Determinant 3.4 Application of Determinants
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1 Determinant 3.4 Application of Determinants September 25
2 Goals We will discuss three applications of determinants. We we give formula to compute the inverse A 1 of an invertible matrix A. We will give the (formula) to solve systems of linear equations. Give formulas to compute area of a triangle and volume of a tetrhedron.
3 Science is a wonderful thing if one does not have to earn one s living at it. - Albert Einstein
4 Definition of Adjoint Adjoint and Definition. Let A be a square matrix of order n. In section 3.1 we define cofactors C ij. Define the cofactor matrix C as C = C 11 C 12 C 13 C 1n C 21 C 22 C 13 C 2n C 31 C 32 C 33 C 3n C n1 C n2 C n3 C nn
5 Adjoint and Continued: Definition of Adjoint The Adjoint matrix of A is defined to be the transpose C T of the cofactor matrix C: C 11 C 21 C 31 C n1 C 12 C 22 C 32 C n2 Adj(A) = C T = C 13 C 23 C 33 C n3 C 1n C 2n C 3n C nn
6 Theorem 3.10 Adjoint and Theorem 3.10 Let A be a square matrix. Then, A[Adj(A)] = [Adj(A)]A = det(a)i n So, det(a) 0 = A 1 = 1 det(a) Adj(A). Proof. Expanding by first row, we have: a 11 C 11 +a 12 C a 1n C 1n = det(a) which is the (1, 1)-entry in the following product:
7 Continued Adjoint and A[Adj(A)] = = a 11 a 12 a 1n a 21 a 22 a 2n a n1 a n2 a nn A???????? = C 11 C 21 C n1 C 12 C 22 C n2 C 1n C 2n C nn A??? A??? A The i th -diagonal entry is obtained by expanding det(a), along i th -row.
8 Continued Adjoint and Now I want to compute the element at the (1,2) position. It is = a 11 C 21 +a 12 C a 1n C 2n Consider the following matrix: a 11 a 12 a 1n a 11 a 12 a 1n B = a 31 a 32 a 3n and A = a n1 a n2 a nn a 11 a 12 a 1n a 21 a 22 a 2n a n1 a n2 a nn Here B is obtained from A by replacing the second row of A by its first row.
9 Continued Adjoint and So, the cofactors of entries of the second row of both matrices are same, which are that of A: C 21,C 22,C 23,...,C 2n. So, the (2,1) entry of the product A[Adj(A)] is = a 11 C 21 +a 12 C a 1n C 2n a 11 a 12 a 1n a 11 a 12 a 1n = a 31 a 32 a 3n = 0 (first 2 rows are equal). a n1 a n2 a nn
10 Continued Adjoint and A[Adj(A)] = A 0?? A??? A Similarly, all the other off-diagonal entries are zero. So A 0 0 A[Adj(A)] = 0 A 0 = A I n. 0 0 A
11 Continued Adjoint and So, ( ) 1 A A Adj(A) = I n Similarly, using column expansion for determiant, we have ( ) 1 A Adj(A) A = I n So, The proof is complete. A 1 = 1 A Adj(A)
12 Example 1 Adjoint and Compute [ the cofactor ] matrix, Adjoint and inverse of 7 12 A =. The Cofactor Matrix [ 3 33 ] [ ] ( 1) C = ( 1) ( 1) ( 1) 2+2 =. [ 7 ] Adj(A) = C T = Also the determinant 3 7 A = = 195. So A 1 = 1 A Adj(A) = 1 [ ] [ = ]
13 Example 2 Adjoint and Find the cofactor matrix, adjoint and inverse of A = First, we compute all the 9 cofactors : C 11 = ( 1) = 4,C 12 = ( 1) = 6, C 13 = ( 1) = 2.
14 Continued Adjoint and C 21 = ( 1) C 23 = ( 1) C 32 = ( 1) = 6,C 22 = ( 1) = 26, = 12,C 31 = ( 1) = 2 = 12,C 33 = ( 1) = 6.
15 Continued Adjoint and So, cofactor matrix and adjoint of A is: C = ,Adj(A) = C T = So, A 1 = 1 A Adj(A) = =
16 Foreword: Adjoint and Cramer s rule gives a formula to solve systems of n linear equations in n variables when the system has exactly one solution. Recall that such a system has exactly one solution if and only if the determinant of the coefficient matrix A 0.
