6.4 inverse matrices t
- 1. Example A: Find the inverse matrix A-1 if Inverse Matrices a. A = 1 2 1 3 Extend 1 2 1 3 to 1 2 1 0 1 3 0 1 Apply row operations to tranform it so the identity matrix I is on the left side: 1 2 1 0 1 3 0 1 (-1)R1 Add to R2 -1 -2 -1 0 1 2 1 0 0 1 -1 1 0 -2 2 -2 1 0 3 -2 0 1 -1 1 (-2)R2 Add to R1 = identity matrix Hence A-1 = 3 -2 -1 1 . One checks easily that AA-1 = A-1A = 1 0 0 1 .
- 2. Inverse Matrices b. A = 0 2 0 3 Extend 0 2 0 3 to 0 2 1 0 0 3 0 1 Since the first column consists of 0's only, there is no way we may introduce a 1 at the a11 position by row operations. Hence A-1 doesn't exist. c. A = 0 1 -1 1 -1 2 1 1 0 Extend 0 1 -1 1 -1 2 1 1 0 0 1 -1 1 0 0 1 -1 2 0 1 0 1 1 0 0 0 1
- 3. Inverse Matrices Apply row operations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 (-1)R2 add to R1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 -1 -1 -2 0 -1 0 0 1 1 1 0 0 1 1 2 0 1 0 0 0 -2 0 -1 1 0 -1 -1 -1 0 0 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 -2 0 -1 1 (-1)R1 add to R3 0 1 0 1 -1/2 1/2 1 0 0 -1 1/2 1/2 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 1 0 1/2 -1/2 0 0 -1 0 -1/2 1/2 (-1)R3 + R1 (-1/2)R3 R3 + R2 0 0 1 0 1/2 -1/2 = A-1We check that A-1*A = A*A-1 = I. Let’s use an inverse matrix to solve a matrix equation.
- 4. Example B: a. Write the following system as a matrix equation If we let A = 1 2 1 3 x + 2y = 4 x + 3y = -1 , X = x y , The system may be expressed as 1 2 1 3 x y = 4 -1 then the matrix equation is simply AX = B. and B = 4 -1 . An Application of Inverse Matrices To solve for the matrix X, we use the following inverse matrix method.
- 5. To solve the equation AX = B for the matrix X, we multiply to the left of both sides by A-1 (if it exists) AX = B A-1AX = A-1B IX = A-1B or that X = A-1B b. Solve the matrix equation AX = B from part a. by the inverse matrix method. from the previous example we've that Given that A = 1 2 1 3 3 -2 -1 1 . A-1 = An Application of Inverse Matrices
- 6. Multiply A-1 to the left on both sides of the equation Inverse Matrices 1 2 1 3 x y = 4 -1 we get 3 -2 -1 1 1 2 1 3 x y = 3 -2 -1 1 4 -1 0 1 1 0 x y = 14 -5 Hence x y = 14 -5 or that x = 14, and y = -5. With inverse matrices, we may do matrix algebra that is similar to the “traditional” algebra.
- 7. Inverse Matrices Exercise A. Find the inverse matrix of each of the following matrices, if it exits 1 0 0 1 1) 2) 3) 4)1 5 0 1 a 5 0 𝑏 0 𝑏 𝑎 5 3 5 –2 1 5) 6) 7) 8) 1 –2 7 –1 𝑎 𝑏 𝑏 𝑎 𝑎 𝑏 𝑎 𝑏 9) 10) 11) 12) 1 0 0 0 1 0 0 0 1 1 2 3 0 1 4 0 0 1 a 0 0 0 b 0 0 0 c a 2 3 0 b 4 0 0 c 13) 14) 1 0 2 0 1 0 −1 0 1 1 2 3 0 1 4 −1 2 1
- 8. 2. Verify the following inverse formula for 2x2 matrices. 𝑎 𝑏 𝑐 𝑑 –1 = 1 ad – bc 𝑑 –𝑏 –𝑐 𝑎 (Hint: you don’t have to compute the inverse, just check that their product is I so it must be the one and only inverse.) 3. Verify the that (AB) –1 = B –1 A–1. So if A and B are invertible then AB is invertible. Ex B. A matrix that has an inverse is said to be invertible. 1. (A–1 is unique) Verify that if A is invertible, then there is only one A–1. This means to show that if there are matrices B and C that invert A so that AB = BA = I and AC = CA = I, then B = C. 4. Verify that if AB is invertible, then A and B must be invertible. (Hint: Let C is (AB) –1, then A–1 or B –1 can be “solved”.) So if A or B is not invertible then AB is not invertible. Inverse Matrices
- 9. (Answers to the odd problems) Exercise A. 𝐴−1 = 1 0 0 1 1) 𝐴−1 = 1/𝑎 −5/𝑎𝑏 0 1/𝑏 𝐴−1 = 1 13 1 −5 2 3 5) 𝐴−1 = 1 𝑎2 − 𝑏2 𝑎 −𝑏 −𝑏 𝑎 9) 𝐴−1 = 1 0 0 0 1 0 0 0 1 𝐴−1 = 1/a 0 0 0 1/b 0 0 0 1/c 13) 𝐴−1 = 1 3 1 0 −2 0 3 0 1 0 1 3) 7) 11) Inverse Matrices