A) no solution
B) unique solution
C) infinitely many solution
D) None of these
Correct Answer: B
Solution :
[b] Given, \[ax+3y=1\]and \[-12x+ay=2\] |
Here, \[{{a}_{1}}=a,\,\,\,{{b}_{1}}=3,\,\,{{c}_{1}}=1\] |
and \[{{a}_{2}}=-12,\,{{b}_{2}}=a,\,\,{{c}_{2}}=2\] |
Now, \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{a}{-12},\,\,\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{3}{a}\]and \[\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{1}{2}\] |
Condition for unique solution, |
\[\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}\Rightarrow \frac{a}{-12}\ne \frac{3}{a}\Rightarrow {{a}^{2}}\ne -36\] |
So, the given equations has unique solution for all real values of a. |
You need to login to perform this action.
You will be redirected in
3 sec