MA 138 Calculus 2 with Life Science Applications Matrices (Section 9.2)

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1 MA 38 Calculus 2 with Life Science Applications Matrices (Section 92) Alberto Corso albertocorso@ukyedu Department of Mathematics University of Kentucky Friday, March 3, 207

2 Identity Matrix and Inverse of a Matrix For any n, the identity matrix is an n n matrix, denoted by I n, with s on its diagonal line and 0 s elsewhere; that is, I n = 0 0 Property of the Identity Matrix Ṣuppose that A is an m n matrix Then I m A = A = AI n Inverse of a Matrix Suppose that A is an n n square matrix If there exists an n n square matrix B such that AB = I n = BA then B is called the inverse matrix of A and is denoted by A

3 Example (Part I)Checking Verify that: 3 5 A = 2 4 and B = = 2 5/2 3/2 are inverses of each other That is A B = I 2 = B A A 2 = and B 2 = are inverses of each other That is A 2 B 2 = I 3 = B 2 A 2

4 Matrix Representation of Linear Systems We observe that the system of linear equations a x + a 2 x a n x n = b a 2 x + a 22 x a 2n x n = b 2 a m x + a m2 x a mn x n = b m can be written in matrix form as AX = B, where a a 2 a n x b a 2 a 22 a 2n x 2 b 2 = } a m a m2 {{ a mn } x n }{{} } b m {{ } A X B

5 The Guiding Light A simple key observation: To solve 5x = 0 for x, we just divide both sides by 5 ( multiply both sides by /5 = 5 ) That is, 5x = 0 5 5x = 5 0 x = 2 as 5 5 = and 5 0 = 2 We have learnt how to write a system of n linear equations in n variables in the matrix form AX = B To solve AX = B, we therefore need an operation that is analogous to multiplication by the reciprocal of A We have defined, whenever possible, a matrix A that serves this function (ie, A A = Identity Matrix) Then, whenever possible, we can write the solution of AX = B as AX = B A AX = A B X = A B

6 Example (Part II) Using the results verified in Example (Part I) and our Guiding Light ( Principle), solve the following systems of linear equations by transforming them into matrix form { 3x + 5y = 7 2x + 4y = 6 3x + 5y z = 0 2x y + 3z = 9 4x + 2y 3z =

7 Properties of Matrix Inverses The following properties of matrix inverses are often useful Properties of Matrix Inverses Suppose A and B are both invertible n n matrices then A is unique; (A ) = A; (AB) = B A ; (A T ) = (A ) T

8 How do we find the inverse (if possible) of a matrix? First of all the matrix has to be a square matrix! 3 5 Suppose n = 2 For example, A = 2 4 x y We need to find a matrix B = such that AB = I 2 = BA z w 3x + 5z 3y + 5w 0 AB = I 2 = 2x + 4z 2y + 4w 0 { { 3x + 5z = 3y + 5w = 0 and 2x + 4z = 0 2y + 4w = /2 row reduce /2

9 Warning (using the other condition) 3 5 Consider again the matrix A = 2 4 x y We need to find a matrix B = such that AB = I 2 = BA z w Suppose we impose instead the condition BA = I 2 3x + 2y 5x + 4y 0 BA = I 2 = 3z + 2w 5z + 4w 0 { { 3x + 2y = 3z + 2w = 0 and 5x + 4y = 0 5z + 4w = row reduce /2 3/2 Morale: We work with the transpose of A and of A

10 General Method for finding (if possible) the inverse Let A be an n n matrix Finding a matrix B with AB = I n results in n linear systems, each consisting of n equations in n unknowns The corresponding augmented matrices have the same matrix A on their left side and a column of 0 s and a single on their right side By solving these n systems simultaneously, we can speed up the process of finding the inverse matrix To do so, we construct the augmented matrix A I n We row reduce to obtain, if possible, the augmented matrix I n B A row reduce B The matrix B, if it exists, is the inverse A of A

11 Example 2 Find the inverse of the 3 3 matrix A =

12 General Formula for a 2x2 Matrix a b a a 2 For simplicity we write A = instead of A = c d a 2 a 22 a b 0 Construct the augmented matrix c d 0 Perform the Gaussian Elimination Algorithm Set = ad bc a R b a a 0 b a a 0 a R 2 c d 0 b a 0 c a 0 a R 2 cr R b a R 2 0 d bc a c a 0 d 0 c b a

13 The Inverse of a 2 2 Matrix a b Let A = be a 2 2 matrix c d We define det(a) = ad bc A is invertible ( nonsingular) if and only if det(a) 0 In particular, A = d b det(a) c a Looking back at the formula for A, where A is a 2 2 matrix whose determinant is nonzero, we see that, to find the inverse of A we divide by the determinant of A, switch the diagonal elements of A, change the sign of the off-diagonal elements If the determinant is equal to 0, then the inverse of A does not exist

14 Example 3 Find the inverse of the matrix 5 A = 2 7 B = 0

15 The determinant can be defined for any n n matrix The general formula is computationally complicated for n 3 We mention the following important result Part (2) below will be of particular interest to us in the near future Theorem Suppose that A is an n n matrix, and X and 0 are n matrices Then A is invertible ( nonsingular) if and only if det(a) 0 The matrix equation ( system of linear equations) AX = 0 has a nontrivial solution A is singular det(a) = 0

16 Example 4 Find the solution of the following matrix equations ( systems of linear equations) 2 x 0 = 3 5 y 0 4 x y = 0