# How to prove that these two matrices have the same determinant?

I want to prove that the following 2 matrices have the same determinant (which they do and it is equal to 288) Apologies for quality: [Matrix one] [Matrix two]  I know that the first matrix is a Vandermonde matrix, and it is easy to calculate its determinant. The second matrix looks similar, but I can't find the connection. Any ideas would be helpful.

• Matrix 2 is obtained by doing some column operations namely C2-C1, C3-C1,.... These preserve the determinant. May 5 at 0:35
• How difficult is it to compute the determinants with software? May 5 at 2:10

Let $$A$$ be the first matrix, and $$B$$ the second matrix , then

$$B = A C$$

where

$$C = \begin{bmatrix} 1 && -1 && 0 && 0 && 0 \\0 && 1 && -1 && 0 && 0 \\ 0 && 0 && 1 && -1 && 0 \\ 0 && 0 && 0 && 1 && -1 \\ 0 && 0 && 0 && 0 && 1 \end{bmatrix}$$

From the well-known property of determinants we know that

$$\det(B)= \det(A) \det(C)$$

Since $$C$$ is upper-triangular, then its determinant is the product of its diagonal elements, which is $$1$$, hence

$$\det(B) = \det(A)$$

• Thank you, this helps a lot! May 5 at 1:02
• Can you explain how you arrived at $B=AC$? May 5 at 2:11
• Just by examining the matrices $A$ and $B$, you can see that the first column of $B$ is the same as $A$, and the $k$-th column of $B$ is the $k$-th column of $A$ minus the $(k-1)$-th column of $A$, for $k = 2,3,4,5$. May 5 at 5:31