The task is to calculate the following determinant by using the properties of a determinant: $$\begin{vmatrix} n^2 & (n+1)^2 & (n+2)^2 \\ (n+1)^2 & (n+2)^2 & (n+3)^2 \\ (n+2)^2 & (n+3)^2 & (n+4)^2 \\ \end{vmatrix}$$ Hint: The result is $-8$.
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what are you looking for? – user123124 Jun 26 '15 at 14:03
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@Johan The way to calculate it. – Jun 26 '15 at 14:04
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If I had to do this, I would begin with the $2\times 2$ leading principal minor, which gives the other corners by change of variable, then appy Dodgson's method of condensation. – hardmath Jun 26 '15 at 14:05
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Wouldt it be dependent on n? – user123124 Jun 26 '15 at 14:05
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@Johan No, it isn't. I have checked it on Wolfram Alpha. – Jun 26 '15 at 14:09
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ah maybe it cancels out.. – user123124 Jun 26 '15 at 14:09
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Use the formula for the 3x3 determinant...I know the algebra is unpleasant, but at least you'll be able to calculate any variable determinant. All these tricks do, is teach you how to handle rare exceptions. For instance, the Dodgson's usually ok to use, but it becomes inefficient if there are zeros in the matrix. – Zach466920 Jun 26 '15 at 14:32
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@Zach466920 Using algebra will give you a solution, provided you are cautious of course, but I'm not entirely sure if that's the purpose of the exercise. – Jun 26 '15 at 14:44
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@JimVerr then the Op should specify if that's the case, because he says "using obviously the properties of a determinant"... – Zach466920 Jun 26 '15 at 14:46
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this problem has several answers at the link in my previous comment – Brian Fitzpatrick Jun 26 '15 at 17:54
1 Answers
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Expanding all the terms isn't so bad in this example.
ROWS-TRANSFORMATIONS $$A=\begin{pmatrix} n^2&n^2+2n+1&n^2+4n+4\\ n^2+2n+1&n^2+4n+4&n^2+6n+9\\ n^2+4n+4&n^2+6n+9&n^2+8n+16 \end{pmatrix}\sim \begin{pmatrix} n^2&n^2+2n+1&n^2+4n+4\\ n^2+2n+1&n^2+4n+4&n^2+6n+9\\ 2n+3&2n+5&2n+7 \end{pmatrix}\sim\begin{pmatrix} n^2&n^2+2n+1&n^2+4n+4\\ 2n+1&2n+3&2n+5\\ 2n+3&2n+5&2n+7 \end{pmatrix}\sim\begin{pmatrix} n^2&n^2+2n+1&n^2+4n+4\\ 2n+1&2n+3&2n+5\\ 2&2&2 \end{pmatrix}$$ COLUMN-TRANSFORMATIONS $$\begin{pmatrix} n^2&n^2+2n+1&n^2+4n+4\\ 2n+1&2n+3&2n+5\\ 2&2&2 \end{pmatrix}\sim\begin{pmatrix} n^2&n^2+2n+1&2n+3\\ 2n+1&2n+3&2\\ 2&2&0 \end{pmatrix}\sim\begin{pmatrix} n^2&2n+1&2n+3\\ 2n+1&2&2\\ 2&0&0 \end{pmatrix}\sim\begin{pmatrix} n^2&2n+1&2\\ 2n+1&2&0\\ 2&0&0 \end{pmatrix}$$ So $\mathrm{det}(A)=-8$.
user246336
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1I calculated it using row operations as in your example more or less but I was hoping if I could get a more neat answer by adding a column to another or something like that, but your way might be the only one in the end. Thanks for the effort. – Jun 26 '15 at 14:42