Question

For what value of $\alpha $, the system of equations $
  \alpha x + 3y = \alpha - 3 \\
  12x + \alpha y = \alpha \\
 $ will have no solution.

Answer 1

Hint- Here, we will proceed by comparing the given pair of linear equations with any general pair of linear equations i.e., ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$. Then using the condition for having no solution i.e., $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$.

Complete Step-by-Step solution:
The given system of linear equations are $
  \alpha x + 3y = \alpha - 3 \\
   \Rightarrow \alpha x + 3y - \left( {\alpha - 3} \right) = 0{\text{ }} \to {\text{(1)}} \\
 $ and $
  12x + \alpha y = \alpha \\
   \Rightarrow 12x + \alpha y - \alpha = 0{\text{ }} \to {\text{(2)}} \\
 $
As we know that for any pair of linear equations ${a_1}x + {b_1}y + {c_1} = 0{\text{ }} \to {\text{(3)}}$ and ${a_2}x + {b_2}y + {c_2} = 0{\text{ }} \to {\text{(4)}}$ to have no solution (inconsistent solution), the condition which must be satisfied is that the ratio of the coefficients of x should be equal to the ratio of the coefficients of y which further should not be equal to the ratio of the constant terms in the pair of linear equations.
The condition is $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}{\text{ }} \to (5{\text{)}}$
By comparing equations (1) and (3), we get
${a_1} = \alpha ,{b_1} = 3,{c_1} = - \left( {\alpha - 3} \right)$
By comparing equations (2) and (4), we get
${a_2} = 12,{b_2} = \alpha ,{c_2} = - \alpha $
For the given pair of linear equations to have no solution, equation (5) must be satisfied
By equation (5), we can write
$
  \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}} \\
   \Rightarrow \dfrac{\alpha }{{12}} = \dfrac{3}{\alpha } \ne \dfrac{{ - \left( {\alpha - 3} \right)}}{{ - \alpha }}{\text{ }} \to {\text{(6)}} \\
 $
By equation (6), we can write
\[
   \Rightarrow \dfrac{\alpha }{{12}} = \dfrac{3}{\alpha } \\
   \Rightarrow {\alpha ^2} = 3 \times 12 = 36 \\
   \Rightarrow \alpha = \pm \sqrt {36} \\
   \Rightarrow \alpha = \pm 6 \\
 \]
Now, putting \[\alpha \]= 6 in equation (6), we have
$
   \Rightarrow \dfrac{6}{{12}} = \dfrac{3}{6} \ne \dfrac{{ - \left( {6 - 3} \right)}}{{ - 6}} \\
   \Rightarrow \dfrac{1}{2} = \dfrac{1}{2} \ne \dfrac{{ - 3}}{{ - 6}} \\
   \Rightarrow \dfrac{1}{2} = \dfrac{1}{2} \ne \dfrac{1}{2} \\
 $
which is not true, so \[\alpha \]= 6 is neglected.
Now, putting \[\alpha \] = -6 in equation (6), we have
$
   \Rightarrow \dfrac{{ - 6}}{{12}} = \dfrac{3}{{ - 6}} \ne \dfrac{{ - \left( { - 6 - 3} \right)}}{{ - \left( { - 6} \right)}} \\
   \Rightarrow \dfrac{{ - 1}}{2} = \dfrac{1}{{ - 2}} \ne \dfrac{9}{6} \\
   \Rightarrow - \dfrac{1}{2} = - \dfrac{1}{2} \ne \dfrac{3}{2} \\
 $
which is always true so we can say \[\alpha \] = -6
Therefore, the required value of \[\alpha \] is -6.

Note- A pair of linear equations which are given by ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ can also have unique solution (consistent solution) if the condition $\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}$ is satisfied. Also. For these pair of linear equations to have infinitely many solutions, the condition $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$ should always be satisfied.