Interchanging Rows Of Matrix Changes Sign Of Determinants! Now a days I am learning about matrix and determinants and I confused on one
properties of determinants which is: interchanging two rows/Columns of a
determinant changes the sign of the determinant.
My question is what is the logic(reason) that -ve sign is places outside the
determinants while interchanging rows/Columns but no sign is places outsides in
gaussian elimination (OR more specific in matrix)
I don't understand the logic behind this. I Google it a lot but found no answer.
Can anybody please explain why we do this.
 A: This is simple. Note that
$$\det(PA) = \det(P)\det(A).$$
If you want $P$ to swap rows $k$ and $l$, then
$$P = \begin{bmatrix}
1 \\
& \ddots \\
& & 1 \\
& & & 0 & 0 & \dots & 0 & 1 \\
& & & 0 & 1 & \dots & 0 & 0 \\
& & & \vdots & \vdots & \ddots & \vdots & \vdots \\
& & & 0 & 0 & \dots & 1 & 0 \\
& & & 1 & 0 & \dots & 0 & 0 \\
& & &   &   &   &   &   & 1 \\
& & &   &   &   &   &   &   & \ddots \\
& & &   &   &   &   &   &   &   & 1
\end{bmatrix}.$$
In other words, $P$ is constructed by swapping rows (or, equivalently, columns) $k$ and $l$ of the identity matrix.
Now, check that $\det(P) = -1$, and you have what you asked about.
A: I will assume that we already know the following results:


*

*If two rows of a square matrix are the same, then determinant is zero. (This can be shown by induction using Laplace's expansion. See also ProofWiki.)

*If we have three square matrices $A$, $B$, $C$ which only differ in one row and the remaining row of $C$ is sum of the corresponding rows of $A$ and $B$, then $|C|=|A|+|B|$.
(This can be shown using Leibniz formula, which many texts take as a definition of a determinant. It is also a consequence of multilinearity of determinant. See also ProofWiki
Let us denote by $\vec\alpha_1,\dots,\vec\alpha_n$ the rows of the matrix $A$.
Then we have
$$
\begin{vmatrix}
\vec\alpha_1 \\
\vec\alpha_i+\vec\alpha_j \\
\vec\alpha_i+\vec\alpha_j \\
\vec\alpha_n
\end{vmatrix}
=0
$$
since this matrix has repeated rows.
At the same time we have
$$
0=\begin{vmatrix}
\vec\alpha_1 \\
\vec\alpha_i+\vec\alpha_j \\
\vec\alpha_i+\vec\alpha_j \\
\vec\alpha_n
\end{vmatrix}=
\begin{vmatrix}
\vec\alpha_1 \\
\vec\alpha_i+\vec\alpha_j \\
\vec\alpha_i \\
\vec\alpha_n
\end{vmatrix}+
\begin{vmatrix}
\vec\alpha_1 \\
\vec\alpha_i+\vec\alpha_j \\
\vec\alpha_j \\
\vec\alpha_n
\end{vmatrix}=
\begin{vmatrix}
\vec\alpha_1 \\
\vec\alpha_i\\
\vec\alpha_i \\
\vec\alpha_n
\end{vmatrix}+
\begin{vmatrix}
\vec\alpha_1 \\
\vec\alpha_j \\
\vec\alpha_i \\
\vec\alpha_n
\end{vmatrix}+
\begin{vmatrix}
\vec\alpha_1 \\
\vec\alpha_i \\
\vec\alpha_j \\
\vec\alpha_n
\end{vmatrix}+
\begin{vmatrix}
\vec\alpha_1 \\
\vec\alpha_j \\
\vec\alpha_j \\
\vec\alpha_n
\end{vmatrix}=
\begin{vmatrix}
\vec\alpha_1 \\
\vec\alpha_j \\
\vec\alpha_i \\
\vec\alpha_n
\end{vmatrix}+
\begin{vmatrix}
\vec\alpha_1 \\
\vec\alpha_i \\
\vec\alpha_j \\
\vec\alpha_n
\end{vmatrix}$$
which implies
$$\begin{vmatrix}
\vec\alpha_1 \\
\vec\alpha_j \\
\vec\alpha_i \\
\vec\alpha_n
\end{vmatrix}=-
\begin{vmatrix}
\vec\alpha_1 \\
\vec\alpha_i \\
\vec\alpha_j \\
\vec\alpha_n
\end{vmatrix}.$$
