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Transcript
Chapter 2 Section 4 The Inverse Matrix Problem • Find X in the a matrix equation: AX=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! 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) Restriction / Caution • Restriction: – A matrix has to be a square matrix in order to have an inverse!!! • Caution: – Not all square matrices have inverses!!! 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 Example of an Inverse Matrix A= 4 8 7 A– 1 = –2 –3 –2 3 5 4 4 –2 8 –3 7 –2 Then the inverse of A is: 3 5 4 –1 = –2 2 –1 3 –5 4 5 –6 4 Example of a Square Matrix that Does Not have an Inverse A= 1 0 1 3 1 5 2 4 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 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. Exercise 21 (page 98) Use the fact that: 9 0 2 – 20 – 9 – 5 4 0 1 –4 –2 –1 0 5 0 1 –1 Solve: 9x + 2z = 1 – 20x – 9y – 5z + 5w = 0 4x + z = 0 – 4x – 2y – z + w = – 1 Using a matrix equation! = 1 0 –4 0 0 1 0 2 –2 0 0 –5 9 0 1 –9 Exercise 21 continued (a) 9x + 2z = 1 – 20x – 9y – 5z + 5w = 0 4x + z = 0 – 4x – 2y – z + w = – 1 9 – 20 4 –4 0 2 –9 –5 0 1 –2 –1 0 5 0 1 x y z w = 1 0 0 –1 Exercise 21 continued (b) x y z w x y z w = 9 – 20 4 –4 0 –9 0 –2 = 1 0 –4 0 0 1 0 2 2 –5 1 –1 0 5 0 1 –2 0 0 –5 9 0 1 –9 –1 1 0 0 –1 1 0 0 –1 Exercise 21 continued (c) x y z w = 1 5 –4 9 Thus x = 1, y = 5, z = – 4, and w = 9 Exercise 31 (page 98) Using a matrix equation, solve: 2x – 4z + 7z = 11 x + 3y – 5z = – 9 3x – y + 3z = 7 2 1 3 –4 3 –1 A 7 –5 3 x y z • X = = 11 –9 7 B Exercise 31continue (a) So A·X=B X = A– 1 · B = = –1 2 –4 7 1 3 –5 3 –1 3 – 0.8 5.6 5 · = So x = – 4/5, y = 28/5, and z = 5 11 –9 7 – 4/5 28/5 5 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.