Today, at my linear algebra exam, there was this question that I couldn't solve.
Prove that
$$\det \begin{bmatrix}
n^{2} & (n+1)^{2} &(n+2)^{2} \\
(n+1)^{2} &(n+2)^{2} & (n+3)^{2}\\
(n+2)^{2} & (n+3)^{2} & (n+4)^{2}
\end{bmatrix} = -8$$
Clearly, calculating the determinant, with the matrix as it is, wasn't the right way. The calculations went on and on. But I couldn't think of any other way to solve it.
Is there any way to simplify $A$, so as to calculate the determinant?
Best Answer
Here is a proof that is decidedly not from the book. The determinant is obviously a polynomial in n of degree at most 6. Therefore, to prove it is constant, you need only plug in 7 values. In fact, -4, -3, ..., 0 are easy to calculate, so you only have to drudge through 1 and 2 to do it this way !