# How to find the determinant of this $3 \times 3$ Hankel matrix?

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?

• Have you tried to utilize the determinant's variability to certain line/column actions? Feb 14 '14 at 17:53
• Won't I have to expand the $^{2}$ in order to do that? Feb 14 '14 at 17:55
• Am I allowed to assign a number to n? Feb 14 '14 at 17:57
• You're not allowed to do that, no, unless you've proven that it's constant. And regarding your earlier question, find a way to use the power of $2$. Feb 14 '14 at 17:58

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 !

• The book or THE book? Feb 14 '14 at 21:03
• This is an excellent gimmick, worth remembering when you have some combinatorial identity to prove. Feb 14 '14 at 21:04
• Very nice idea! Feb 14 '14 at 21:17
• I might be overlooking something obvious here, but is there a "nice" way to extend this idea to "check" if $\det(A) = -8$? (i.e. not just to prove this is constant, but also to prove the value holds?) Feb 15 '14 at 3:56
• @anorton -8 is the value you get when you plug in one of the numbers Feb 15 '14 at 9:46

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.

That's not the nicest way to do it but it's quite simple and works.

To avoid developping the squares, I used $a^2-b^2 =(a+b)(a-b)$ (and $a-b$ was always $1$ when I used it). Then, once I had enough $0$s, I simply exapended by the last column.

$$\begin{array}{l} \begin{vmatrix} n^{2} & (n+1)^{2} &(n+2)^{2} \\ (n+1)^{2} &(n+2)^{2} & (n+3)^{2}\\ (n+2)^{2} & (n+3)^{2} & (n+4)^{2} \end{vmatrix} &= \begin{vmatrix} n^{2} & (n+1)^{2} &(n+2)^{2}-(n+1)^{2} \\ (n+1)^{2} &(n+2)^{2} & (n+3)^{2}-(n+2)^{2}\\ (n+2)^{2} & (n+3)^{2} & (n+4)^{2}-(n+3)^{2} \end{vmatrix}\\ &= \begin{vmatrix} n^{2} & (n+1)^{2} &2n+3 \\ (n+1)^{2} &(n+2)^{2} & 2n+5\\ (n+2)^{2} & (n+3)^{2} & 2n+7 \end{vmatrix}\\ &= \begin{vmatrix} n^{2} & (n+1)^{2}-n^{2} &2n+3 \\ (n+1)^{2} &(n+2)^{2}-(n+1)^{2} & 2n+5\\ (n+2)^{2} & (n+3)^{2}-(n+2)^{2} & 2n+7 \end{vmatrix}\\ &= \begin{vmatrix} n^{2} & 2n+1 &2n+3 \\ (n+1)^{2} &2n+3 & 2n+5\\ (n+2)^{2} & 2n+5 & 2n+7 \end{vmatrix}\\ &= \begin{vmatrix} n^{2} & 2n+1 &2n+3 \\ (n+1)^{2} &2n+3 & 2n+5\\ (n+2)^{2}-(n+1)^{2} & 2n+5-(2n+3) & 2n+7-(2n+5) \end{vmatrix}\\ &= \begin{vmatrix} n^{2} & 2n+1 &2n+3 \\ (n+1)^{2} &2n+3 & 2n+5\\ 2n+3 & 2 & 2 \end{vmatrix}\\ &= \begin{vmatrix} n^{2} & 2n+1 &2n+3 \\ (n+1)^{2}-n^{2} &2n+3-(2n+1) & 2n+5-(2n+3)\\ 2n+3 & 2 & 2 \end{vmatrix}\\ &= \begin{vmatrix} n^{2} & 2n+1 &2n+3 \\ 2n+1 &2 & 2\\ 2n+3 & 2 & 2 \end{vmatrix}\\ &= \begin{vmatrix} n^{2} & 2n+1 &2n+3-(2n+1) \\ 2n+1 &2 & 2-2\\ 2n+3 & 2 & 2-2 \end{vmatrix}\\ &= \begin{vmatrix} n^{2} & 2n+1 &2 \\ 2n+1 &2 & 0\\ 2n+3 & 2 & 0 \end{vmatrix}\\ &= \begin{vmatrix} n^{2} & 2n+1 &2 \\ 2n+1 &2 & 0\\ 2n+3-(2n+1) & 2-2 & 0-0 \end{vmatrix}\\ &= \begin{vmatrix} n^{2} & 2n+1 &2 \\ 2n+1 &2 & 0\\ 2 & 0 & 0 \end{vmatrix}\\ &= 2\begin{vmatrix} 2n+1 &2\\ 2 & 0 \end{vmatrix}\\ &= -8 \end{array}$$

• how come your first equation holds? Feb 14 '14 at 18:25
• column 3 <- column 3 - column 2 Feb 14 '14 at 18:42
• Thanks , i can't believe i didn't see that. Feb 14 '14 at 19:55
• I don't know if this would fit in the box allocated to the question on the exam :p Feb 14 '14 at 21:47
• Think outside the box. Feb 15 '14 at 9:36