} & Ax=b \\ & \left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {} & {{a}_{1n}} \\ {{a}_{21}} & {{a}_{22}} & {} & {{a}_{2n}} \\ {} & {} & {} & {} \\ {{a}_{m1}} & {{a}_{m2}} & {} & {{a}_{mn}} \\ \end{matrix} \right] \\ & \\ & x=\left[ \begin{matrix} {{x}_{1}} \\ {{x}_{2}} \\ {} \\ {{x}_{n}} \\ \end{matrix} \right] \\ & \\ & b=\left[ \begin{matrix} {{b}_{1}} \\ {{b}_{2}} \\ {} \\ {{b}_{n}} \\ \end{matrix} \right] \\ & \frac{A}{b}=\frac{\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {} & {{a}_{1n}} \\ {{a}_{21}} & {{a}_{22}} & {} & {{a}_{2n}} \\ {} & {} & {} & {} \\ {{a}_{m1}} & {{a}_{m2}} & {} & {{a}_{mn}} \\ \end{matrix} \right]}{\left[ \begin{matrix} {{b}_{1}} \\ {{b}_{2}} \\ {} \\ {{b}_{n}} \\ \end{matrix} \right]} \\ & \frac{A}{b}=\left\{ \begin{matrix} {{\grave{a}}_{11}}{{x}_{1}}+{{a}_{12}}{{x}_{2}}+...+{{a}_{1n}}{{x}_{n}}=0 \\ {{a}_{21}}{{x}_{1}}+{{a}_{22}}{{x}_{2}}+...+{{a}_{2n}}{{x}_{n}}=0 \\ {} \\ {{a}_{m1}}{{x}_{1}}+{{a}_{m2}}{{x}_{2}}+..+{{a}_{mn}}{{x}_{n}}=0 \\ \end{matrix} \right. \\ \end{align}\ \)

b)  Mostraremos agora que a situação contrária ao item A , não pode ocorrer:

\(\begin{align} & \frac{A}{b}=\left\{ \begin{matrix} {{\grave{a}}_{11}}{{x}_{1}}+{{a}_{12}}{{x}_{2}}+...+{{a}_{1n}}{{x}_{n}}=0 \\ {{a}_{21}}{{x}_{1}}+{{a}_{22}}{{x}_{2}}+...+{{a}_{2n}}{{x}_{n}}=0 \\ {} \\ {{a}_{m1}}{{x}_{1}}+{{a}_{m2}}{{x}_{2}}+..+{{a}_{mn}}{{x}_{n}}=0 \\ \end{matrix} \right. \\ & {{\frac{A}{b}}^{-1}}=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {} & {{a}_{1n}} \\ {{a}_{21}} & {{a}_{22}} & {} & {{a}_{2n}} \\ {} & {} & {} & {} \\ {{a}_{m1}} & {{a}_{m2}} & {} & {{a}_{mn}} \\ \end{matrix} \right]\cdot \left[ \begin{matrix} {{x}_{1}} \\ {{x}_{2}} \\ {} \\ {{x}_{n}} \\ \end{matrix} \right]\cdot \left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right] \\ & {{\frac{A}{b}}^{-1}}\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right]=0\left[ \begin{matrix} {{x}_{1}} \\ {{x}_{2}} \\ {} \\ {{x}_{n}} \\ \end{matrix} \right] \\ & {{\frac{A}{b}}^{-1}}=0 \\ \end{align}\ \)

Portanto, obtemos que \(\boxed{\frac{{{A^{ - 1}}}}{b} = 0}\).