Copyright © 2009 Pearson Education, Inc. CHAPTER 9: Systems of Equations and Matrices 9.1 Systems of Equations in Two Variables 9.2 Systems of Equations in Three Variables 9.3 Matrices and Systems of Equations 9.4 Matrix Operations 9.5 Inverses of Matrices 9.6 Determinants and Cramer’s Rule 9.7 Systems of Inequalities and Linear Programming 9.8 Partial Fractions

Copyright © 2009 Pearson Education, Inc. 9.5 Inverses of Matrices  Find the inverse of a square matrix, if it exists.  Use inverses of matrices to solve systems of equations.

Slide Copyright © 2009 Pearson Education, Inc. The Identity Matrix

Slide Copyright © 2009 Pearson Education, Inc. Example For find each of the following. a) AI b) IA

Slide Copyright © 2009 Pearson Education, Inc. Inverse of a Matrix For an n  n matrix A, if there is a matrix A  1 for which A  1 A = I = A A  1, then A  1 is the inverse of A. Verify that is the inverse of. We show that BA = I = AB.

Slide Copyright © 2009 Pearson Education, Inc. Finding an Inverse Matrix To find an inverse, we first form an augmented matrix consisting of A on the left side and the identity matrix on the right side. Then we attempt to transform the augmented matrix to one of the form The 2  2 identity matrix The 2  2 matrix A

Slide Copyright © 2009 Pearson Education, Inc. Example Find A  1, where A =.

Slide Copyright © 2009 Pearson Education, Inc. Example continued Thus, A  1 =.

Slide Copyright © 2009 Pearson Education, Inc. Notes If a matrix has an inverse, we say that it is invertible, or nonsingular. When we cannot obtain the identity matrix on the left using the Gauss-Jordan method, then no inverse exists.

Slide Copyright © 2009 Pearson Education, Inc. Solving Systems of Equations Matrix Solutions of Systems of Equations For a system of n linear equations in n variables, AX = B, if A is an invertible matrix, then the unique solution of the system is given by X = A  1 B. Since matrix multiplication is not commutative in general, care must be taken to multiply on the left by A  1.

Slide Copyright © 2009 Pearson Education, Inc. Example Use an inverse matrix to solve the following system of equations: 3x + 4y = 5 5x + 7y = 9 We write an equivalent matrix, AX = B: In the previous example we found A  1 =

Slide Copyright © 2009 Pearson Education, Inc. Example continued We now have X = A  1 B. The solution of the system of equations is (  1, 2).