Theorem. Let A be an n by n matrix. Then the following conditions hold.
where Cnn is the cofactor of entry A(n,n) that is the determinant of the matrix obtained by deleting the last row and the last column of matrix A.
Proof.
1. Suppose A has zero i-th row. Multiply this row by 2. On the one
hand the determinant must increase by a factor of 2 (see the first
2. Again we prove the statement for rows only.
Consider the
expression of det(A). Take the typical term in this expression:
where p runs over all permutations of numbers 1,...,n-1. This sum is equal to
the determinant of the matrix obtained by removing the last row and the
last column of A.
of A as the sum of the following n rows:
By the first theorem
about determinants, part 2,
the determinant of the matrix A is equal to the
sum of determinants of n matrices Bj obtained by replacing the i-th row of A
by one of these n rows. Since each of these rows contains exactly one non-zero
entry, we can apply the previous statement (statement 3) of our theorem.
Notice that the co-factors of (i,j)-entry in the matrix Bj is the same as the
co-factor of the (i,j)-entry of matrix A because we remove the i-th row when we
calculate the co-factor and all other rows in A and Bj are the same.
5. Proof is left as exercise.Hint: use the previous statement.
This term contains exactly
one entry from the last row. By our assumption there is only one
non-zero entry in this row, namely A(n,n). Therefore if the term is not
equal to 0, p1 must be equal to n. Therefore the non-zero terms in the expression of det(A) correspond to permutations p with pn=n. So these
permutations actually permute numbers from 1 to n-1. If we consider this p
as a permutation of numbers from 1 to n-1, its sign does not change.
Thus if we take A(n,n)
out of each of the non-zero terms in the expression of det(A) we obtain
the sum of terms of the form
3. Hint. Swap the i-th row and the n-th row, the j-th column and the n-th
column of the matrix A. Then A(i,j) becomes the (n,n)-entry of the resulting
matrix. Then one can apply the previous statement and the first theorem about determinants, part e); this theorem is responsible to the sign (-1)i+j
in the definition of determinants).
The complete proof is left as an exercise.
4. Represent the i-th row
[ A(i,1)
0
0
...
0 ]
[ 0
A(i,2)
0
...
0 ]
.......................
[ 0
0
0
...
A(i,n) ]