Is this true or false: Switching two columns of $A \in \mathbb{R}^{n \times n}$ doesn't change $det(A)$

Question

Let $n \in \mathbb{N}.$

True or false? Switching two columns of $A \in \mathbb{R}^{n \times n}$ doesn't change $det(A)$. State why.

Let $A=\begin{pmatrix} 1 & 1 & 1\\ 1 & 0 & 0\\ 1 & 0 & 1 \end{pmatrix}$, then $det(A)=-1$

Now switching first column with second column, we have $A'=\begin{pmatrix} 1 & 1 & 1\\ 0 & 1 & 0\\ 0 & 1 & 1 \end{pmatrix}$, then $det(A')=1$

In this example, we saw that the determinant changed, thus the statement is wrong.

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Did I do it correctly and is the reasoning correct? Is there a better, more general / valid reasoning?

Answer 1

It is false in general. It would be enough to think about the identity matrix.

However, if $A$ is not of full rank, then switching two columns would not change the determinant.

Answer 2

To be more general there is a result that says $\det (A\cdot B) = \det A \cdot \det B$.

You can write a column switch as a matrixmultiplication.

small example: \begin{align} \left( \begin{matrix} a&b&c \\ d&f&g \\ h&i&j \end{matrix} \right) \cdot \left( \begin{matrix} 1&0&0 \\ 0&0&1 \\ 0&1&0 \end{matrix} \right) = \left( \begin{matrix} a&c&b \\ d&g&f \\ h&j&i \end{matrix} \right) \end{align} the determinate of the switching matrix is always $-1$ if you want to switch more than $1$ column then the determinate will be $\pm 1$ but the absolute value of the determinate of a switching matrix will alway be $+1$.

Answer 3

Think of cross product of vectors, as position of vectors are interchanged direction of product vector is opposite.