- #1

Perrry

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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- Thread starter Perrry
- Start date

In summary, we have discussed how to construct a matrix A_n with n \geq 3 and how the columns of this matrix are linearly dependent. This information can be used to determine the determinant of A_n.

- #1

Perrry

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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- #2

marcmtlca

- 16

- 0

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?

- #3

tacman

- 1,874

- 0

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=

The n-determinant, also known as the determinant of an nxn matrix, is a mathematical value that is calculated using the elements of an nxn matrix. It is used to determine whether a square matrix has an inverse or not, and is also used in solving systems of linear equations.

The n-determinant for an nxn matrix can be found using various methods, such as the cofactor expansion method, the row reduction method, or the use of the Leibniz formula. Each method involves different steps and calculations, but they all result in the same value for the n-determinant.

The n-determinant of an nxn matrix has many important applications in mathematics, physics, and engineering. It is used to determine whether a matrix has an inverse, to solve systems of linear equations, to find areas and volumes of geometric shapes, and to calculate work and energy in physics problems.

Yes, the n-determinant for an nxn matrix can be both positive and negative. The sign of the n-determinant depends on the arrangement of the elements in the matrix. A positive n-determinant indicates that the matrix has an inverse, while a negative n-determinant indicates that the matrix does not have an inverse.

Yes, there are some shortcuts and tricks that can be used to find the n-determinant for an nxn matrix, such as using properties of determinants, using symmetry in the matrix, or using patterns in the matrix. However, these methods may not always be applicable and may not work for all types of matrices.

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