Show that

$$\det \begin{bmatrix} n^2 & (n+1)^2 & (n+2)^2 \\ (n+1)^2 & (n+2)^2 & (n+3)^2 \\ (n+2)^2 & (n+3)^2& (n+4)^2\\ \end{bmatrix} = -8$$

I'm pretty sure I'll have to use row/ column operations to simplify the matrix, but I absolutely don't know where to start. Even a hint will be useful, thanks!

(The field wasn't specified, but I'm assuming it's $\mathbb{R}$? This was an exercise for my linear algebra class 2 years ago and I still don't know how to prove this.)


Let $r_k$ and $c_k$, with $k\in\{1,2,3\}$, the $k$-th row and column of the involved matrix.

  1. Perform $r_3\leftarrow r_3-2r_2+r_1$
  2. Perform $r_2\leftarrow r_2-r_1$

The determinant in unchanged, and it equals: $$\det\begin{pmatrix}n^2&(n+1)^2&(n+2)^2\\ 2n+1 & 2n+3 & 2n+5 \\ 2 & 2 & 2\end{pmatrix}=\det\begin{pmatrix}2n^2&2(n+1)^2&2(n+2)^2\\ 2n+1 & 2n+3 & 2n+5 \\ 1 & 1 & 1\end{pmatrix}$$ or: $$\det\begin{pmatrix}-n&n+2&3n+8\\ 2n+1 & 2n+3 & 2n+5 \\ 1 & 1 & 1\end{pmatrix}=\det\begin{pmatrix}-n&2n+2&2n+6\\ 2n+1 & 2 & 2 \\ 1 & 0 & 0\end{pmatrix}$$ that is $(2n+2)\cdot 2-(2n+6)\cdot 2=\color{red}{-8}$ as wanted.


You may play with various instances of the identity $$ (k + 1)^{2} - k^{2} = 2 k + 1 $$ until the matrices collapses to a form that is easy to calculate.


I think this is a question from an entrance exam!!!

So to solve this type of matrix equation in a short span of time, In these type of equation the value of value of n doesn't matter


Substitute $n=0$, we get, $$det\left[ \begin{array}{ccc} 0& (1)^2 & (2)^2 \\ (1)^2 & (2)^2 & (3)^2 \\ (2)^2 & (3)^2& (4)^2\\ \end{array} \right]=-8$$


Substitute $n=1$,we get, $$det\left[ \begin{array}{ccc} 1^2& 2^2 & 3^2 \\ 2^2 & 3^2 & 4^2 \\ 3^2 & 4^2& 5^2\\ \end{array} \right]=-8$$

In both the case you will get -8 as answer!!!!

Now check with n=2,3,4,5,6,7.........

If my answer has worth for you then give me positive feedback 👍

  • $\begingroup$ This would be an alright answer if the OP has stated that this is for multiple choice problem where all the answer choices are simply constants. However, without knowing that the determinant of the matrix is constant ahead of time, this answer doesn't show that the determinant of the matrix is $-8$. $\endgroup$ – benguin Mar 8 '17 at 18:24
  • $\begingroup$ Ok. Thanks for the Negativity !!! $\endgroup$ – Creepy Creature Mar 8 '17 at 19:15
  • $\begingroup$ It has nothing to do with negativity and everything to do with providing correct and accurate information (especially on a site dedicated to mathematics). If I see incorrect or inaccurate information, then (1) downvote and (2) provide a comment giving the reason why I downvoted so that the answerer has the feed necessary to provide higher quality answers in the future and/or to revise their answer to a quality where I can change my vote. As your answer stands right now, the OP might incorrectly think that your method is a valid way of finding the determinant. $\endgroup$ – benguin Mar 8 '17 at 21:14
  • $\begingroup$ If you revise your answer to state that your method works in the case that it is known ahead of time that the determinant is constant (for example if all the options on a multiple choice test are all constants), then your answer will be accurate and correct and I will change my vote accordingly. $\endgroup$ – benguin Mar 8 '17 at 21:16

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