To obtain the condition for vanishing of the given determinant

Question

If $a,b,c$ are distinct real numbers obtain the condition under which the following determinant vanishes. $$\left| \begin{array}{cc} a & a^2 & 1+a^3\\ b & b^2 & 1+b^3\\ c & c^2 & 1+c^3\\ \end{array} \right|$$

My answer: After a little calculation I was able to show that $D=0$ reduces to $abc=-1$.

Is there a simpler one line answer to this? especially since this matrix looks suspiciously similar to the Vandermonde matrix?

Answer 1

This is NOT a one line solution you are expecting but it is an idea which is very useful in solving such determinants which may remind one of Vandermonde matrix.

I'm replacing $a$ by $x$ (for the sake of clarity). Using the linearity of determinants we get $$\det= \begin{vmatrix} x & x^2 & 1+x^3\\ b & b^2 & 1+b^3\\ c & c^2 & 1+c^3\\ \end{vmatrix} =\begin{vmatrix} x & x^2 & 1\\ b & b^2 & 1\\ c & c^2 & 1\\ \end{vmatrix}+\begin{vmatrix} x & x^2 & x^3\\ b & b^2 & b^3\\ c & c^2 & c^3\\ \end{vmatrix} =A(x)+B(x). $$ The determinant on the left can be thought of as a third degree polynomial in $x$. Let us call the first determinant (on right) as $A(x)$ (a second degree polynomial in $x$) and the second determinant (on right) as $B(x)$ (a third degree polynomial in $x$).

First consider $A(x)$. Observe that for $x=b$ or $x=c$, this determinant is $0$. Thus both $x-b$ and $x-c$ are factors of this polynomial. Thus $$A(x)=K(x-b)(x-c).$$ Moreover $$A(0)=bc(c-b) = K(bc).$$ Thus $K=c-b.$ This means $$A(x)=(c-b)(x-b)(x-c).$$

Likewise $$B(x)=bc(c-b)x(x-b)(x-c).$$ Thus the given determinant is $$\det=(c-b)(x-b)(x-c)[1+bcx].$$ Putting back everything in terms of $a$, we get $$\det=(c-b)(a-b)(a-c)[1+abc].$$ Now you get all the conditions when this can be $0$, namely $$a=b \quad \text{ or } \quad b=c \quad \text{ or } \quad c=a \quad \text{ or } \quad abc=-1.$$

Answer 2

There is a pretty easy way to obtain the determinant:

First, use the linearity of the determinant to split up the matrix: $$\det (A)= \begin{vmatrix} a & a^2 & 1+a^3\\ b & b^2 & 1+b^3\\ c & c^2 & 1+c^3\\ \end{vmatrix} =\begin{vmatrix} a & a^2 & 1\\ b & b^2 & 1\\ c & c^2 & 1\\ \end{vmatrix}+\begin{vmatrix} a & a^2 & a^3\\ b & b^2 & b^3\\ c & c^2 & c^3\\ \end{vmatrix}$$ Then, use $\det(A \cdot B) = \det(A) \cdot \det(B)$ to split up the right matrix; also, switch the colums in the left matrix. What you get are two identical Vandermonde matrices and a diagonal matrix: $$\det (A)=\begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \\ \end{vmatrix}+\begin{vmatrix} a & 0 & 0\\ 0 & b & 0\\ 0 & 0 & c\\ \end{vmatrix}\begin{vmatrix} 1 & a & a^2\\ 1 & b & b^2\\ 1 & c & c^2\\ \end{vmatrix}$$

Now, you can get the determinants with the Vandermonde formula and the formula for diagonal matrices:

$$\det (A)=(b-a)(c-a)(c-b)+abc((b-a)(c-a)(c-b))$$ $$=(b-a)(c-a)(c-b)(abc+1)$$

Now you can easily see when $\det (A)$ vanishes:

$$\det (A) = 0 \iff (b = a) \lor (c = a) \lor (c = b) \lor (abc = 1) $$

Answer 3

Let $V$ be the Vandermonde matrix in $a,b,c\;$ with the top row being $(1 \; a \; a^2).$ Your determinant is $E=\det(V'')+a b c \det (V)$ where $V''$ is obtained from $V$ by interchanging the first column of $V$ with the second, giving $V',$ and interchanging the second and third columns of $V',$ giving $V''.$ Therefore $\det (V'')=-\det (V')=-(-\det (V)=\det (V)$ and $E=(1+a b c)\det (V)=(1+a b c)(a-b)(b-c)(a-c).$