Question: 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?

Answer: 
Recall that $a^2-b^2=(a+b)(a-b)$. Subtracting $\operatorname{Row}\_1$ from $\operatorname{Row}\_2$ and from $\operatorname{Row}\_3$ gives 
$$
\begin{bmatrix}
n^2 & (n+1)^2 & (n+2)^2 \\
2n+1 & 2n+3 & 2n+5 \\
4n+4 & 4n+8 & 4n+12
\end{bmatrix}
$$
Then subtracting $2\cdot\operatorname{Row}\_2$ from $\operatorname{Row}\_3$ gives
$$
\begin{bmatrix}
n^2 & (n+1)^2 & (n+2)^2 \\
2n+1 & 2n+3 & 2n+5 \\
2 & 2 & 2
\end{bmatrix}
$$
Now, subtracting $\operatorname{Col}\_1$ from $\operatorname{Col}\_2$ and $\operatorname{Col}\_3$ gives
$$
\begin{bmatrix}
n^2 & 2n+1 & 4n+4 \\
2n+1 & 2 & 4 \\
2 & 0 & 0
\end{bmatrix}
$$
Finally, subtracting $2\cdot\operatorname{Col}\_2$ from $\operatorname{Col}\_3$ gives 
$$
\begin{bmatrix}
n^2 & 2n+1 & 2 \\
2n+1 & 2 & 0 \\
2 & 0 & 0
\end{bmatrix}
$$
Expanding the determinant about $\operatorname{Row}\_3$ gives
$$
\det A
=
2\cdot\det
\begin{bmatrix}
2n+1 & 2\\
2 & 0
\end{bmatrix}
=2\cdot(-4)=-8
$$
as advertised.
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 !