3.2 - 3.2 Properties of Determinants Theorem Let A be a square matrix E1 be a replacement elementary matrix E2 be an interchange elementary matrix

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 3.2 Properties of Determinants Theorem : Let A be a square matrix, E1 be a replacement elementary matrix E2 be an interchange elementary matrix E3 be a scaling elementary matrix then detE1A = detA detE2A = −detA detE3A = k detA EXs: (1) 1 3 30 52 2 4 1 R2 ← R2 + 2R1 R3 ← R3 − 3R1 R4 ← R4 − R1 → 13 01 00 0 7 2 8 27 = −30 1 0 30 17 2 8 −2 −5 7 0 −4 2 −5 0 −4 2 −6 1 −1 2 −4 R4 ← R4 − R3 R3 ← R3 + 4R2 → 0 0 30 0 −1 1 (2) 24 11 6 0 12 11 3 0 1 1 −1 4 = −2 1 00 1 1 −1 = 2 1 1 −1 1 1 −1 = −2 1 2 11 (3) 111 123 111 123 0 3 = −2 0 1 111 000 1 1 1 = 0 0 0 =− 1 2 3 =0 Theorem A n × n (square) matrix A is invertible ⇔ detA = 0 Question: Why? Answer: Because taking A to RE form U , suppose there are r interchanges. Then detA = (−1)r detU = (−1)r · product of diagonal entries So detA = 0 iff detU = 0 iff there are r pivots iff A is invertible. Question: What about detE for an elementary matrix? 2 Answer: E1 replacement matrix E2 scaling matrix E3 interchange matrix =⇒ E1 triangular so detE1 = product of diagonal entries = 1 E2 diagonal matrix so detE2 = k (the scaling factor) E3 identity matrix with 2 rows switched so detE3 = −1 so detEi A = detEi · detA ( ∗) EX: −1 2 3 0 3430 5466 4243 R3 ← R3 − 2R4 → −1 2 34 42 30 30 43 −3 0 −2 0 3 −1 2 = 3C44 = 3(−1)4+4 34 3 3 −3 0 −2 R2 ← R2 − 2R1 → −3 0 −2 ˆ = 3 · 2C12 = −6 5 −3 −3 −2 = −6(−10 − 9) = 114 −1 2 =3 3 5 0 −3 Theorem detA = detAT Many ways to show this. We’ll just assume it is true. Multiplicative Property: detAB = detA · detB Why is this true? Case 1: n × n A is not invertible Then AB is not invertible, so 0 = det(AB ) = 0 · detB = detA · detB. Case 2: A is invertible, then its RRE is U = In where −1 − − − A = Ep 1 Ep−1 · · ·E2 1 E1 1 for E1 E2 · · ·Ep A = U = I , so by using (*) for p times −1 − − − detAB = det(Ep 1 Ep−1 · · ·E2 1E1 1 )B −1 − − − = detEp 1 detEp−1 · · ·detE2 1 detE1 1detB = detA · detB 4 Note: For invertible A, detA−1 detA = detI = 1 ⇒ detA−1 = 1/detA Warning: det(A + B ) = detA + detB in general! Note: If one can form the row echelon form U for A using only replacements (that is, without interchanges and without scaling any equations), then detA = detU = product of pivot elements. If one needs r interchanges to get U , then detA = (−1)r product of pivot elements. This is how detA can be computed when needed, which 2n3 flops for n × n it rarely is. The computation cost is ∼ 3 matrix A. EX: Let A and B be 4 × 4 matrices, with detA = −1 and detB = 2. Compute: a. detAB e. detB −1AB b. detB 5 c. det2A d. detAT A Answer: a. By the multiplicative property of determinants, detAB = (detA)(detB ) = −1 × 2 = −2. b. Similarly, detB 5 = (detB )5 = 25 = 32. 5 c. Since det2A = 24 detA = 16 × (−1) = −16. d. detAT A = (detAT )(detA) = (detA)(detA) = (−1) × (−1) = 1. e. detB −1 AB = (detB −1)(detA)(detB ) = (1/detB )(detA)(detB ) = detA = −1. . 20 0 2 2A = (2I )A = 0 0 00 00 0 0 A, 2 0 02 6 Geometric Interpretation in R2 Determinants of 2 × 2 matrices A = area of parallelogram = |ad − bc|. y (a + c, b + d) ab cd ) (a, b (c, d) x ∴ detA = 0 iff A is singular iff there is no parallelogram. EX: : A = 11 10 −1 −1 1 1 01 10 A= A= 7 Geometric Interpretation in R3 z y r1 r1 + r2 + r3 r3 r2 x r 1 A = r2 r3 Volume of parallelepiped = detA When is detA = 0? Some applications of determinants • to tell when A is invertible (detA = 0) • to solve Ax = b (Cramer’s Rule - only practical in very special cases) • to find eigenvalues • to use for solving differential equations 8 ...
View Full Document

This note was uploaded on 04/09/2011 for the course MATH 232 taught by Professor Russel during the Spring '10 term at Simon Fraser.

Ask a homework question - tutors are online