Lemma:
If two rows coincide then det M = 0. For example det of the 3-by-3 matrix 1, 1, 0; 1, 1, 0; 2, 4, 5 equals 0.
25. Properties of determinants
25. Properties of determinants
Recall that det M equals the sum over permutations sigma of 1 up to n of the sign of sigma times the product M_{1 sigma(1)} times dot dot dot times M_{n sigma(n)} We'll use this formula to prove some nice properties of the determinant.
If row i and row j coincide then M_{i k} = M_{j k} for all k. Consider one of the terms sign of sigma M_{1 sigma(1)} times dot dot dot times M_{i sigma(i)} times dot dot dot times M_{j sigma(j)} times dot dot dot times M_{n sigma(n)} in the determinant. Since M_{i k} = M_{j k} for all k, we have M_{i sigma(i) = M_{j sigma(i)} and M_{j sigma(j)} = M_{i sigma(j)}. Therefore sign of sigma times M_{1 sigma(1)} times dot dot dot times M_{i sigma(i)} times dot dot dot times M_{j sigma(j)} times dot dot dot times M_{n sigma(n)} equals sign of sigma times M_{1 sigma(1)} times dot dot dot times M_{i sigma(j)} times dot dot dot times M_{j sigma(i)} times dot dot dot times M_{n sigma(n)} This is another term in our determinant, corresponding to the permutation sigma prime, which is defined by sigma prime (i) = sigma(j), sigma prime (j) equals sigma(u) and sigma prime (k) equals sigma(k) for all k not equal to i or j. In other words: M_{1 sigma(1)} times dot dot dot times M_{i sigma(j)} times dot dot dot times M_{j sigma(i)} times dot dot dot times M_{n sigma(n)} = M_{1 sigma prime (1)} times dot dot dot times M_{i sigma prime (i)} times dot dot dot times M_{j sigma prime (j)} times dot dot dot times M_{n sigma prime (n)} Since sigma prime differs from sigma by a single transposition, the sign of sigma prime is opposite to the sign of sigma. Therefore these two terms contribute with opposite signs to det M, and cancel out when you sum them together. Therefore all terms cancel in pairs, and the determinant is zero.
Type I row operations
Lemma:
If M prime is obtained from M by the row operation R_i maps to R_i + lambda R_j then det M prime equals det M.
We have det of M prime equals the sum over sigma of sign sigma times M prime_{1, sigma(1)} times dot dot dot times M prime_{i sigma(i)} times dot dot dot times M prime _{n, sigma(n)}, which equals the sum over sigma of sign sigma times M_{1 sigma(1)} times dot dot dot times (M_{i sigma(i)} + lambda M_{j sigma(i)} times dot dot dot times M_{n sigma(n)}, which equals the sum of sign sigma M_{1 sigma(1) times dot dot dot M_{i sigma(i)} times dot dot dot times M_{n sigma(n)}, plus lambda times the sum of signa sigma M_{1 sigma(1)} times dot dot dot times M_{j sigma(i)} times dot dot dot times M_{n sigma(n)}. The first term here is det M. The second term is the determinant of the matrix obtained from M by replacing the ith row with the jth row; in other words, the jth row now appears twice. By the previous lemma, this second term vanishes, so det M prime equals det M.
We can put any matrix into echelon form using type I row operations without changing the determinant. Once you're in echelon form (since you're then upper triangular), the determinant is just the product of diagonal entries.
Type III row operations
Lemma:
If M prime is obtained from M by switching two rows (say R_i and R_j then det M prime equals minus det M.
Every term in det M has the form sign sigma M prime_{1 sigma(1)} times dot dot dot times M prime_{i sigma(i)} times dot dot dot times M prime_{j sigma(j)} times dot dot dot times M prime_{n sigma(n)}. Since M prime and M are related by switching rows i and j, this is just sign sigma M_{1 sigma(1)} times dot dot dot times M_{i sigma(j)} times dot dot dot times M_{j sigma(i)} times dot dot dot times M_{n sigma(n)}, which is equal to sign sigma M_{1 sigma prime(1)} times dot dot dot times M_{i sigma prime(i)} times dot dot dot times M_{j sigma prime(j)} times dot dot dot times M_{n sigma prime(n)}, where sigma prime is (as before): sigma prime (i) = sigma(j), sigma prime (j) equals sigma(u) and sigma prime (k) equals sigma(k) for all k not equal to i or j. Since sign of sigma prime equals minus sign of sigma, this is equal to minus sign sigma prime M_{1 sigma prime(1)} times dot dot dot times M_{i sigma prime(i)} times dot dot dot times M_{j sigma prime(j)} times dot dot dot times M_{n sigma prime(n)}, so for every term in det M prime there's a corresponding term in det M which differs only by a sign. Therefore det M prime equals minus det M.
Type II row operations
Lemma:
If M prime is obtained from M by R_i maps to lambda R_i then det M prime equals lambda times det M.
Exercise.