A determinant problem

Question

If $f(n)=\alpha^n+\beta^n$ and

$$A=\left| \begin{array}{ccc} 3 & 1+f(1) & 1+f(2) \\ 1+f(1) & 1+f(2) & 1+f(3) \\ 1+f(2) & 1+f(3) & 1+f(4) \end{array} \right|$$

$=k(1-\alpha)^2(1-\beta)^2(\alpha-\beta)^2$

then $k=$ $a) 1\:\:\:$ $b)-1\:\:\:$ $c) \alpha\beta\:\:\:$ $d)\alpha\beta\gamma$

I have done the sum, but an answer isn't provided, so please see if I'm correct.

$$A=\left| \begin{array}{ccc} 3 & 1+\alpha+\beta & 1+\alpha^2+\beta^2 \\ 1+\alpha+\beta & 1+\alpha^2+\beta^2 & 1+\alpha^3+\beta^3 \\ 1+\alpha^2+\beta^2 & 1+\alpha^3+\beta^3 & 1+\alpha^4+\beta^4 \end{array} \right|$$

$=\left| \begin{array}{ccc} 1 & 1 & 1+\alpha^2+\beta^2 \\ 1 & 1 & 1+\alpha^3+\beta^3 \\ 1 & 1 & 1+\alpha^4+\beta^4 \end{array} \right|$

$+\left| \begin{array}{ccc} 1 & \alpha & 1+\alpha^2+\beta^2 \\ 1 & \alpha^2 & 1+\alpha^3+\beta^3 \\ 1 & \alpha^3 & 1+\alpha^4+\beta^4 \end{array} \right|$

$+ \left| \begin{array}{ccc} 1 & \beta & 1+\alpha^2+\beta^2 \\ 1 & \beta^2 & 1+\alpha^3+\beta^3 \\ 1 & \beta^3 & 1+\alpha^4+\beta^4 \end{array} \right|$

$+ \left| \begin{array}{ccc} 1 & 1 & 1+\alpha^2+\beta^2 \\ \alpha & 1 & 1+\alpha^3+\beta^3 \\ \alpha^2 & 1 & 1+\alpha^4+\beta^4 \end{array} \right|$

$+ \left| \begin{array}{ccc} 1 & \alpha & 1+\alpha^2+\beta^2 \\ \alpha & \alpha^2 & 1+\alpha^3+\beta^3 \\ \alpha^2 & \alpha^3 & 1+\alpha^4+\beta^4 \end{array} \right|$

$+ \left| \begin{array}{ccc} 1 & \beta & 1+\alpha^2+\beta^2 \\ \alpha & \beta^2 & 1+\alpha^3+\beta^3 \\ \alpha^2 & \beta^3 & 1+\alpha^4+\beta^4 \end{array} \right|$

$+ \left| \begin{array}{ccc} 1 & 1 & 1+\alpha^2+\beta^2 \\ \beta & 1 & 1+\alpha^3+\beta^3 \\ \beta^2 & 1 & 1+\alpha^4+\beta^4 \end{array} \right|$

$+ \left| \begin{array}{ccc} 1 & \alpha & 1+\alpha^2+\beta^2 \\ \beta & \alpha^2 & 1+\alpha^3+\beta^3 \\ \beta^2 & \alpha^3 & 1+\alpha^4+\beta^4 \end{array} \right|$

$+ \left| \begin{array}{ccc} 1 & \beta & 1+\alpha^2+\beta^2 \\ \beta & \beta^2 & 1+\alpha^3+\beta^3 \\ \beta^2 & \beta^3 & 1+\alpha^4+\beta^4 \end{array} \right|$

Now, $\left| \begin{array}{ccc} 1 & \beta & 1+\alpha^2+\beta^2 \\ \beta & \beta^2 & 1+\alpha^3+\beta^3 \\ \beta^2 & \beta^3 & 1+\alpha^4+\beta^4 \end{array} \right|=0$

$\left| \begin{array}{ccc} 1 & 1 & 1+\alpha^2+\beta^2 \\ 1 & 1 & 1+\alpha^3+\beta^3 \\ 1 & 1 & 1+\alpha^4+\beta^4 \end{array} \right|=0$

$ \left| \begin{array}{ccc} 1 & \alpha & 1+\alpha^2+\beta^2 \\ \alpha & \alpha^2 & 1+\alpha^3+\beta^3 \\ \alpha^2 & \alpha^3 & 1+\alpha^4+\beta^4 \end{array} \right|=0$

Putting that and taking common, we ultimately get:

$A= (\alpha\beta^2-\beta\alpha^2-\beta^2+\beta+\alpha^2-\alpha) \cdot B$

where $B= \left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & \alpha^2 & \beta^2 \end{array} \right|$ $=-(1-\alpha)(1-\beta)(\alpha-\beta)$

So finally, factorizing, $A= (1-\alpha)^2 (1-\beta)^2 (\alpha-\beta)^2$

So, $k=1$

Am I correct?

Answer 1

Let $g(x) = (x-1)(x-\alpha)(x-\beta)$, a monic polynomial with the roots $1, \alpha, \beta$. Then the Vandermonde matrix of roots to g is

$$M = \left(\begin{array}{ccc}1 & 1 & 1\\1 & \alpha & \alpha^2\\ 1 & \beta & \beta^2\end{array}\right)$$

with the known determinant being product of differences of roots $(1-\alpha)(1-\beta)(\alpha-\beta)$. By considering the matrix $M^t M$ you will get the matrix of power sums of the roots, which is the matrix you want to compute the determinant of, and $\det(M^t M) = (\det(M))^2 = (1-\alpha)^2(1-\beta)^2(\alpha-\beta)^2$.

This generalizes the result to larger matrices too.

Answer 2

Yes, you are correct. But assuming that one of the given options is correct (which is safe to do in, say, an exam), it would have been enough to test for a properly chosen pair, say $\alpha=0,\beta=-1$. Then the matrix is

$$\left(\begin{array}{ccc}3 & 0 & 2\\ 0 & 2 & 0\\ 2 & 0 & 2\end{array}\right)$$

and its determinant is easily computed to be $4$. The only answer which fits is therefore (a), $k=1$.