Step 1
Given Matrix (A) and Vector ($\bf{b}$):
\[\begin{array}{l}{\bf{b}} = \left[ {\begin{array}{*{20}{c}}{{b_1}}\\{{b_2}}\end{array}} \right]\\\\A = \left[ {\begin{array}{*{20}{c}}2&{ - 1}\\{ - 6}&3\end{array}} \right]\end{array}\]We have to prove that the the Equation $A{\bf{x}} = {\bf{b}}$ does not have a solution for all possible values of $\bf{b}$. Also we have to find the values for which the solution is possible.

Step 2: System of Equations
\[\left[ {\begin{array}{*{20}{c}}2&{ - 1}\\{ - 6}&3\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{b_1}}\\{{b_2}}\end{array}} \right]\]

Step 3: The Augmented Form

\[M = \left[ {\begin{array}{*{20}{c}}2&{ - 1}&{{b_1}}\\{ - 6}&3&{{b_2}}\end{array}} \right]\]

Step 4: The Echelon Form

\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}2&{ - 1}&{{b_1}}\\{ - 6}&3&{{b_2}}\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}2&{ - 1}&{{b_1}}\\0&0&{{b_2} + 3{b_1}}\end{array}} \right]\,\,\,\,::\{ {R_3} = {R_3} + 2{R_1}\} \end{array}\]For the matrix shown above, if the last row is completely zero then the system has infinite solutions. Otherwise, the equation has no solution.

Thus the equation does not always have a solution.

Step 5: Possible Solution

To have a poss solution last row should be completely zero. \[\begin{array}{l}{b_2} + 3{b_1} = 0\\\\{b_2} = - 3{b_1}\end{array}\]

ANSWERS
Possible Solution only if $${b_2} = - 3{b_1}$$