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

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.

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

• The solution is perfect. You thought the statement was false, and you showed this by finding an explicit counterexample. – Alex Mathers Mar 11 '17 at 16:32
• note that you can't change the absolute value of $\det A$ by switching columns or rows. – Nathanael Skrepek Mar 11 '17 at 16:34
• Also, for what it's worth I'll add that it's usually easier to try to find the simplest counterexamples. For instance, the matrix $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ would have worked, and its determinant is much easier to compute – Alex Mathers Mar 11 '17 at 16:35
• In fact, swapping any two rows or any two columns multiplies the determinant by $-1$. It might be helpful to compute the determinant of a few permutation matrices. – Couchy Mar 11 '17 at 16:40

However, if $A$ is not of full rank, then switching two columns would not change the determinant.
To be more general there is a result that says $\det (A\cdot B) = \det A \cdot \det B$.
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 $+1$.