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- 1. The Inverse of a Matrix ✒ ✏ ✑7.4 Introduction In number arithmetic every number a = 0 has a reciprocal b written as a−1 or 1 a such that ba = ab = 1. Similarly a square matrix A may have an inverse B = A−1 where AB = BA = I. We develop a rule for finding the inverse of a 2 × 2 matrix (where it exists) and we look at two methods of finding the inverse of a 3 × 3 matrix (where it exists). Non-square matrices do not possess inverses. # ✧ ✥ ✦ Prerequisites Before starting this Block you should . . . x be familiar with the algebra of matrices y be able to calculate a determinant z know what a cofactor is Learning Outcomes After completing this Block you should be able to . . . know the condition for the existence of an inverse matrix use the formula for finding the inverse of a 2 × 2 matrix find the inverse of a 3 × 3 matrix using row operations and using the determinant method Learning Style To achieve what is expected of you . . . allocate sufficient study time briefly revise the prerequisite material attempt every guided exercise and most of the other exercises
- 2. 1. The inverse of a square matrix We know that any non-zero number k has an inverse; for example 2 has an inverse 1 2 or 2−1 . The inverse of the number k is usually written 1 k or, more formally, by k−1 . This numerical inverse has the property that k × k−1 = k−1 × k = 1 We now show that an inverse of a matrix can, in certain circumstances, also be defined. Given an n × n square matrix A, then an n × n square matrix B is said to be the inverse matrix of A if AB = BA = I where I is, as usual, the identity matrix of the appropriate size. Example Show that the inverse matrix of A = −1 1 −2 0 is B = 1 2 0 −1 2 −1 Solution All we need do is to check that AB = BA = I. AB = −1 1 −2 0 × 1 2 0 −1 2 −1 = 1 2 −1 1 −2 0 × 0 −1 2 −1 = 1 2 2 0 0 2 = 1 0 0 1 The reader should check that BA = I also. We make a number of important remarks. • Non-square matrices do not have an inverse. • The inverse of A is usually written A−1 . • Not all square matrices have an inverse. Now do this exercise Consider A = 1 0 2 0 , and let B = a b c d be a candidate for the inverse of A. Find AB and BA. Answer Now do this exercise Equate the elements of AB to those of I = 1 0 0 1 and solve the resulting equations. Answer Hence, we have a contradiction (as would also have been obtained by equating the elements of BA to those of I). The matrix A therefore has no inverse and is said to be a singular matrix. A matrix which has an inverse is said to be non-singular. • If a matrix has an inverse then that inverse is unique. Suppose B and C are both inverses of A. Then, by definition, AB = BA = I and AC = CA = I Consider the two ways of forming the product CAB Engineering Mathematics: Open Learning Unit Level 1 7.4: Matrices 2
- 3. 1. CAB = C(AB) = CI = C 2. CAB = (CA)B = IB = B. Hence B = C and the inverse is unique. • There is no such operation as division in matrix algebra. In arithmetic we have 3−1 × 6 = 1 3 × 6 = 2 = 6 × 1 3 = 6 × 3−1 but in matrix algebra we must write either A−1 B or BA−1 , depending on the order required. • Assuming that the square matrix A has an inverse A−1 then the solution of the system of equations AX = B is found by pre-multiplying both sides by A−1 . A−1 (AX) = A−1 B, changing brackets (A−1 A)X = A−1 B i.e IX = A−1 B, using A−1 A = I finally X = A−1 B which is the solution we seek. 2. The inverse of a 2×2 matrix In this section we show how (if it exists) the inverse of a 2 × 2 matrix can be obtained. Now do this exercise Form the matrix products AB and BA where A = a b c d and B = d −b −c a . Answer You will see that had we chosen C = 1 ad−bc d −b −c a instead of B then both products AC and CA will be equal to I. Hence this matrix C is the inverse of A. Further, if ad − bc = 0 then A has no inverse. (Note that for the matrix A = 1 0 2 0 , which occurred in the last guided exercise, ad − bc = 1 × 0 − 0 × 2 = 0 confirming, as we found, that A has no inverse.) Key Point The Inverse of a 2 × 2 Matrix If ad − bc = 0 then the 2 × 2 matrix A = a b c d has a (unique) inverse given by A−1 = 1 ad − bc d −b −c a Note that ad − bc = |A|, the determinant of the matrix A. In words: to find the inverse of a 2×2 matrix A we effectively interchange the diagonal elements a and d, change the sign of the other two elements and then divide by the determinant of A. 3 Engineering Mathematics: Open Learning Unit Level 1 7.4: Matrices