17 Reading assignment Adjoint and Reading Assignment: 3.4 Example 1, 2
18 Theorem 3.11: Adjoint and Theorem: Suppose Ax = B is a system of n equations in n variables. Assume the A 0. Then, x 1 = A 1 A,x 2 = A 2 A,,x n = A n A where A i is obtained from A by replacing the i th row of A by the right hand side B, For example, b 1 a 12 a 1n a 11 b 1 a 1n b 1 a 12 a 1n a 11 b 1 a 1n A 1 = b 3 a 32 a 3n,A 2 = a 31 b 3 a 3n, b n a n2 a nn a n1 b n a nn
19 Proof. Adjoint and Since A 0, the coefficient matrix A is invertible and A 1 = 1 Adj(A). We have Ax = B implies A A 1 Ax = A 1 B = x = A 1 B = 1 A Adj(A)B (Read = as implies ) So, x 1 X = x 2 = 1 A Adj(A)B x n
20 Continued Adjoint and x 1 x 2 x n = 1 A C 11 C 21 C n1 C 12 C 22 C n2 C 1n C 2n C nn b 1 b 2 b n = 1 A A 1 A 3 A n Last equality is obtained by column expansion of determinants. The proof is complete Reading Assignment: 3.4 Example 3, 4
21 Example 3. Adjoint and Use Cramer s rule to solve the following linear system 2x 1 3x 2 +2x 3 = 9 x 1 +x 2 +x 3 = 3 4x 1 +x 2 x 3 = 11 The determiant of the coefficient matrix A = = 25
22 Continued Adjoint and x 1 = x 2 = A A = = 2.8 = = 3
23 Continued Adjoint and x 3 = A = = 2.8
24 Example 4 Adjoint and Use Cramer s rule to solve the following linear system { 2x 3y = 9 x +y = 3 The determiant of the coefficient matrix A = = 5
25 Continued Adjoint and x = A y = A = 0 5 = 0 = 15 5 = 3
26 Example 5. Adjoint and Use Cramer s rule to solve the following linear system x 1 +4x 2 +6x 3 = 3 x 1 +2x 2 +2x 3 = 1 x 1 +2x 2 +4x 3 = 2 The determiant of the coefficient matrix A = = 4
27 Continued Adjoint and x 1 = x 2 = A A = 0 4 = 0 = 0 4 = 0
28 Continued Adjoint and x 3 = A = 2 4 =.5
29 Area of Adjoint and Determiant can be used to compute area of triangles and volume of tetrahedorns. (x 3,y 3 ) (x 2,y 2 ) 4 (x 1,y 1 )
30 Formula for Area of Adjoint and Formula: Area of a triangle whose vertices are (x 1,y 1 ),(x 2,y 2 ),(x 3,y 3 ) is give by Area = ± 1 1 x 1 y x 2 y 2 1 x 3 y 3 Final answer is obviously non-negative, and we adjust ± accordingly. Remark. These three points are colinear, if the above determinant is zero. Reading Assignment: 3.4 Example 3, 4
31 Exercise 6 Adjoint and Find the area of the triangle with vertices at (1, 9),(2,3)(5, 7). Solution: Area = ± x 1 y 1 1 x 2 y 2 1 x 3 y 3 = ± = ±1 ( 46) = 23(unit)2 2
32 Adjoint and Formula for volume of tetrahedron Formula: Recall a tetrhedron had four sides and four vertices. The volume of a tetrahedron whose vertices are (x 1,y 1,z 1 ),(x 2,y 2,z 2 ),(x 3,y 3,z 3 ),(x 4,y 4,z 4 ) is give by Volume = ± 1 1 x 1 y 1 z 1 1 x 2 y 2 z x 3 y 3 z 3 1 x 4 y 4 z 4 Final answer is obviously non-negative, and we adjust ± accordingly.
33 Continued Adjoint and Remark. These four points are coplaner, if the above determinant is zero. Reading Assignment: 3.4 Example 7, 8
34 Exercise 7 Adjoint and Find the volume of the tetrahedron with vertices at (1, 7,2),(2,2,3)( 4, 7,1),(1, 1,2). Solution: Vo; ume = ± x 1 y 1 z 1 1 x 2 y 2 z 2 1 x 3 y 3 z 3 1 x 4 y 4 z 4 = ± = ± 1 ( 24) = 4 (unit)3 6
35 : 3.4 Exercise 5, 19, 25, 26, 27, 29, 38, 39, 50, 51
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