Can $x$ take every value here?

Question

To find $x$ such that this determinant is zero, a,b,c are distinct and real constants. $$ \begin{vmatrix} x-a & a^2 & a^3 \\ x-b & b^2 & b^3 \\ x-c & c^2 & c^3 \\ \end{vmatrix} =0$$

I get that it can be done by row operations and further simplification. But I don't understand why my 300 IQ move does not get the right answer.

$$ \begin{vmatrix} \frac xa-1 & a^3 & a^3 \\ \frac xb-1 & b^3 & b^3 \\ \frac xc-1 & c^3 & c^3 \\ \end{vmatrix} $$

Two columns now being the same implies the determinant is zero. So x can take any number ??

You might have to explain what this operation is, and why you think the determinant below is related to the determinant above. — Theo Bendit, Nov 03 '21 at 08:04
@TheoBendit I took "a,b,c" from one column and gave it to another — Linkin, Nov 03 '21 at 08:05
Btw, there is a simple way to sort out such a problem: Choose concrete values for $a, b, c$ and evaluate both determinants. Then compare the results. — Martin R, Nov 03 '21 at 08:07
@Linkin That's perfectly fine for one constant. You could divide the first column by $a$, and multiply the second by $a$, but you can't do this for $3$ different constants, selectively by row, all at the same time. — Theo Bendit, Nov 03 '21 at 08:08
I still don't understand your operation. If we look at the first row, it seems like you have divided the first element and multiplied the rest of the elements by $a$. — Matti P., Nov 03 '21 at 08:08

score 2 · Accepted Answer · answered Nov 03 '21 at 08:18

There is no row operation that can get you from

$$ \begin{vmatrix} x-a & a^2 & a^3 \\ x-b & b^2 & b^3 \\ x-c & c^2 & c^3 \\ \end{vmatrix} $$

To

$$ \begin{vmatrix} \frac xa-1 & a^3 & a^3 \\ \frac xb-1 & b^3 & b^3 \\ \frac xc-1 & c^3 & c^3 \\ \end{vmatrix}. $$

For one, the first determinant is defined for all values $a,b,c$, while the second one is only defined for nonzero values of $a,b,c$, so the two expression are clearly not equal.

But even on regions where both expressions are defined, they are not equal. You can verify this by plugging in $x=0, a=1, b=2, c=3$ in which case you get

$$ \begin{vmatrix} x-a & a^2 & a^3 \\ x-b & b^2 & b^3 \\ x-c & c^2 & c^3 \\ \end{vmatrix} = \begin{vmatrix} -1 & 1 & 1 \\ -2 & 4 & 8 \\ -3 & 9 & 27 \\ \end{vmatrix} = -12 $$

and

$$ \begin{vmatrix} \frac xa-1 & a^3 & a^3 \\ \frac xb-1 & b^3 & b^3 \\ \frac xc-1 & c^3 & c^3 \\ \end{vmatrix} = 0 $$

and since $0\neq 12$, the two determinants are clearly not the same.

I don't know what "rule" you used to get from one expression to the other, but what I can tell you is that you either did not use the rule correctly, or you used a rule that is not really a rule.

Can $x$ take every value here?

1 Answers1