# Determinant Problem: Show $\det B = \det A$

• geoffrey159
In summary, the given matrix B is defined as a matrix with elements equal to those of A, except with alternating signs in each row. Using recursion and the property that the determinant of a matrix is equal to the sum of its elements multiplied by their cofactors, it can be shown that the determinants of A and B are equal. Additionally, by considering the permutations of the matrix, it can also be shown that the determinants are equal.
geoffrey159

## 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!

## 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##.

geoffrey159 said:

## 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!

## 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##.
I think the basic idea is correct, although it's a bit difficult to follow as the cofactors aren't very clear to see, resp. analyze. Do you apply the induction hypothesis on them?

Anyway, if you write the determinant as single formula, then the entire proof boils down to ##|\,\mathcal{Sym}(n)\, : \,\mathcal{A}(n)\,|=2##.

The problem with the determinants in the cofactors is that neighbor columns have the same sign and break the 'chessboard' structure.
That's why I reverse the sign of the ##k-1## first columns, in order to recover this structure, which allows me to apply ##{\cal P}(n-1)##.

Have you considered to use a compact formula without recursion for the determinant?

I have received an explanation in terms of permutations, but I didn't get it. That's why I tried to find another way. What do you mean ?

geoffrey159 said:
I have received an explanation in terms of permutations, but I didn't get it. That's why I tried to find another way. What do you mean ?
The determinant is the weighted sum over all diagonals, i.e. permutations: ##\det A = \sum_{\sigma \in \mathcal{Sym}(n)} (-1)^{\operatorname{sgn}\sigma}a_{k\,\sigma(k)}## Here you have only to analyze a single permutation ##\sigma##. The question is: can there be an odd number of pairs ##(k\,,\, \sigma(k)) ## with ##k + \sigma(k) \equiv 1 (2)\,##?

Edit: Corrected errors.

Last edited:
I think I get it, but it is a little difficult:

##
\begin{align*} \det B &= \sum_{\sigma\in S_n} \epsilon(\sigma) b_{\sigma(1)1}\cdots b_{\sigma(n)n}\\
&= \sum_{\sigma\in S_n} \epsilon(\sigma) a_{\sigma(1)1}\cdots a_{\sigma(n)n} (-1)^{\#S_\sigma}
\end{align*}##

where set ##S_\sigma = \{\text{couples }(k,\sigma(k)) \text{ such that } k+\sigma(k) \text{ is odd}, 1\le k \le n \} ##

But for any permuatition ##\sigma## of ##\{1,...,n\}##: ##\sum_k (k + \sigma(k)) = \underbrace{n(n+1)}_\text{even}##

So ## \#S_\sigma## must be even and ##\det B =
\sum_{\sigma\in S_n} \epsilon(\sigma) a_{\sigma(1)1}\cdots a_{\sigma(n)n} = \det A##EDITED: exhausted :-)

Last edited:
It must be true for every single diagonal, for otherwise it might not compensate in the sum. Furthermore doesn't the signature of the permutation change, so it has to be true for every single permutation: ##\begin{bmatrix}1&2&3&\ldots &n\\k_1&k_2&k_3&\ldots &k_n\end{bmatrix}##. Now we change the sign of the diagonal in total by $$\prod_{i=1}^n (-1)^{i+k_i}=\prod_{i=1}^n (-1)^{i+\sigma(i)}$$.

Thank you for your help !

geoffrey159 said:

## 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!

## 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##.

Another way is to change the signs of every second column (so all columns look the same), then multiply the resulting matrix by
$$\pmatrix{1 & 0 & 0 & \ldots & 0 \\ 0 & -1 & 0 & \ldots & 0 \\ 0 & 0 & 1 & \ldots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \ldots & \pm 1}$$
The first move changes the sign of the determinant of ##B## in a predictable way, and the second move changes it by another predictable way and gets you back to ##\det(A)##.

StoneTemplePython
piggybacking here, use Ray's matrix

##\mathbf S:=
\pmatrix{1 & 0 & 0 & \ldots & 0 \\
0 & -1 & 0 & \ldots & 0 \\
0 & 0 & 1 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & \pm 1}##

verify the involution ##\mathbf {SS} = \mathbf I \to \mathbf S = \mathbf S^{-1}##

recognize that the two matrices ##\mathbf A## and ##\mathbf B## are similar and hence have same determinant, characteristic polynomial, etc.

i.e. confirm
##\mathbf B = \mathbf S \mathbf A \mathbf S = \mathbf S \mathbf A \mathbf S^{-1}##

that's it.

StoneTemplePython said:
piggybacking here, use Ray's matrix

##\mathbf S:=
\pmatrix{1 & 0 & 0 & \ldots & 0 \\
0 & -1 & 0 & \ldots & 0 \\
0 & 0 & 1 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & \pm 1}##

verify the involution ##\mathbf {SS} = \mathbf I \to \mathbf S = \mathbf S^{-1}##

recognize that the two matrices ##\mathbf A## and ##\mathbf B## are similar and hence have same determinant, characteristic polynomial, etc.

i.e. confirm
##\mathbf B = \mathbf S \mathbf A \mathbf S = \mathbf S \mathbf A \mathbf S^{-1}##

that's it.

Much nicer!

## 1. What is a determinant?

A determinant is a mathematical concept used to determine the properties and behavior of a matrix. It is a numerical value that can be calculated from the elements of a square matrix and is used to solve various equations and problems.

## 2. How is the determinant of a matrix calculated?

The determinant of a matrix is calculated by using a specific formula that involves the elements of the matrix and their positions within the matrix. This formula varies depending on the size of the matrix, but it typically involves multiplying certain elements and adding or subtracting them in a specific manner.

## 3. What does it mean for two matrices to have equal determinants?

If two matrices have equal determinants, it means that they have the same numerical value when their determinants are calculated. This indicates that the two matrices have similar properties and behaviors, and can potentially be used interchangeably in certain equations and problems.

## 4. How can you prove that two matrices have equal determinants?

To prove that two matrices have equal determinants, you can use various mathematical techniques such as row reduction, cofactor expansion, or Laplace expansion. These techniques involve manipulating the elements of the matrices in a specific way to show that their determinants have the same numerical value.

## 5. Why is showing $\det B = \det A$ important?

Showing that $\det B = \det A$ is important because it allows us to establish a similarity or equivalence between two matrices. This can be useful in various applications, such as solving systems of equations, calculating eigenvalues and eigenvectors, and determining the invertibility of a matrix.

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