Show the value of determinant is 0

Question

I dont have any idea how to show the value of the following determinant is 0 without expanding the determinants

$$ \begin{vmatrix} 1 & a & a^2-bc \\ 1 & b & b^2-ca \\ 1 & c & c^2-ab \\ \end{vmatrix} $$

Any ideas?

Answer 1

Subtract the first row from each of the other rows: $$ \begin{vmatrix} 1 & a & a^2-bc \\ 1 & b & b^2-ca \\ 1 & c & c^2-ab \\ \end{vmatrix} =\begin{vmatrix} 1 & a & a^2-bc \\ 0 & b-a & (b-a)(b+a+c) \\ 0 & c-a & (c-a)(c+a+b) \\ \end{vmatrix}\\ $$ Note that the second and third rows are both multiples of $\begin{bmatrix}0&1&a+b+c\end{bmatrix}$; therefore, they are linearly dependent.

Answer 2

Brother in questions like these (in which you see 1 in all entries on same row or columm), you have to subtract proper rows so as to form couple of zeroes so that you can easily expand determinant using that couple of zeroes you formed( in either that row or columm wise). in your case you need to subtract row 1st from row 2 and row 2 from row 3rd and then expand determinant along columm 1.

Answer 3

$\Delta=\begin{vmatrix} 1 & a & a^2-bc \\ 1 & b & b^2-ca \\ 1 & c & c^2-ab \\ \end{vmatrix}=\begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\\ \end{vmatrix}-\begin{vmatrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \\ \end{vmatrix}=\Delta_1-\Delta_2$, say.

Now, $\Delta_2=\begin{vmatrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \\ \end{vmatrix}=\frac{1}{abc}\begin{vmatrix} a & a^2 & abc \\ b & b^2 & abc \\ c & c^2 & abc \\ \end{vmatrix}$ $\quad(a,b,c\ne0)$

$\qquad\qquad=\begin{vmatrix} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \\ \end{vmatrix}=-\begin{vmatrix} a & 1 & a^2 \\ b & 1 & b^2 \\ c & 1 & c^2 \\ \end{vmatrix}=\begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\\ \end{vmatrix}=\Delta_1$

The properties used are self-explanatory.

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Alternatively, to show $\Delta_1=\Delta_2$ we can use the following argument:

Let $f(a,b,c)=\Delta_2-\Delta_1$

Then, $f(a,b,c)=k(a-b)(b-c)(a-c)$ for some constant $k$ as $f(b,b,c)=f(a,b,a)=f(a,b,b)=0$.

The coefficient of $a^2b$ in $f$ is $0\Rightarrow k=0.$