Determining the n-Determinant for an nxn Matrix with a Specific Matrix Element

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SUMMARY

The discussion focuses on determining the n-determinant \(D_n\) of an nxn matrix \(A_n\) defined by the element \(a_{ik} = i + k\) for \(i, k = 1, \ldots, n\). It is established that for \(n \geq 3\), the columns of \(A_n\) exhibit linear dependence, specifically \(c_3 + c_1 = 2c_2\), which directly implies that the determinant \(D_n = \text{det}(A_n) = 0\). The case for \(n = 1\) is also noted, indicating that the determinant must be evaluated for all specified values of \(n\).

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Perrry
Let \begin{gather*}A_n\end{gather*} be an nxn matrix with the matrixelement \begin{gather*}a_ik\end{gather*}=i+k, i, k = 1, ... ,n. Decide for every value the n-determinant \begin{gather*}D_n\end{gather*} = det(\begin{gather*}A_n\end{gather*}). Don´t forget the value of n=1.

We are two guys here at home that don´t get it right. What shall we start with? We are both newbies on this!

Thanks in advance

Perrry
 
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Consider a matrix A_n such as the one you described with n \geq 3.

Look at the columns of A_n:

the first column, c_1 looks like: c_1=(2,...,n+1)^T the next column looks like c_2=(3,...,n+2)^T and the third column looks like c_3=(4,...,n+3)^T. Since we took n \geq 3 we know that we can always get c_1,c_2,c_3.

Observe that: c_3-c_2=c_2-c_1=(1,...,1)^T Therefore, since c_3+c_1=2c_2 we have that the columns are linearly dependent. Does this say anything about the determinant?
 

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