Determining a determinant using recurrence relations

In summary, the conversation discusses determining a determinant by using elementary transformations and Laplace expansion. The determinant can be rewritten recursively as $D_n=(-1)^{n+1} \cdot (n-2) \cdot D_{n-1}$, and the conversation further explores finding the exponent of -1 in this recursion. It is determined that the exponent is related to the number of swaps needed to change the matrix to one where the anti-diagonal becomes the diagonal. The final conclusion is that the exponent of -1 is $n(n-1)/2$.
  • #1
karseme
15
0
I'm a little stuck here. I should determine the following determinant. I first tried to simplify it a little by using elemntary transformations. And then I did Laplace expansion on the last row.

$\begin{vmatrix}2 & 2 & \cdots & 2 & 2 & 1 \\ 2 & 2 & \cdots & 2 & 2 & 2 \\ 2 & 2 & \cdots & 3 & 2 & 2 \\ \vdots & \vdots & \quad & \vdots & \vdots & \vdots \\ 2 & n-1 & \cdots & 2 & 2 & 2 \\ n & 2 & \cdots & 2 & 2 & 2\end{vmatrix}=\begin{vmatrix}1 & 1 & \cdots & 1 & 1 & 1 \\ 0 & 0 & \cdots & 0 & 0 & 2 \\ 0 & 0 & \cdots & 1 & 0 & 2 \\ \vdots & \vdots & \quad & \vdots & \vdots & \vdots \\ 0 & n-3 & \cdots & 0 & 0 & 2 \\ n-2 & 0 & \cdots & 0 & 0 & 2\end{vmatrix} \\ =(-1)^{n+1} \cdot (n-2) \cdot \begin{vmatrix}1 & 1 & \cdots & 1 & 1 & 1 \\ 0 & 0 & \cdots & 0 & 0 & 2 \\ 0 & 0 & \cdots & 1 & 0 & 2 \\ \vdots & \vdots & \quad & \vdots & \vdots & \vdots \\ 0 & n-4 & \cdots & 0 & 0 & 2 \\ n-3 & 0 & \cdots & 0 & 0 & 2\end{vmatrix} + (-1)^{2n} \cdot 2 \cdot \begin{vmatrix}1 & 1 & \cdots & 1 & 1 & 1 \\ 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 0 & \cdots & 0 & 1 & 0 \\ \vdots & \vdots & \quad & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & n-3 & \cdots & 0 & 0 & 0\end{vmatrix}$.

The second determinant is equal to 0. And the first determinant is of the same form as the initial, just one degree smaller.
We have recursion here. I used labels:

$D_n=\begin{vmatrix}2 & 2 & \cdots & 2 & 2 & 1 \\ 2 & 2 & \cdots & 2 & 2 & 2 \\ 2 & 2 & \cdots & 3 & 2 & 2 \\ \vdots & \vdots & \quad & \vdots & \vdots & \vdots \\ 2 & n-1 & \cdots & 2 & 2 & 2 \\ n & 2 & \cdots & 2 & 2 & 2\end{vmatrix}=\begin{vmatrix}1 & 1 & \cdots & 1 & 1 & 1 \\ 0 & 0 & \cdots & 0 & 0 & 2 \\ 0 & 0 & \cdots & 1 & 0 & 2 \\ \vdots & \vdots & \quad & \vdots & \vdots & \vdots \\ 0 & n-3 & \cdots & 0 & 0 & 2 \\ n-2 & 0 & \cdots & 0 & 0 & 2\end{vmatrix}$

I have:

$D_n=(-1)^{n+1} \cdot (n-2) \cdot D_{n-1}$.

I wasn't sure how to continue this recursion though. I've got something like:

$2 \cdot (-1)^? \cdot (n-2)!$

I am not sure what I should get as exponent of -1. I've tried a few things but everything that I've got was different from the solution that my tutor gave us. Yet he said that the solution might be incorrect.
Those recursions are just so confusing to me. I would appreciate it if someone could explain me how to determine that exponent.
 
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  • #2
karseme said:
I'm a little stuck here. I should determine the following determinant. I first tried to simplify it a little by using elemntary transformations. And then I did Laplace expansion on the last row.

