Mathematics - Determinant Question with Solution | TestHub

The system of linear 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 infinitely many solutions, 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 $D_3=\left|\begin{array}{ccc}{a}_1& b_1& d_1\\ a_2& b_2& d_2\\ a_3& b_3& d_3\end{array}\right|=0.$

Hence, for the system of equations $x+y+z=5, x+2y+3z=9$ and $x+3y+\alpha z=\beta$ have infinitely many solution, if

$D=\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 3& \alpha \end{array}\right|=0$

$\Rightarrow \left(2\alpha -9\right)-\left(\alpha -3\right)+\left(3-2\right)$

$\Rightarrow 2\alpha -9-\alpha +3+1=0$

$\Rightarrow \alpha =5.$

And, $D_1=\left|\begin{array}{ccc}5& 1& 1\\ 9& 2& 3\\ \beta & 3& \alpha \end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}5& 1& 1\\ 9& 2& 3\\ \beta & 3& 5\end{array}\right|=0$

$\Rightarrow 5\left(10-9\right)-\left(45-3\beta \right)+\left(27-2\beta \right)=0$

$\Rightarrow 5-45+3\beta +27-2\beta =0$

$\Rightarrow \beta =13.$

Hence, $\beta -\alpha =13-5=8.$