Here is the theorem.

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

  1. If A has a zero row (column) then det(A)=0.
  2. If the last row (column) of A contains exactly one non-zero number A(n,n) then

    det(A)=A(n,n)*Cnn

    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.

  3. If the i-th row (column) of A contains exactly one non-zero entry A(i,j) (i, j are between 1 and n) then
    det(A)=A(i,j)*Cij
    where Cij is the cofactor of entry A(i,j) that is (-1)i+j multiplied by the determinant of the matrix obtained from A by deleting the i-th row and the j-th column (the determinant of this smaller matrix has a special name also; it is called the minor of entry A(i,j)).
  4. For every i=1,...,n we have
    det(A)=A(i,1) Ci1 + A(i,2) Ci2 +...+ A(i,n) Cin
    This is called the cofactor expansion of det(A) along the i-th row (a similar statement holds for columns).
  5. The determinant of a triangular matrix is the product of the diagonal entries.

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 theorem about determinants, part 1 ). On the other hand the matrix does not change (zero vector times 2 is zero vector), so the determinant must stay the same. There is only one number, 0, which remains the same after we multiply it by 2. Thus det(A)=0.

2. Again we prove the statement for rows only. Consider the expression of det(A). Take the typical term in this expression:

(-1) p A(p1,1) A(p2,2)... A(pn,n)


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

(-1) p A(p1,1) A(p2,2)... A(pn-1,n-1)

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.

Example:
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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), A(i,2),...,A(i,n)]

of A as the sum of the following n rows:

[ A(i,1) 0 0 ... 0 ]
[ 0 A(i,2) 0 ... 0 ]
.......................
[ 0 0 0 ... A(i,n) ]

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.