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

[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

 

[marcmtlca]

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?

 

[tacman]

Hello Perrry,

Determining the n-determinant for an nxn matrix can seem daunting at first, but with some practice and understanding of the process, it becomes easier. Let's break down the steps to find the n-determinant for the given matrix \begin{gather*}A_n\end{gather*}.

Step 1: When n=1, the matrix \begin{gather*}A_n\end{gather*} is just a 1x1 matrix with a single element \begin{gather*}a_{11}\end{gather*}. Therefore, the n-determinant \begin{gather*}D_n\end{gather*} is simply the value of \begin{gather*}a_{11}\end{gather*}. In this case, \begin{gather*}D_1 = a_{11} = 1 + 1 = 2\end{gather*}.

Step 2: For n > 1, we can use the cofactor expansion method to find the n-determinant. This involves choosing a row or column and multiplying each element in that row or column by its corresponding cofactor, then adding or subtracting these products to get the n-determinant.

Let's choose the first row for our example. We can write the n-determinant as:

\begin{gather*}D_n = \sum_{j=1}^{n} (-1)^{1+j}a_{1j}M_{1j}\end{gather*}

where \begin{gather*}a_{1j}\end{gather*} is the element in the first row and jth column, and \begin{gather*}M_{1j}\end{gather*} is the cofactor of \begin{gather*}a_{1j}\end{gather*}.

Step 3: To find the cofactor \begin{gather*}M_{1j}\end{gather*}, we need to find the determinant of the submatrix formed by removing the first row and jth column. In our case, this submatrix will be an (n-1)x(n-1) matrix. We can use the same process to find its determinant, and continue recursively until we reach a 1x1 matrix, as we did in Step 1.

Step 4: Let's work through an example for n=