1. Chapter 2 Section 4 The Inverse Matrix

2. Problem Find X in the a matrix equation: A X = B (recall that A is the matrix of the coefficients, X is the matrix of the variables, and B is the matrix of the constants). There is no division when dealing with matrices! We will use the inverse matrix!

3. Definition Given a matrix A, then A – 1 is the inverse matrix of matrix A if and only if: A  A – 1 = A – 1  A = I (where I is the identity matrix)

4. Restriction / Caution Restriction: –A matrix has to be a square matrix in order to have an inverse!!! Caution: –Not all square matrices have inverses!!!

5. How to Find an Inverse Matrix We will let the calculator do all the work for us. Enter the matrix using the matrix editor, and then exit to the home screen. Call up the matrix that you want the inverse for, then hit the [ x -1 ] key. The resulting matrix (if there is one) is the inverse matrix

6. Example of an Inverse Matrix A = 4– 2 3 8– 3 5 7 – 2 4 Then the inverse of A is: A – 1 = 4– 2 3 8– 3 5 7 – 2 4 – 1 = – 2 2 – 1 3 – 5 4 5 – 6 4

7. Example of a Square Matrix that Does Not have an Inverse A = 1 3 2 0 1 4 1 5 10 does not have an inverse! The following message will appear on your calculator if you try to take the inverse of the above matrix: [A] – 1. ERR: SINGULAR MAT 1:Quit 2:Goto

8. WARNING The inverse matrix will solve a matrix equation ONLY when there is a unique solution to the system of equations. If there is a possibility that there is either an infinite number of solutions or no solution, then use the Gauss-Jordan elimination method (rref) as described in sections 1 and 2.

9. Exercise 21 (page 98) Use the fact that: 9 0 2 0 – 20 – 9 – 5 5 4 0 1 0 – 4 – 2 – 1 1 1 0 – 2 0 0 1 0 – 5 – 4 0 9 0 0 2 1 – 9 –1 = Solve: 9x + 2z = 1 – 20x – 9y – 5z + 5w = 0 4x + z = 0 – 4x – 2y – z + w = – 1 Using a matrix equation!

10. Exercise 21 continued (a) 9 0 2 0 – 20 – 9 – 5 5 4 0 1 0 – 4 – 2 – 1 1 = 9x + 2z = 1 – 20x – 9y – 5z + 5w = 0 4x + z = 0 – 4x – 2y – z + w = – 1 xyzwxyzw 1 0 – 1

11. Exercise 21 continued (b) 9 0 2 0 – 20 – 9 – 5 5 4 0 1 0 – 4 – 2 – 1 1 = xyzwxyzw 1 0 – 1 xyzwxyzw = 1 0 – 2 0 0 1 0 – 5 – 4 0 9 0 0 2 1 – 9 1 0 – 1

12. Exercise 21 continued (c) = xyzwxyzw 1 5 – 4 9 Thus x = 1, y = 5, z = – 4, and w = 9

13. Exercise 31 (page 98) 2 – 4 7 1 3 – 5 3 – 1 3 = xyzxyz 11 – 9 7 2x – 4z + 7z = 11 x + 3y – 5z = – 9 3x – y + 3z = 7 Using a matrix equation, solve: AX = B

14. Exercise 31continue (a) So A · X = B X = A – 1 · B = 2– 4 7 13 – 5 3 – 1 3 – 1 · 11 – 9 7 = – 0.8 5.6 5 = – 4/5 28/5 5 So x = – 4/5, y = 28/5, and z = 5

15. A Repeat of the WARNING If the system of equations has the either an infinite number of solutions or no solutions, then there will not be an inverse matrix ( i.e. The ERR: SINGULAR MAT error in the calculator). Unfortunately the error message will not tell you which of the two situations caused the error message; for that you will have to use the Gauss-Jordan elimination method to make the determination.