Theorem. Let A be an n by n matrix. Then the following conditions hold.

1. If we multiply a row (column) of A by a number, the determinant of A will be multiplied by the same number.
2. If the i-th row (column) in A is a sum of the i-th row (column) of a matrix B and the i-th row (column) of a matrix C and all other rows in B and C are equal to the corresponding rows in A (that is B and C differ from A by one row only), then det(A)=det(B)+det(C).
3. If two rows (columns) in A are equal then det(A)=0.
4. If we add a row (column) of A multiplied by a scalar k to another row (column) of A, then the determinant will not change.
5. If we swap two rows (columns) in A, the determinant will change its sign.
6. det(A)=det(A_T).

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 ]

1. Add the j-th row to the i-th row:
   

                        [ r1               ]
                          ...
                        [ ri+rj ]
                         ...
                        [ rj               ]
                          ...
                        [ rn               ]
   

2. Subtract the i-th row of the resulting matrix from the j-th row:

                        [ r1               ]
                          ...
                        [ ri+rj ]
                          ...
                        [ -ri              ]
                          ...
                        [ rn               ]
   

3. Add the j-th row of the resulting matrix to the i-th row:
   

                        [ r1  ]
                          ...
                        [ rj  ]
                          ...
                        [ -ri ]
                          ...
                        [ rn  ]
   

4. Multiply the j-th row of the resulting matrix by -1:
   

                        [ 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.

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