Calculus and linear algebra for biomedical engineering Week 4: Inverse matrices and determinants

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1 Calculus and linear algebra for biomedical engineering Week 4: Inverse matrices and determinants Hartmut Führ Lehrstuhl A für Mathematik, RWTH Aachen October 30, 2008

2 Overview 1 Matrix-by-matrix multiplication 2 Inverse matrices 3 Computing inverses 4 Determinants

3 Motivation Suppose that we have variables x 1,..., x m, y 1,..., y n, z 1,..., z k, which are related by the following linear equations: with suitable matrices A, B. x = Ay, y = B z, Is there a way to directly express the dependence of x on z as x = C z, in terms of a matrix C? The answer is yes, and the matrix C is the matrix product of A and B.

4 Matrix multiplication Denition. Let A R m n and B R n k. Denote the columns of B by b1,..., b k. Then the matrix AB R m k is dened as AB = (Ab1 Ab2... Ab k). Hence the jth column of AB is obtained by multiplying A with the jth column of B.

5 Properties of matrix multiplication Theorem. Let A, A R m n and B, B R n k. 1 If 0m,n denotes the zero m n matrix, then 0 k,ma = 0 k,n, A0 n,k = 0 m,k. 2 Distributive laws: A(B + B ) = AB + AB and (A + A )B = AB + A B. 3 Associativity: For all C R k l : A(BC) = (AB)C. 4 Note: Matrix multiplication is not commutative! If we regard vectors x R n as n 1-matrices, matrix-by-vector multiplication is a special case of matrix-by-matrix multiplication. Recall motivation: If x = Ay and y = B z, how does x depend on z? By associativity: x = A(B z) = (AB)z

6 Identity matrix Denition. For m N let the (m m-) identity matrix I m be dened by I m =... R m m Then I m x = x, for all x R m. As a consequence, for arbitrary A R n m : A = I n A = AI m.

7 Inverse matrix Denition. Let A R n n. Then B R n n is called inverse of A, denoted B = A 1, if AB = BA = I n. If A has an inverse, A is called invertible or regular. Theorem. Let A, B R n n. (a) If AB = I n or BA = I n, then B = A 1. (b) If A and B are invertible, then AB is invertible with (AB) 1 = B 1 A 1.

8 Invertible matrices and linear systems of equations Theorem. Let A R n n. Then the following are equivalent: (a) A is invertible. (b) For every vector y R n, the system Ax = y has a unique solution. (c) The linear system Ax = 0 has rank n. Sketch of proof: (a) (b): Dene x = A 1 y. Then Ax = A(A 1 y) = I n y = y. (b) (a): For i = 1,..., n, nd the solution x i of Ax i = e i, with e i = (0,..., 0, 1, 0,..., 0) T (entry 1 at ith position), and let X = (x1... xn). Then AX = (Ax1... Axn) = (e1... en) = I n.

9 Matrix inversion and systems of linear equations Recall: Inverting a matrix A R n n amounts to simultaneously solving n linear systems of equations Ax i = e i. Here the coecient matrix is always the same, only the right hand side changes. Linear systems are systematically solved by Gauss elimination. The steps in the Gauss elimination only depend on the coecient matrix. The Gauss algorithm should be useful for the computation of inverse matrices.

10 Matrix inversion via Gauss algorithm Let A R n n be given. 1 Introduce the extended matrix à = (A I n ) R n 2n. 2 Using Gauss elimination, transform the extended matrix to à 1 = (B C), where B R n n has upper triangular form. Make sure to apply all basic transformations to the right hand side also. 3 If B has rank < n, the matrix A is not invertible. 4 If B has rank n, apply basic transformations to à 1 to obtain Then C = A 1. à 2 = (I n C ).

