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
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 Spring '10
 Russel
 Linear Algebra, Algebra, Determinant, Matrices, Invertible matrix, Diagonal matrix, Det, detA

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