We will prove this theorem using mathematical induction on the size
\(n\) of the matrix. The proof has two main parts: showing expansion along any row gives the same result, and showing expansion along any column gives the same result.
Part 1: Expansion along any row
Base case (\(n = 2\)): For a
\(2 \times 2\) matrix
\(A = \begin{bmatrix} a_{11} \amp a_{12} \\ a_{21} \amp a_{22} \end{bmatrix}\text{,}\) we verify both row expansions:
Row 1 expansion (by definition):
\(\det(A) = a_{11}a_{22} - a_{12}a_{21}\)
Row 2 expansion:
\begin{align*}
\sum_{j=1}^{2} a_{2j}C_{2j} \amp= a_{21}C_{21} + a_{22}C_{22}\\
\amp= a_{21}(-1)^{2+1}M_{21} + a_{22}(-1)^{2+2}M_{22}\\
\amp= a_{21}(-1)(a_{12}) + a_{22}(1)(a_{11})\\
\amp= -a_{21}a_{12} + a_{22}a_{11}\\
\amp= a_{11}a_{22} - a_{12}a_{21}
\end{align*}
Thus both row expansions give the same result for
\(n = 2\text{.}\)
Inductive step: Assume the theorem holds for all
\((n-1) \times (n-1)\) matrices. We need to show that expanding an
\(n \times n\) matrix
\(A\) along row
\(i\) gives the same result as expanding along row 1.
Let \(A = [a_{kl}]\text{.}\) By definition:
\begin{equation*}
\det(A) = \sum_{j=1}^{n} a_{1j}C_{1j}
\end{equation*}
We want to show:
\begin{equation*}
\det(A) = \sum_{j=1}^{n} a_{ij}C_{ij}
\end{equation*}
for any row \(i\text{.}\)
Key idea: We can transform row
\(i\) to row 1 by performing
\(i-1\) consecutive row swaps. Each row swap changes the sign of the determinant. After
\(i-1\) swaps, row
\(i\) becomes row 1, and we can apply the definition.
Let \(B\) be the matrix obtained from \(A\) by moving row \(i\) to row 1 (through \(i-1\) swaps). Then:
\begin{equation*}
\det(A) = (-1)^{i-1}\det(B)
\end{equation*}
Now \(B\) has \(a_{i1}, a_{i2}, \ldots, a_{in}\) as its first row. By definition:
\begin{equation*}
\det(B) = \sum_{j=1}^{n} a_{ij}C_{1j}^{(B)}
\end{equation*}
where \(C_{1j}^{(B)}\) denotes the \((1,j)\)-cofactor of \(B\text{.}\)
The crucial observation is that \(C_{1j}^{(B)} = (-1)^{i-1}C_{ij}\text{,}\) because:
-
The minor \(M_{1j}^{(B)}\) (obtained by deleting row 1 and column \(j\) of \(B\)) is essentially the same as \(M_{ij}\) (from \(A\)), up to \(i-1\) row swaps
-
The sign factor \((-1)^{1+j}\) for \(B\) differs from \((-1)^{i+j}\) for \(A\) by exactly \((-1)^{i-1}\)
Therefore:
\begin{align*}
\det(A) \amp= (-1)^{i-1}\det(B)\\
\amp= (-1)^{i-1} \sum_{j=1}^{n} a_{ij}C_{1j}^{(B)}\\
\amp= (-1)^{i-1} \sum_{j=1}^{n} a_{ij}(-1)^{i-1}C_{ij}\\
\amp= \sum_{j=1}^{n} a_{ij}C_{ij}
\end{align*}
This completes the proof that expansion along any row gives the same result.
Part 2: Expansion along any column
The proof for column expansion is analogous. We can transform column \(j\) to column 1 by performing \(j-1\) consecutive column swaps. The argument proceeds similarly, showing that:
\begin{equation*}
\det(A) = \sum_{i=1}^{n} a_{ij}C_{ij}
\end{equation*}
for any column \(j\text{.}\)
Alternatively, we can appeal to the fact that
\(\det(A) = \det(A^T)\) (which we will prove later). Since row expansion for
\(A^T\) corresponds to column expansion for
\(A\text{,}\) the column expansion formula follows from the row expansion formula applied to
\(A^T\text{.}\)