11 Inverting a 4 4 matrix We wish to invert A via the Gauss algorithm, where A = Following the usual scheme, we get à =

12 Inverting a 4 4 matrix Now the entries above the diagonal are also removed using basic transformations

13 Inverting a 4 4 matrix Hence: A 1 =

14 General denition of determinant Denition. Let A R (n+1) (n+1), and 1 j n + 1. Then the submatrix A j R n n is obtained by deleting the rst column and jth row of A. Denition. Let A = (a i,j ) i,j be a matrix. 1 For A = (a) R 1 1, we dene det(a) = a. 2 Let det : R n n R be already dened. We then dene det : R (n+1) (n+1) R as follows: det(a) = a 1,1 det(a 1 ) a 2,1 det(a 2 ) ±... + ( 1) n a n+1,1 det(a n+1 n+1 = ( 1) j 1 a j,1 det(a j ) j=1

15 Example: Computing a 2 2-determinant Let A = ( a c b d ) R 2 2 Then, in the notation of the denition, the submatrices A 1 and A 2 are given as A 1 = (d) R 1 1, A 2 = (b) R 1 1 and thus det(a) = det ( a c b d ) = ad cb.

16 Example: Computing a 3 3 determinant Let A = Using the denition, and the formula for 2 2 determinants, ( ) ( ) det(a) = 1 det ( ) ( 1) 3 2 = 1 (15 12) 3(35 24) 1(14 12) = 32.

17 Determinants and invertibility Theorem. (Cramer's rule) Let A = (a1... an) R n n, with det(a) 0. Let y R n. For i = 1,..., n, dene B i by replacing the ith column of A by y, B i = (a1... a i 1 y a i+1 an) R n n. Then the vector x = (x 1,..., x n ) T, with x i = det(b i) det(a), is the unique solution of the linear equation Ax = y. Consequence. Let A R n n. Then A is invertible if and only if det(a) 0.

18 Example: Solving a 2 2 system with Cramer's rule We want to solve the equation ( ) x = ( 8 4 ). The coecient matrix has determinant 4, thus there is a unique solution x = (x 1, x 2 ) T, with ( ) 8 3 det 4 8 x 1 = = = 19 and ( ) 2 8 det 4 4 x 2 = = 40 = Note: For high dimensions, Cramer's rule is not an ecient way to solve linear systems!

19 Computational rules for determinants Theorem. Let A R n n, A = (a i,j ) i,j=1...,n (a) Let A T = (b i,j ) i,j=1...n, where b i,j = a j,i. Then det(a T ) = det(a). (b) If A = BC for suitable matrices B, C R n n, then det(a) = det(b)det(c). (c) Let A be of the form A = ( A 1 B 0 n k,k A 2 with A 1 R k k, A 2 R (n k) (n k), and C R k (n k). Then det(a) = det(a 1 )det(a 2 ). (d) If B is upper triangular with diagonal entries d 1,..., d n, then det(b) = d 1 d 2... d n. ),

20 Determinants via basic transformations Theorem. Let A, B R n be such that B is obtained from A by a basic transformation. (a) If B is obtained by swapping two rows, then det(b) = det(a). (b) If B is obtained by multiplication of a row with a constant c, then det(b) = c det(a). (c) If B is obtained by adding a multiple of a column to another column, then det(b) = det(a). The same rules apply if B is obtained by applying a basic transformation to columns of A.

21 Example: Using basic transformations Recall the matrix A from above A = Using basic transformations, we can compute det(a) as ( ) = det 13 9 = = 32. det(a) = det

22 Example: 4 4 determinant Consider the matrix A = Then, after swapping rows det(a) = det = det ( using det(b) = det(b T ))

23 Example: 4 4 determinant... = 2det = 2det = 8det ( ) = 8(9 2) = 56. (denition of det) (swapping columns) (denition of det)

24 Example: 4 4-determinant using block matrices Again consider the matrix A = Swapping columns transforms A into a block matrix: det(a) = det = det ( ) det = (9 2) ( 8) = 56 ( )

25 Summary Matrix-by-matrix multiplication and its properties. Identity matrix Invertible matrices Computing inverse matrices by the Gauss algorithm Denition and basic properties of determinants Determinants and invertibility. Cramer's rule. Computing determinants: Gauss algorithm, block matrices, etc.

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