Determinants Before we can get to the definition of the determinant of a matrix, we first need to understand permutations. Let S = {1,2,...,n} then a permutation is a 11 function from S to S. We can think of a permutation on n elements as a reordering of the elements.
Example (2,1,3) is a permutation on 3 elements. We have f(1) = 2 f(2) = 1 f(3) = 3 There are exactly 6 permutations on 3 elements. They are (1,2,3) (1,3,2) (2,1,3) (2,3,1) (3,1,2) (3,2,1) The identity permutation is the permutation that keeps the elements in numerical order. For example e_{3} = (1,2,3) We define a transposition of two elements the permutation that switches the elements. For example (2,1,3) is a transposition that switches 1 and 2. We can compose two permutations since they are functions. Given a permutation, how many transpositions does it take in order to get to the identity permutation? It turns out that this is not a well defined question, since there are many ways of getting back to the identity. However given any permutation, the parity of the number of transpositions to get back to the identity is independent of how it is done. A permutation is called even, if it takes an even number of transpositions to get back to the identity and odd of it takes an odd number.
Example (3,4,2,1) is an odd permutation since we can get back to the identity via (3,4,2,1) > (1,4,2,3) > (1,2,4,3) > (1,2,3,4)
Definition of the Determinant We are now ready to define the determinant of the matrix. The definition will be quite difficult to understand as it is written. We strongly encourage you to read through the examples and try some of your own. Definition Let A be an n x n matrix. Then the determinant of A is the number given by
where the sum is taken over all possible permutations on n elements and the sign is positive if the permutation is even and negative if the permutation is odd.
Example Find the det A where
Solution We use all the permutations on 3 elements. The permutation (1,2,3) is even and corresponds with the product a_{11}a_{22}a_{33} = (2)(3)(0) = 0 The permutation (1,3,2) is odd and corresponds with the product a_{11}a_{23}a_{32} = (2)(5)(1) = 10 Notice that the i subscripts are 1, 2, and 3 and the j subscripts are 1, 3, 2. The permutation (2,1,3) is odd and corresponds with the product a_{12}a_{21}a_{33} = (0)(1)(0) = 0 The permutation (2,3,1) is even and corresponds with the product a_{12}a_{23}a_{31} = (0)(5)(10) = 0 The permutation (3,1,2) is even and corresponds with the product a_{13}a_{21}a_{32} = (4)(1)(1) = 4 The permutation (3,2,1) is odd and corresponds with the product a_{13}a_{22}a_{31} = (4)(3)(10) = 120 Finally, we add these all up to get 0 + 10 + 0 + 0  4  120 = 114 Example Find det A where
Solution There are 24 permutations, however we only need those corresponding to products that are nonzero. Since the last row has only one nonzero entry (third column), the last number in the permutation must be a 3. Also the first column has only one nonzero entry, hence the one must be in the second entry. For the first number in the permutation, we can only have a 4, since the first and second entries are zeros and 3 is already taken up by the fourth number. That leaves only 2 for the third number since 1 and 3 correspond with zero entries and 3 is reserved for the fourth number. Hence the only permutation that leaves us with a nonzero product is (4,1,2,3) This permutation is odd and corresponds with the product (3)(1)(5)(2) = 30 Since all other permutations lead to a zero product, the determinant is 30. Properties of the Determinant The payoff of this definition of the determinant is that we can quickly prove many properties of the determinant. The list is long and we will prove only a few of them. Theorem If A has a row of zeros, then det A = 0 Proof Each term of det A includes one factor that contains each row, hence each term has a zero factor. The sum of zeros is zero. Theorem If A has two identical rows, then det A = 0 Proof We will prove this for the first two rows. The general case is similar. The parity of (s_{1},s_{2},...,s_{n}) is the opposite of the parity of (s_{2},s_{1},...,s_{n}). Hence each term is repeated twice once with a positive coefficient and once with a negative coefficient. The terms cancel each other out and add up to zero. The next three theorems explore what happens to the determinant after row operations have occurred.
Theorem Let A be a square matrix and B be the matrix after a row of A has been multiplied by a constant. Then det B = c det A Proof Assume that the first row row has been multiplied by c. The general case is similar. The terms in each are identical except that the first factor for A is a_{1j1} and the first factor for B is ca_{1j1}. Since each term for the B determinant has an extra factor of c, the c can be factored out. Theorem Let A be a square matrix and B be the matrix after two rows of A have been switched. Then det A =  det B Proof This theorem comes directly from the definition, since switching two elements of a permutation changes the parity of the permutation. Hence switching two rows changes the sign of each term. Theorem Let A be a square matrix and B be the matrix after a row of A have been replaced by that row plus a multiple of another row. Then det A = det B We will not prove this one here. Theorem Let A be a triangular matrix then det A is equal to the product of the diagonal entries.
Proof The terms of the determinant of A will only be nonzero when each of the factors are nonzero. If A is lower triangular, then the only nonzero element in the first row is also in the first column. For the second row, we have already used the first column, hence the only nonzero element is the second column. Continuing this way we obtain the product of the diagonals. This permutation is the identity (1,2,...,n) which is even. The next two theorems we will state without proof. Theorem det A = det A^{T}
Theorem det (AB) = (det A)(det B) Theorem If A is nonsingular, then
1 Proof We have AA^{1} = I hence det(AA^{1}) = det I But det(AA^{1}) = (det A)(det A^{1}) and det I = 1 since the identity is triangular. Hence (det A)(det A^{1}) = 1 Divide and the result follows. Back to the Matrices and Applications Home Page Back to the Linear Algebra Home Page Back to the Math Department Home Page email Questions and Suggestions