$\begin{vmatrix}2 & 2 & \cdots & 2 & 2 & 1 \\ 2 & 2 & \cdots & 2 & 2 & 2 \\ 2 & 2 & \cdots & 3 & 2 & 2 \\ \vdots & \vdots & \quad & \vdots & \vdots & \vdots \\ 2 & n-1 & \cdots & 2 & 2 & 2 \\ n & 2 & \cdots & 2 & 2 & 2\end{vmatrix}=\begin{vmatrix}1 & 1 & \cdots & 1 & 1 & 1 \\ 0 & 0 & \cdots & 0 & 0 & 2 \\ 0 & 0 & \cdots & 1 & 0 & 2 \\ \vdots & \vdots & \quad & \vdots & \vdots & \vdots \\ 0 & n-3 & \cdots & 0 & 0 & 2 \\ n-2 & 0 & \cdots & 0 & 0 & 2\end{vmatrix} \\ =(-1)^{n+1} \cdot (n-2) \cdot \begin{vmatrix}1 & 1 & \cdots & 1 & 1 & 1 \\ 0 & 0 & \cdots & 0 & 0 & 2 \\ 0 & 0 & \cdots & 1 & 0 & 2 \\ \vdots & \vdots & \quad & \vdots & \vdots & \vdots \\ 0 & n-4 & \cdots & 0 & 0 & 2 \\ n-3 & 0 & \cdots & 0 & 0 & 2\end{vmatrix} + (-1)^{2n} \cdot 2 \cdot \begin{vmatrix}1 & 1 & \cdots & 1 & 1 & 1 \\ 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 0 & \cdots & 0 & 1 & 0 \\ \vdots & \vdots & \quad & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & n-3 & \cdots & 0 & 0 & 0\end{vmatrix}$.

The second determinant is equal to 0. And the first determinant is of the same form as the initial, just one degree smaller.
We have recursion here. I used labels:

$D_n=\begin{vmatrix}2 & 2 & \cdots & 2 & 2 & 1 \\ 2 & 2 & \cdots & 2 & 2 & 2 \\ 2 & 2 & \cdots & 3 & 2 & 2 \\ \vdots & \vdots & \quad & \vdots & \vdots & \vdots \\ 2 & n-1 & \cdots & 2 & 2 & 2 \\ n & 2 & \cdots & 2 & 2 & 2\end{vmatrix}=\begin{vmatrix}1 & 1 & \cdots & 1 & 1 & 1 \\ 0 & 0 & \cdots & 0 & 0 & 2 \\ 0 & 0 & \cdots & 1 & 0 & 2 \\ \vdots & \vdots & \quad & \vdots & \vdots & \vdots \\ 0 & n-3 & \cdots & 0 & 0 & 2 \\ n-2 & 0 & \cdots & 0 & 0 & 2\end{vmatrix}$

I have:

$D_n=(-1)^{n+1} \cdot (n-2) \cdot D_{n-1}$.

I wasn't sure how to continue this recursion though. I've got something like:

$2 \cdot (-1)^? \cdot (n-2)!$

I am not sure what I should get as exponent of -1. I've tried a few things but everything that I've got was different from the solution that my tutor gave us. Yet he said that the solution might be incorrect.
Those recursions are just so confusing to me. I would appreciate it if someone could explain me how to determine that exponent.

I did this by changing the matrix to the one where the anti diagonal becomes the diagonal. The number of swaps needed to do this is n/2 if n is even and (n-1)/2 if n is odd. Each swap contributes -1 so we want the sign to be + when n or n-1 is a multiple of 4 this means the power of -1 is n(n-1)/2. Hope this helps
 

FAQ: Determining a determinant using recurrence relations

What is a determinant?

A determinant is a mathematical concept used to describe the properties and behavior of a square matrix. It is a numerical value that can be calculated from the elements of a matrix and is used in various mathematical and scientific applications.

What are recurrence relations?

Recurrence relations are mathematical equations that define a sequence of numbers or values based on previous terms in the sequence. They are often used to solve problems in various fields, including mathematics, physics, and computer science.

How is a determinant calculated using recurrence relations?

A determinant can be calculated using recurrence relations by breaking down the original matrix into smaller submatrices, applying the recurrence relation to each submatrix, and then combining the results to find the final determinant. This process is repeated until the determinant can be easily evaluated using basic mathematical operations.

What are the advantages of using recurrence relations to determine a determinant?

One advantage of using recurrence relations to determine a determinant is that it can be used for matrices of any size, whereas other methods may only work for specific matrix sizes. Additionally, recurrence relations can often provide a more efficient and elegant solution compared to other methods.

Can recurrence relations be used for non-square matrices?

No, recurrence relations can only be used to determine the determinant of square matrices. Non-square matrices do not have a determinant and therefore cannot be solved using recurrence relations.

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