det(ka) = k n det A.

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1 Properties of determinants Theorem. If A is n n, then for any k, det(ka) = k n det A. Multiplying one row of A by k multiplies the determinant by k. But ka has every row multiplied by k, so the determinant gets multiplied by k n times. Theorem. If A contains a row (or column) of zeros, then det A = 0. Simply expand the determinant along that row or column. Every term in the sum will be zero. M1300 Vector Geometry and Linear Algebra 1

2 Theorem. If A has two equal rows, or two equal columns, then det A = 0. To see this, suppose row i and row j of A are equal. Then apply the row operation of switching them: A R i R j A, and det(a ) = det(a). But A = A, so det(a) = det(a). This means det(a) = 0. Corollary. If one row of A is a multiple of another row (two rows are proportional), then det(a) = 0. For if row j is k times row i where k 0, then we can apply A R j 1/k B. Then det(b) = det(a), and now B has two equal rows, so det(b) = 0 by the theorem. M1300 Vector Geometry and Linear Algebra 2

3 Corollary. If any row (column) of A is a linear combination (see pp. 27, 31) of other rows (columns) of A, then det(a) = 0. Proof. Let the rows of A be a 1, a 2,..., a n (see (9) on p. 30). If, for example, a n = c 1 a c n 1 a n 1, then apply the elementary row operations R j c i R i for i = 1, 2,..., n 1. This will make no change in the determinant, but the resulting matrix A will have a row of zeros. So det(a) = det(a ) = 0. M1300 Vector Geometry and Linear Algebra 3

4 Theorem. A and A T have the same determinant: det(a T ) = det A. The theorem is obvious for 2 2 matrices:» T! a b det = c d a c a b d = ad cb = ad bc = c b d M1300 Vector Geometry and Linear Algebra 4

5 For a 3 3 matrix, 2 a b 3 c det 4d e f5 = g h i and 2 a d 3 g det 4b e h5 = c f i a b c d e f g h i a d g b e h c f i = +aei + bfg + cdh ceg afh bdi = +aei + dhc + gbf gec ahf dbi In general, because the transpose changes rows into columns, expansion along row i of A gives the same sum as expansion down column i of A T. (You have to check that the cofactor C ij for A T is the equal to the cofactor C ji for A). M1300 Vector Geometry and Linear Algebra 5

6 Theorem. A is invertible if and only if det A 0. Proof. We know from Section 2.1 that A adj(a) = det(a)i, so if det(a) 0, we have 1 A det(a) adj(a) = I So A 1 1 = det(a) adj(a). For the reverse, i.e., to show that if A is invertible, then det(a) 0, we use Gaussian elimination. From Chapter 1, we know that if A is invertible, then A can be transformed to I with elementary row operations: A I. However, recall the effects of row operations on determinants: if A B, then det(b) is equal to cdet(a), where c is either a nonzero multiplier (in the case R i c), or is 1 (in the case R i R j ), or is 1 (in the case R i + kr j ). So if det(a) = 0, we would have det(b) = 0. But det(i) = 1 0, so det(a) 0. M1300 Vector Geometry and Linear Algebra 6

7 Combining this with Theorem ( Equivalent Statements ) we now have seven equivalent ways to say invertible : Theorem. If A is an n n matrix, the following are equivalent (that is, if any of them is true, they are all true, and if any of them is false, they are all false): (a) A is invertible. (b) Ax = 0 has exactly one solution (the trivial one). (c) The RREF of A is I n. (d) A is a product of elementary matrices. (e) Ax = b has a solution for every possible b. (I.e., the system is consistent). (f) Ax = b has exactly one solution for every possible b. (g) det(a) 0. M1300 Vector Geometry and Linear Algebra 7

8 Theorem. det(ab) = (det A)(det B). This is one of the most important properties for determinants. To prove it, we proceed to first prove a very simple case, namely when A is elementary. Lemma. If E is an elementary matrix, then det(ea) = (det E)(det A). This is just a matter of using the known effects of EROs on determinants, from Section 2.2. Since det(i) = 1, we have: If I R i k E, then det(e) = k det(i) = k, and A R i k EA, so det(ea) = k det(a) = det(e) det(a). If I R i R j E, then det(e) = det(i) = 1, and A R i R j EA, so det(ea) = det(a) = det(e) det(a). M1300 Vector Geometry and Linear Algebra 8

9 If I R i +kr j E, then det(e) = det(i) = 1, and A R i +kr j EA, so det(ea) = det(a) = det(e) det(a). Now, to prove the theorem for all matrices A, we have two cases. If A is singular, then also AB is singular, by Theorem Then det(ab) = 0 and det(a) = 0, so det(a) det(b) = 0. Thus, both sides are 0. If A is invertible, then we can write it as a product of elementary matrices: A = E 1 E 2 E 3 E n M1300 Vector Geometry and Linear Algebra 9

10 Then we apply the lemma repeatedly, starting on the left: det(a) = det(e 1 E 2 E 3 E n ) = det(e 1 )det(e 2 E 3 E n ) = det(e 1 ) det(e 2 ) det(e 3 E n ) = = det(e 1 ) det(e 2 ) det(e 3 ) det(e n ) Similarly, det(ab) = det(e 1 E 2 E n B) = det(e 1 ) det(e 2 E 3 E n B) = det(e 1 )det(e 2 )det(e 3 E n B) = = det(e 1 )det(e 2 )det(e 3 ) det(e n ) det(b) = det(a) det(b) M1300 Vector Geometry and Linear Algebra 10

11 When B = A, we have det(a 2 ) = (det(a)) 2, det(a 3 ) = (det(a)) 3, etc. If A is invertible, then det(a) 0, so we have the following. Theorem. If A is invertible, then det(a 1 ) = (det(a)) 1 = 1 det(a) Proof. Since AA 1 = I and det(i) = 1, so 1 = det(aa 1 ) = det(a) det(a 1 ), and since det(a 1 ) 0, we can divide both sides by det(a 1 ). M1300 Vector Geometry and Linear Algebra 11

12 Since A n = (A 1 ) n, we can summarize these facts as follows. Theorem. For any integer n, det(a n ) = ` det A n provided that if n < 0 then we assume A is invertible. M1300 Vector Geometry and Linear Algebra 12

13 Example: Find all possible values of k such that the matrix is invertible. A = k k 4 We need to determine all k for which det(a) is not zero. Expanding by the first column: 1 1 k det(a) = k 4 = (1) 2 1 k k 4 (0) + (3) = (8 k) + 3(1 2k) = 7 11k So A is invertible if and only if 7 11k 0, i.e., if k M1300 Vector Geometry and Linear Algebra 13

14 Example: Given A, let R = A ki. The values of k for which R is singular (= not invertible) are called eigenvalues of A. Find the eigenvalues of R = A ki =» » » k 0 0 k =» 1 k k Then det(r) = (1 k) 2 4 = 1 2k + k 2 4 = k 2 2k 3. So we need to find all k that are solutions of x 2 2x 3 = 0. Factoring, we have x 2 2x 3 = (x 3)(x + 1), so the roots are x = 3, 1. Thus, the eigenvalues of A are 3 and 1. M1300 Vector Geometry and Linear Algebra 14

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