Mathematics - Determinant Question with Solution | TestHub

For the system of equations $a_{1}x+b_{1}y+c_{1}z=d_1, a_{2}x+b_{2}y+c_{2}z=d_2$ and $a_{3}x+b_{3}y+c_{3}z=d_3,$

We have $D=\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|, D_x=\left|\begin{array}{ccc}{d}_1& b_1& c_1\\ d_2& b_2& c_2\\ d_3& b_3& c_3\end{array}\right|, D_y=\left|\begin{array}{ccc}{a}_1& d_1& c_1\\ a_2& d_2& c_2\\ a_3& d_3& c_3\end{array}\right|$ and $D_z=\left|\begin{array}{ccc}{a}_1& b_1& d_1\\ a_2& b_2& d_2\\ a_3& b_3& d_3\end{array}\right|.$

And, the system has no solution, if $D=0$ and atleast one of $D_x, D_y \& D_z$ is non-zero.

Thus, for the given system of equations,

$x+y+z=6,$ $3x+5y+5z=26$ and $\mathrm{x}+2\mathrm{y}+\lambda \mathrm{z}=\mu$

We have $D=\left|\begin{array}{ccc}1& 1& 1\\ 3& 5& 5\\ 1& 2& \lambda \end{array}\right|=0$

$\Rightarrow 1\left(5\lambda -10\right)-1\left(3\lambda -5\right)+1\left(6-5\right)=0$

$\Rightarrow 5\lambda -10-3\lambda +5+1=0$

$\Rightarrow 2\lambda =4$

$\Rightarrow \lambda =2.$

And, $D_z=\left|\begin{array}{ccc}1& 1& 6\\ 3& 5& 26\\ 1& 2& \mu \end{array}\right|\ne 0$

$\Rightarrow 1\left(5\mu -52\right)-1\left(3\mu -26\right)+6\left(6-5\right)\ne 0$

$\Rightarrow 5\mu -52-3\mu +26+6\ne 0$

$\Rightarrow 2\mu \ne 20$

$\Rightarrow \mu \ne 10.$

$\therefore$ For no solution $\lambda =2$ and $\mu \ne 10.$