Theorem. Let A be an n by n matrix. Then the following conditions hold.
2. We have that A(i,j)=B(i,j)+C(i,j) for every j=1,2,...,n. Consider the expression for det(A). Each term in this expression contains exactly one factor from the i-th row of A. Conside one of these terms:
(A(i,j) is the factor from i-th row in this product).
Replace A(i,j) by B(i,j)+C(i,j):
Multiply through:
If we add only the terms containing B's, we get the determinant of B;
if we add all terms containing C's, we get the determinant of C. Thus
det(A)=det(B)+det(C).
3. Suppose that the i-th row in A is equal to the j-th row of A,
that is A(i,k)=A(j,k) for every k=1,2,...,n. Consider an arbitrary product
in the
expression of det(A):
(we use the fact that this product contains one factor from the i-th row
and one factor from the j-th row, and we assume that i occurs before j;
the case when i occurs further than j is similar).
Consider also the product corresponding to the permutation p' obtained from p
by switching i and j:
Now since
these terms are equal. But they occur in the expression of det(A) with opposite signs (remember that p' is obtained from p by one transposition). Thus these products kill each other in det(A). Therefore each term in det(A) gets killed when we combine like terms in det(A), so det(A)=0.
4. Let matrix B be obtained from matrix
A by adding the j-th row multiplied by k
to the i-th row. Let us represent A as a column of rows:
[ r1 ] ... [ ri ] ... [ rj ] ... [ rn ]
Then B has the following form:
[ r1 ] ... [ ri+krj ] ... [ rj ] ... [ rn ]
By property 2 we can conclude that det(B) is equal to the sum of determinants
of two matrices:
[ r1 ] ... [ ri ] ... [ rj ] ... [ rn ]
and
[ r1 ] ... [ krj ] ... [ rj ] ... [ rn ]
The first of these matrices is A. Let us denote the second one by C. So
det(B)=det(A)+det(C). By property 1
det(C) is k times the determinant of the
following matrix:
[ r1 ] ... [ rj ] ... [ rj ] ... [ rn ]
But this matrix has two equal rows, therefore its determinant is equal to 0 (property 3). Thus det(C)=0 and det(B)=det(A). The proof is complete.
5. Suppose that we swap the i-th row and the j-th row of matrix A. Represent A as a column of rows:
[ r1 ] ... [ ri ] ... [ rj ] ... [ rn ]In order to swap ri and rj we can do the following procedure:
[ r1 ] ... [ ri+rj ] ... [ rj ] ... [ rn ]
[ r1 ] ... [ ri+rj ] ... [ -ri ] ... [ rn ]
[ r1 ] ... [ rj ] ... [ -ri ] ... [ rn ]
[ r1 ] ... [ rj ] ... [ ri ] ... [ rn ]
By the properties that we already proved all operations of this procedure except the very last one do not change the determinant. The last operation changes the sign of the determinant. The proof is complete. 6. This is left as an exercise.