Mathematics - Determinant Question with Solution | TestHub

The system of equations $a_{1}x+b_{1}y+c_{1}z+d_1=0, a_{2}x+b_{2}y+c_{2}z+d_2=0$ and $a_{3}x+b_{3}y+c_{3}z+d_3=0$ have a infinitely many solution, if

$D=\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|=0, D_1=\left|\begin{array}{ccc}{d}_1& b_1& c_1\\ d_2& b_2& c_2\\ d_3& b_3& c_3\end{array}\right|=0, D_2=\left|\begin{array}{ccc}{a}_1& d_1& c_1\\ a_2& d_2& c_2\\ a_3& d_3& c_3\end{array}\right|=0$ and $D_3=\left|\begin{array}{ccc}{a}_1& b_1& d_1\\ a_2& b_2& d_2\\ a_3& b_3& d_3\end{array}\right|=0.$

Thus, for infinitely many solutions, we have $D=0$

$\Rightarrow \left|\begin{array}{ccc}1& 1& 1\\ 4& \lambda & -\lambda \\ 3& 2& -4\end{array}\right|=0$

$\Rightarrow 1\left(-4\lambda +2\lambda \right)-1\left(-16+3\lambda \right)+1\left(8-3\lambda \right)=0$

$\Rightarrow -2\lambda +16-3\lambda +8-3\lambda =0$

$\Rightarrow 24-3\lambda =0$

$\Rightarrow \lambda =3$

And, $\lambda =3$ satisfies the equation ${\lambda }^2-\lambda -6=0.$