Find the values of $x$ which makes $\det (A)=0$ without expending determinant Find the values of $x$ which makes $\det(A)=0$ without expending determinant: Let $A$ :
$$\begin{bmatrix}1 & -1 & x \\2 & 1 & x^2\\ 4 & -1 & x^3  \end{bmatrix} $$
How can I solve this?
 A: Solution. $x=0$ or $-1$ or $2$ - The determinant of a Vandermonde matrix $A=(a_j^{i})$ vanishes iff $a_j=a_k$, for some $j\ne k$. So the first columns are powers of $2$, the second, powers of $-1$, and clearly, for $x=0$, vanishes the third colunm.
A: Hint: The determinant of a matrix is $0$ if and only if its rows are not linearly independent. So, you need to find $x$ so that there is no non-trivial solution the system of equations
$$
a\langle1,-1,x\rangle+b\langle2,1,x^2\rangle+c\langle4,-1,x^3\rangle=\langle0,0,0\rangle.
$$
(Note that you could also use the column space for this; however, I think that the solution will be faster this way.)
Based on the first and second coordinates, you can get necessary conditions on $a,b,c$ in order for there to be solutions; then leverage this to find information about $x$.
A: For example, if x=-1 det(A)=0, because the 2nd and 3d columns will be the same.
A: Directly from definition (or from Laplace expansion) you can see that this determinant is a polynomial of degree (at most) 3 in variable $x$.
If you know 3 roots of this polynomial (and the roots $x=0,-1,2$ are relatively easy to guess), then you know that there are no other solutions. (Third degree polynomial can have at most 3 real roots.)
