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This question already has an answer here:

I'm stuck with this one - Any tips?

The Problem:

Let $n \in \mathbb{N}.$ The following $n \times n$ matrix:

$$A = \left( \begin{array}{ccc} 2 & 1 & & & & ...\\ 1 & 2 & 1 & & & ...\\ & 1 & 2 & 1 & & ...\\ & & 1 & 2 & 1 & ...\\ & & & 1 & ... & 1\\ ... & ... & ... & ... & 1 &2 \end{array} \right) $$

e.g. for the main diagonal = 2, the sub- and superdiagonal = 1 .

Show with Induction that $\det(A) = n + 1$.

My solution approach: Laplace Expansion starting with the 2 in the bottom right corner $(a_{n+1,n+1})$. But how can I tell wether its positive or negative? After that I'm stuck with the 1 $(a_{n,n+1})$(the sub matrix matrix becomes ugly and I get a recursively solution). How can I formalize this in a proper manner?

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marked as duplicate by Daniel Fischer Feb 15 '15 at 11:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ A hint for your first problem: is $(n+1)+(n+1)$ odd or even? $\endgroup$ – mdp Nov 18 '13 at 10:09
  • $\begingroup$ Second hint: Why not apply Laplace expansion again to your second submatrix? $\endgroup$ – Casteels Nov 18 '13 at 10:13
  • $\begingroup$ I've done that. Thats why I get a recrusive solution. $\endgroup$ – user109498 Nov 18 '13 at 10:16
  • $\begingroup$ I think you get the $(n-1)\times(n-1)$ version of $A$, for which you know the determinant by induction. $\endgroup$ – Casteels Nov 18 '13 at 10:19
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Why not develop directly wrt the first column? The subindex means the order of the square matrix:

$$\begin{vmatrix}2 & 1 & & & & ...\\ 1 & 2 & 1 & & & ...\\ & 1 & 2 & 1 & & ...\\ & & 1 & 2 & 1 & ...\\ & & & 1 & ... & 1\\ ... & ... & ... & ... & 1 &2\end{vmatrix}_n=2\begin{vmatrix}2 & 1 & & & & ...\\ 1 & 2 & 1 & & & ...\\ & 1 & 2 & 1 & & ...\\ & & 1 & 2 & 1 & ...\\ & & & 1 & ... & 1\\ ... & ... & ... & ... & 1 &2\end{vmatrix}_{n-1}-$$

$$-\begin{vmatrix}1 & 0 & & & & ...\\ 1 & 2 & 1 & & & ...\\ & 1 & 2 & 1 & & ...\\ & & 1 & 2 & 1 & ...\\ & & & 1 & ... & 1\\ ... & ... & ... & ... & 1 &2\end{vmatrix}_{n-1}\stackrel{\text{Ind. Hyp.}}=2\left[(n-1)+1\right]-\left[(n-2)+1\right]=$$

$$=2n-n+1=n+1$$

as the last determinant is easily developed wrt the first entry and we get a $\;(n-2)\times(n-2)\;$ copy of the original matrix!

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    $\begingroup$ ''Why not develop directly wrt the first column?'' I'm feeling so stupid, it's not even funny. Such a simple task and I flop top notch - I can't even tell how the heck I've got a recrusive solution attempt. Thanks anyways. $\endgroup$ – user109498 Nov 18 '13 at 22:22
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$A_n=2A_{n-1}-A_{n-2}$, therefore $A_n=a\cdot n\cdot1^n+b$. The coefficients $a=1,b=1$ can be computed from $A_2=2$ and $A_3=3$.

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