Vectors and Vector Operations



5.2 Properties of Determinants

In this section we look at the algebraic properties of determinants that are the key to computing and applying them.

Property 1 – Determinants of Triangular Matrics. As we saw in the previous section the definition of a n(n involves the sum/difference of n! products. For triangular matrices this sum reduces to a single product.

Definition 1. A matrix A is upper triangular if all the entries below the main diagonal are zero, i.e. Aij = 0 if j < i. A matrix A is lower triangular if all the entries above the main diagonal are zero, i.e. Aij = 0 if j > i.

Example 1.

A = is upper triangular

A = is lower triangular

In particular the transpose of an upper triangular matrix is lower triangular and vice versa.

Proposition 1. If A is upper triangular or lower triangular then det(A) is the product of the elements on the main diagonal.

Proof. det(A) = . If A is upper or lower triangular, then only product that doesn't contain an element below the main diagonal is a11a22(ann. //

Example 2.

= (2)(3)(7) = 42

= (2)(3)(7) = 42

Property 2 – A matrix and its transpose have the same determinant.

Example3. A = and AT = ( det(A) = (3)(4) – (1)(2) = 10 and det(A) = (3)(4) – (2)(1) = 10.

Proposition 2. det(AT) = det(A)

Proof. det(A) is a sum/difference of products where each product is obtained by picking an element from each row of A not repeating the choice of columns. However, this is the same as picking an element from each column of A not repeating the choice of rows. However, this is equivalent to picking an element from each row of AT not repeating the choice of columns. So this is one of the products that goes into det(AT). In order to see that a product has the same sign in both det(A) and det(AT), let's look at a typical product in a 4(4 determinant . One such product is (– 1)3a14a21a32a43. This product is multiplied by (- 1)3 since in three pairs of subscripts the second numbers are in the opposite order as the first numbers. These three pairs are

14 and 21

14 and 32

14 and 43

Note that this product also appears in det(AT) as (- )3a21a32a43a14  =  (- 1)3(AT)12(AT)23(AT)34(AT)41. This product is also multiplied by (- 1)3 since in three pairs of subscripts the second numbers are in the opposite order as the first numbers. These three pairs are

12 and 41

23 and 41

34 and 41

Note that each of the three pairs (i, ((i)) and (j, ((j)) in the product for det(A) where i < j and ((i) > ((j) corresponds to a the pair (((j), j) and (((i), i) in the product for det(AT) where ((j) > ((i) and j > i. So there are the same number of pairs with an order reversal in the product for det(A) and the product for det(AT). So the product has the same sign in each case. //

One consequence of this property of determinants is that other properties that hold for rows also automatically hold for columns.

Property 3 – Linearity of Determinants in any Single Row of Column. In the case of 2(2 determinants we saw that the determinant was a linear function of one of the rows if the other row was held fixed. Similarly it was a linear function of one of the columns if the other column was held fixed. This is true in general.

Proposition 2. det(A) is a linear function of any row if the other rows are fixed. det(A) is a linear function of any column if the other columns are fixed.

Proof. The row part follows from the fact that det(A) is a sum/difference of products where each product is obtained by picking an element from each row of A. The column part follows from the row part and the fact that A and AT have the same determinant. //

Often one uses the linearity in three separate ways.

Corollary. (a) If all the elements of any single row or column are zero, then the determinant is zero.

Example 3.

Property 7 – Cofactor Expansions. Recall that Proposition 1 in the previous section said that a determinant is the sum of the elements of the first row times their cofactors. We shall show now that the determinant is in fact equal to the sum of the elements of any row or column times their cofactors.

Proposition 7. If A is an n(n matrix then for any row i and column j one has

det(A) = =

Proof. The column part follows from the row part by taking transposes. To prove the row part we move row i up to row one by interchanging this row with the one before it i-1 times. Thus det(A) = (-1)i-1det(B) where one of B is row i of A and row k of B is row k-1 of A for k = 2, …, i and row k of B is row k of A for k = i+1, …, n. By Proposition 1 of the previous section one has

det(B) =

where N(1,j) is the 1j-minor of B. However B1j = aij and N(1,j) = M(i,j). So

det(A) = (-1)i-1 =

//

Example 3. Find by expanding in cofactors along the second column.

= - (-1) + (-1) - (1)

= (1)(-10) + (-1)(-11) - (1)(-4) = 5

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