Question

Find the values of $a$ for which the following system of equations has exactly one solution $2x+ay=11,ax+8y=21$

Answer 1

Here $a_1=2,a_2=a,b_1=a$ and $b_2=8$

Now $a_1{b}_2-a_2{b}_1=16-a^2$ .

For the given system to have exactly one solution $16-a^2\ne 0\Rightarrow a^2\ne 16\Rightarrow a\ne \pm 4$

$\therefore a$ can be any real number except $\pm 4$.