Mathematics - Determinant Question with Solution | TestHub

For the system to have no solution, the determinant of the coefficient matrix must be zero, and at least one of $D_x,D_y,D_z$ must be non-zero. The determinant of the coefficient matrix is $D=\left|\begin{array}{ccc}1& 1& 1\\ 1& k+1& 1\\ 1& 1& k+1\end{array}\right|$. Applying $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$, we get $D=\left|\begin{array}{ccc}1& 1& 1\\ 0& k& 0\\ 0& 0& k\end{array}\right|=1\cdot (k\cdot k-0\cdot 0)=k^2$. For no solution, $D=0$, which implies $k^2=0⟹k=0$. If $k=0$, the system becomes:$x+y+z=1$ $x+y+z=2$ $x+y+z=3$ This system is clearly inconsistent as $x+y+z$ cannot simultaneously equal 1, 2, and 3. Thus, for $k=0$, the system has no solution.