Mathematics - Determinant Question with Solution | TestHub

For a homogeneous system of linear equations to have a non-trivial solution, the determinant of the coefficient matrix must be zero.$$\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^3& b^3& c^3\end{array}\right|=0$$ This is a special type of Vandermonde determinant. We can factor it by applying $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$.$$\left|\begin{array}{ccc}1& 0& 0\\ a& b-a& c-a\\ a^3& b^3-a^3& c^3-a^3\end{array}\right|=0$$ Expanding along the first row:$$(b-a)(c^3-a^3)-(c-a)(b^3-a^3)=0$$ $$(b-a)(c-a)(c^2+ca+a^2)-(c-a)(b-a)(b^2+ba+a^2)=0$$ $$(b-a)(c-a)[(c^2+ca+a^2)-(b^2+ba+a^2)]=0$$ $$(b-a)(c-a)[c^2-b^2+ca-ba]=0$$ $$(b-a)(c-a)[(c-b)(c+b)+a(c-b)]=0$$ $$(b-a)(c-a)(c-b)(a+b+c)=0$$ Since $a,b,c$ are distinct, $(b-a)\ne 0$, $(c-a)\ne 0$, and $(c-b)\ne 0$. Therefore, for the determinant to be zero, we must have $a+b+c=0$.