Prove that the determinant of a square matrix can be computed by expansion about the first two columns.

Question

Suppose that $A$ is an $m \times n$ matrix. The $(i, j)$-entry of $A$ is denoted as $[A]_{i,j}$.

Suppose that $A$ is an $m \times n$ matrix. Suppose that $i_1$, $i_2$, $\dots$, $i_s$ are distinct integers less than or equal to $m$, and that $j_1$, $j_2$, $\dots$, $j_t$ are distinct integers less than or equal to $n$. The matrix obtained from $A$ by removing rows $i_1$, $i_2$, $\dots$, $i_s$ and columns $j_1$, $j_2$, $\dots$, $j_t$ is denoted as $A({i_1,\dots,i_s}|{j_1,\dots,j_t})$.

Determinants are defined here.

I will prove, by directly appealing to the definition, that $$ \begin{aligned} \det {(A)} = \sum_{1 \leq i < k \leq n} {\det { \begin{bmatrix} [A]_{i,1} & [A]_{i,2} \\ [A]_{k,1} & [A]_{k,2} \\ \end{bmatrix} } (-1)^{i+k+1+2} \det {(A(i,k|1,2))}}. \end{aligned} $$ for any $n \times n$ matrix $A$ (in which $n > 2$).

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Note that the result is simply an easy corollary of the generalized Laplace expansion. It will be useful in a proof of the alternating property of $\det$ and in a proof of the antisymmetric property of $\det$.

Answer 1

Suppose that $i$, $j$ are two integers. Define $$ \rho(i, j) = \begin{cases} 0, & i < j; \\ 1, & i \geq j. \end{cases} $$

Note that $[A]_{d_2, 2}$ is the $(d_2 - \rho(d_2, d_1), 1)$-entry of $A(d_1|1)$, in which $d_2 \neq d_1$. Hence $$ \begin{aligned} & \det {(A)} \\ = {} & \sum_{d_1 = 1}^{n} {(-1)^{d_1+1} [A]_{d_1,1} \det {(A(d_1|1))}} \\ = {} & \sum_{d_1 = 1}^{n} {(-1)^{d_1+1} [A]_{d_1,1} \sum_{\substack{1 \leq d_2 \leq n \\d_2 \neq d_1}} {(-1)^{d_2 - \rho(d_2,d_1) + 1} [A]_{d_2,2} \det {(A(d_1,d_2|1,2))}}} \\ = {} & \sum_{d_1 = 1}^{n} {\sum_{\substack{1 \leq d_2 \leq n \\d_2 \neq d_1}} {(-1)^{d_1+1} [A]_{d_1,1} (-1)^{d_2 - \rho(d_2,d_1) + 1} [A]_{d_2,2} \det {(A(d_1,d_2|1,2))}}} \\ = {} & \sum_{\substack{1 \leq d_1,d_2 \leq n \\d_1 \neq d_2}} {(-1)^{d_1+1} [A]_{d_1,1} (-1)^{d_2 - \rho(d_2,d_1) + 1} [A]_{d_2,2} \det {(A(d_1,d_2|1,2))}} \\ = {} & \sum_{\substack{1 \leq d_1,d_2 \leq n \\d_1 \neq d_2}} {(-1)^{\rho(d_1,d_2)} [A]_{d_1,1} [A]_{d_2,2} (-1)^{d_1+d_2+1+2} \det {(A(d_1,d_2|1,2))}} \\ = {} & \hphantom{{} + {}} \sum_{\substack{1 \leq d_1,d_2 \leq n \\d_1 < d_2}} {(-1)^{\rho(d_1,d_2)} [A]_{d_1,1} [A]_{d_2,2} (-1)^{d_1+d_2+1+2} \det {(A(d_1,d_2|1,2))}} \\ & + \sum_{\substack{1 \leq d_1,d_2 \leq n \\d_1 > d_2}} {(-1)^{\rho(d_1,d_2)} [A]_{d_1,1} [A]_{d_2,2} (-1)^{d_1+d_2+1+2} \det {(A(d_1,d_2|1,2))}} \\ = {} & \hphantom{{} + {}} \sum_{\substack{1 \leq i,k \leq n \\i < k}} {(-1)^{\rho(i,k)} [A]_{i,1} [A]_{k,2} (-1)^{i+k+1+2} \det {(A(i,k|1,2))}} \\ & + \sum_{\substack{1 \leq k,i \leq n \\k > i}} {(-1)^{\rho(k,i)} [A]_{k,1} [A]_{i,2} (-1)^{k+i+1+2} \det {(A(k,i|1,2))}} \\ = {} & \sum_{1 \leq i < k \leq n} {([A]_{i,1} [A]_{k,2} - [A]_{k,1} [A]_{i,2})\, (-1)^{i+k+1+2} \det {(A(i,k|1,2))}} \\ = {} & \sum_{1 \leq i < k \leq n} {\det { \begin{bmatrix} [A]_{i,1} & [A]_{i,2} \\ [A]_{k,1} & [A]_{k,2} \\ \end{bmatrix} } (-1)^{i+k+1+2} \det {(A(i,k|1,2))}}. \end{aligned} $$