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 [tex]\begin{gather*}A_n\end{gather*}[/tex] be an nxn matrix with the matrixelement [tex]\begin{gather*}a_ik\end{gather*}[/tex]=i+k, i, k = 1, ... ,n. Decide for every value the n-determinant [tex]\begin{gather*}D_n\end{gather*}[/tex] = det([tex]\begin{gather*}A_n\end{gather*}[/tex]). 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 [tex]A_n[/tex] such as the one you described with [tex]n \geq 3[/tex].

Look at the columns of [tex]A_n[/tex]:

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

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

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