We prove each part separately.
Part 1: Multiplying a row by a scalar
Suppose \(B\) is obtained from \(A\) by multiplying row \(i\) by scalar \(k\text{.}\) Using cofactor expansion along row \(i\text{:}\)
\begin{align*}
\det(B) \amp= \sum_{j=1}^{n} b_{ij}C_{ij}\\
\amp= \sum_{j=1}^{n} (ka_{ij})C_{ij}\\
\amp= k\sum_{j=1}^{n} a_{ij}C_{ij}\\
\amp= k\det(A)
\end{align*}
Note that the cofactors
\(C_{ij}\) are the same for both
\(A\) and
\(B\text{,}\) since they depend on minors that do not involve row
\(i\text{.}\)
Suppose
\(B\) is obtained from
\(A\) by adding
\(c\) times row
\(j\) to row
\(i\) (where
\(i \neq j\)). Let’s denote the rows of
\(A\) as
\(\mathbf{r}_1, \mathbf{r}_2, \ldots, \mathbf{r}_n\text{.}\) Then
\(B\) has row
\(i\) equal to
\(\mathbf{r}_i + c\mathbf{r}_j\text{,}\) while all other rows remain unchanged.
Using the linearity property of determinants in a single row (which follows from cofactor expansion):
\begin{equation*}
\det(B) = \det[\mathbf{r}_1, \ldots, \mathbf{r}_i + c\mathbf{r}_j, \ldots, \mathbf{r}_n]
\end{equation*}
\begin{equation*}
= \det[\mathbf{r}_1, \ldots, \mathbf{r}_i, \ldots, \mathbf{r}_n] + c\det[\mathbf{r}_1, \ldots, \mathbf{r}_j, \ldots, \mathbf{r}_n]
\end{equation*}
The first determinant is
\(\det(A)\text{.}\) The second determinant has two identical rows (row
\(i\) and row
\(j\) are both
\(\mathbf{r}_j\)), so it equals zero. Therefore,
\(\det(B) = \det(A)\text{.}\)
We first prove a lemma: if
\(A\) has two identical rows, then
\(\det(A) = 0\text{.}\)
Proof of lemma: Suppose rows
\(i\) and
\(j\) of
\(A\) are identical (with
\(i \lt j\)). If we interchange these two rows to obtain matrix
\(B\text{,}\) then
\(B = A\text{.}\) However, by the row interchange property we’re trying to prove, we would have
\(\det(B) = -\det(A)\text{.}\) Since
\(B = A\text{,}\) we have
\(\det(A) = -\det(A)\text{,}\) which implies
\(2\det(A) = 0\text{,}\) so
\(\det(A) = 0\text{.}\)
Now we prove Part 3. Let \(B\) be obtained from \(A\) by interchanging rows \(i\) and \(j\) (with \(i \lt j\)). Consider the matrix \(C\) obtained from \(A\) by:
By Part 2, \(\det(C) = \det(A)\text{.}\) Now perform these operations on \(C\text{:}\)
-
Subtract row \(i\) from row \(j\text{:}\) row \(j\) becomes \(\mathbf{r}_j - (\mathbf{r}_i + \mathbf{r}_j) = -\mathbf{r}_i\)
-
Multiply row \(j\) by \(-1\text{:}\) row \(j\) becomes \(\mathbf{r}_i\)
-
Subtract row \(j\) from row \(i\text{:}\) row \(i\) becomes \((\mathbf{r}_i + \mathbf{r}_j) - \mathbf{r}_i = \mathbf{r}_j\)
After these operations, we obtain matrix \(B\text{.}\) By Parts 1 and 2:
\begin{equation*}
\det(B) = (-1)\det(C) = -\det(A)
\end{equation*}
Alternatively, a more direct proof uses mathematical induction on
\(n\text{,}\) the size of the matrix, combined with cofactor expansion.
