- #1

geoffrey159

- 535

- 72

## Homework Statement

Given a matrix ##A = (a_{ij})##, we define matrix ##B = \begin{pmatrix} a_{11} & - a_{12} & a_{13} & \cdots \\ - a_{21} & a_{22} & -a_{23} & \cdots \\

a_{31} & - a_{32} & a_{33} & \cdots \\

\vdots & \vdots & \vdots & \vdots \end{pmatrix}##.

Another way to define ##B## is to say that ##b_{ij} = a_{ij}## if ##i+j## is even and ##- a_{ij}## otherwise.

The question is: show that ##\det B = \det A##

I have my idea which I have written below, do you agree with my explanation? Thanks!

## Homework Equations

## The Attempt at a Solution

[/B]

By recursion over property ##{\cal P}(n)##:

*the property is true for matrices of order*##n##.

##{\cal P}(2)## is true, and we assume that that ##{\cal P}(n-1)## is true. Let us show that ##{\cal P}(n)## must be true.

Expanding the determinant according to the first line, I have:

## \det B = \sum_k b_{1k} \text{ cof}(b_{1k}) = \sum_{k \text{ even}} -a_{1k} \text{ cof}(b_{1k}) + \sum_{k \text{ odd}} a_{1k} \text{ cof}(b_{1k})##

But no matter the parity of ##k##, if I reverse the sign of the ##k-1## first columns in the determinant hidden inside ##\text{ cof}(b_{1k})##, which is of order ##n-1##, and then use the recursion hypothesis, I have:

## \text{ cof}(b_{1k}) = (-1)^{k-1} \text{ cof}(a_{1k})##

Replacing the cofactors in the first formula, I have:

## \det B = \sum_{k \text{ even}} -a_{1k} (-\text{ cof}(a_{1k})) + \sum_{k \text{ odd}} a_{1k} \text{ cof}(a_{1k}) = \sum_k a_{1k} \text{ cof}(a_{1k}) = \det A##

This shows that ##{\cal P}(n)## is true, so it must be true for any integer ##n##.