Definition (k-th Order Minor): In an n-th order determinant D, a k-th order minor M is a k×k determinant formed by selecting any k rows and k columns from D (k≤n), and taking the k2 elements at the intersections of these rows and columns in their original order.
When k<n, after removing these k rows and k columns from D, the remaining (n−k)-th order determinant M′ formed by the leftover elements is called the cofactor of the k-th order minor M.
Note: In the above example, the algebraic cofactor of M is
(−1)(1+2+4)+(2+3+5)M′=−M′
Definition (Algebraic Cofactor): Let M be a k-th order minor of D with row and column indices i1,i2,⋯,ik;j1,j2,⋯,jk, respectively. The cofactor M′ of M is multiplied by the sign (−1)(i1+i2+⋯+ik)+(j1+j2+⋯+jk) to form the algebraic cofactor of M.
Lemma: Every term in the product of any minor M of determinant D and its algebraic cofactor A corresponds to a term in the expansion of determinant D, with the same sign.
Proof: First, consider the case where M is located in the upper-left corner of determinant D:
Each term in M can be written as a1α1a2α2⋯akαk, where α1α2⋯αk is a permutation of 1,2,⋯,k. The sign of this term is (−1)τ(α1α2⋯αk).
Each term in M′ can be written as ak+1,βk+1ak+2,βk+2⋯anβn, where βk+1βk+2⋯βn is a permutation of k+1,k+2,⋯,n, and its sign is (−1)τ((βk+1−k)(βk+2−k)⋯(βn−k)).
The product of these two terms is a1α1a2α2⋯akαkak+1,βk+1⋯anβn, and the combined sign is:
Thus, this product is a term in the expansion of determinant D with the same sign.
Now, we will prove the general case.
Let minor M be located in rows i1,i2,⋯,ik and columns j1,j2,⋯,jk of D, where i1<i2<⋯<ik, and j1<j2<⋯<jk.
We rearrange the rows and columns of determinant D such that M is moved to the top-left corner.
First, swap the i1-th row with rows i1−1,i1−2,⋯,2,1, which takes i1−1 swaps to move the i1-th row to the first row. Then, swap the i2-th row i2−2 times to move it to the second row, and so on. In total, this requires
Thus, the expansion of D1 contains the same terms as that of D, but with each term having a sign difference of (−1)i1+i2+⋯+ik+j1+j2+⋯+jk.
Now, M is located in the top-left corner of D1, and the cofactor and algebraic cofactor of M in D1 are both M′. Therefore, each term in the product MM′ corresponds to a term in D1 with the same sign.
The algebraic cofactor A of M in D gives MA=(−1)i1+i2+⋯+ik+j1+j2+⋯+jkMM′, which is a term in D1 with a sign difference of (−1)i1+i2+⋯+ik+j1+j2+⋯+jk. Hence, MA is a term in D with the same sign.
Theorem (Laplace's Theorem): For any k (1≤k≤n−1) rows chosen from determinant D, the sum of the products of all k-th order minors formed from these rows and their respective algebraic cofactors equals the determinant D.
Proof: Let the minors formed from the chosen k rows of D be M1,M2,⋯,Mt, with algebraic cofactors A1,A2,⋯,At. The theorem claims that:
D=M1A1+M2A2+⋯+MtAt
where
t=Cnk=k!(n−k)!n!
By the lemma, each term in MiAi corresponds to a term in D with the same sign, and there are no common terms between MiAi and MjAj (i=j).
The left-hand side of the equation contains n! terms, and the right-hand side contains
i=1∑tk!(n−k)!=k!(n−k)!k!(n−k)!n!=n!
terms. Hence, the theorem is proven.
Example: In the determinant D=10102−10112134131, selecting the first two rows yields six minors:
According to Laplace's theorem, expanding D by the first n rows, all of the minors formed by the first n rows and other elements will be zero except for the minor at the top left, so
Next, we show that D=C. Perform elementary row operations on D: add the a11-multiple of the (n+1)-th row, the a12-multiple of the (n+2)-th row, …, and the a1n-multiple of the (2n)-th row to the first row to get
Similarly, successively add the ak1-multiple of the (n+1)-th row, the ak2-multiple of the (n+2)-th row, …, and the akn-multiple of the (2n)-th row to the k-th row to obtain