- 4. Now do this exercise Which of the following matrices has an inverse? A = 1 0 2 3 , B = 1 1 −1 1 , C = 1 −1 −2 2 , D = 1 0 0 1 Answer Now do this exercise Find the inverse of the matrices A, B and D above. Use the keypoint result Answer It can be shown that the matrix A = cos θ sin θ − sin θ cos θ represents an anti-clockwise rotation through an angle θ of points in an xy-plane about the origin. The inverse matrix B represents a rotation clockwise through an angle θ. It is given therefore by B = cos(−θ) sin(−θ) − sin(−θ) cos(−θ) = cos θ − sin θ sin θ cos θ . Now do this exercise Form the products AB and BA for these ‘rotation matrices’. Answer 3. The inverse of a 3×3 matrix - Gauss elimination method It is true, in general, that if the determinant of a matrix is zero then that matrix has no inverse. If the determinant is non-zero then the matrix has a (unique) inverse. In this section and the next section we look at two ways of finding that inverse using a 3 × 3 matrix; larger matrices can be inverted by the same methods - the process is more tedious and takes longer. The 2 × 2 case could be handled similarly but as we have seen we have a simple formula to use. The method we now describe for finding the inverse of a matrix has many similarities to a technique (introduced in Block 8.3 and known as the Gaussian elimination method) used to obtain solutions of simultaneous equations. This method involves operating on the rows of a matrix in order to reduce it to a unit matrix. The Gauss row operations we shall use are (i) interchanging two rows (ii) multiplying a row by a constant factor (iii) adding a multiple of one row to another. Note that in (ii) and (iii) the multiple could be negative or fractional, or both. The Gauss elimination method is outlined in the following keypoint: Engineering Mathematics: Open Learning Unit Level 1 7.4: Matrices 4
- 5. Key Point Matrix inverse − Gauss elimination method We use the result, quoted without proof, that: if a sequence of row operations applied to a square matrix A reduce it to the identity matrix I of the same size then the same sequence of operations applied to I reduce it to A−1 . Three points to note: • If we cannot reduce A to I then A−1 does not exist. This will become evident by the appearance of a row of zeros. • There is no unique route from A to I and it is experience which selects the optimum route. • It is more efficient to do the two reductions, A to I and I to A−1 , simultaneously. Suppose we wish to find the inverse of the matrix A = 1 3 3 1 4 3 2 7 7 . We first place A and I adjacent to each other. 1 3 3 1 4 3 2 7 7 1 0 0 0 1 0 0 0 1 Now proceed by changing the columns of A left to right to reduce A to the form 1 ∗ ∗ 0 1 ∗ 0 0 1 where ∗ can be any number. This form is called an upper triangular form. First we subtract row 1 from row 2 and twice row 1 from row 3. ‘Row’ refers to both matrices. 1 3 3 1 4 3 2 7 7 1 0 0 0 1 0 0 0 1 R2 − R1 R3 − 2R1 ⇒ 1 3 3 0 1 0 0 1 1 1 0 0 −1 1 0 −2 0 1 Now subtract row 2 from row 3 1 3 3 0 1 0 0 1 1 1 0 0 −1 1 0 −2 0 1 R3 − R2 ⇒ 1 3 3 0 1 0 0 0 1 1 0 0 −1 1 0 −1 −1 1 Phase two of the task consists of continuing the row operations to reduce the elements above the leading diagonal to zero. We proceed right to left. We subtract 3 times row 3 from row 1 (the elements in row 2 column 3 is already zero.) 1 3 3 0 1 0 0 0 1 1 0 0 −1 1 0 1 1 1 R1 − 3R3 ⇒ 1 3 0 0 1 0 0 0 1 4 3 −3 −1 1 0 −1 −1 1 5 Engineering Mathematics: Open Learning Unit Level 1 7.4: Matrices
- 6. Finally we subtract 3 times row 2 from row 1. 1 3 0 0 1 0 0 0 1 4 3 −3 −1 1 0 −1 −1 1 R1 − 3R2 ⇒ 1 0 0 0 1 0 0 0 1 7 0 −3 −1 1 0 −1 −1 1 . Then we claim A−1 = 7 0 −3 −1 1 0 −1 −1 1 . (This can be verified by showing that AA−1 and A−1 A both equal I.) Now do this exercise Consider A = 0 1 1 2 3 −1 −1 2 1 , I = 1 0 0 0 1 0 0 0 1 . Use the Gauss elimination approach to obtain A−1 . First interchange rows 1 and 2, then carry out the operation row 3 + 1 2 row 1 Answer Now do this exercise Now carry out the operation row 3 − 7 2 row 2 followed by row 1 − 1 3 row 3 and row 2 + 1 3 row 3. Answer Now do this exercise Next, subtract 3 times row 2 from row 1 then, divide row 1 by 2 and row 3 by (−3). Finally identify A−1 . Answer Hence A−1 = 5 6 1 6 −2 3 −1 6 1 6 1 3 7 6 −1 6 −1 3 = 1 6 5 1 −4 −1 1 2 7 −1 −2 4. The inverse of a 3×3 matrix - determinant method This method which employs determinants, is of importance from a theoretical perspective. The numerical computations involved are too heavy for matrices of higher order than 3 × 3 and in such cases the Gauss elimination approach is prefered. To obtain A−1 using the determinant approach the steps in the following keypoint are followed: Engineering Mathematics: Open Learning Unit Level 1 7.4: Matrices 6
- 7. Key Point Matrix inverse − the determinant method Given a square matrix A: • find |A|. If |A| = 0 then, as we know, A−1 does not exist. If |A| = 0 we can proceed to find the inverse matrix. • replace each element of A by its cofactor (see Block 7.3). • transpose the result to form the adjoint matrix, adj(A) • then A−1 = 1 |A| adj(A). Try each part of this exercise Find the inverse of A = 0 1 1 2 3 −1 −1 2 1 Part (a) First find |A|. Answer Part (b) Now replace each element of A by its minor. Answer Now do this exercise Now attach the signs from the array + − + − + − − − + (so that where a + sign is met no action is taken and where a − sign is met the sign is changed) to obtain the matrix of cofactors. Then transpose the result to obtain the adjoint matrix. Answer Now do this exercise You can now obtain A−1 . Answer 7 Engineering Mathematics: Open Learning Unit Level 1 7.4: Matrices
- 8. More exercises for you to try 1. Find the inverses of the following matrices (a) 1 2 3 4 (b) −1 0 0 4 (c) 1 1 −1 1 2. Use the determinant method and also the Gauss elimination method to find the inverse of the following matrices (a) A = 2 1 0 1 0 0 4 1 2 (b) B = 1 1 1 0 1 1 0 0 1 Answer Engineering Mathematics: Open Learning Unit Level 1 7.4: Matrices 8
- 9. 5. Computer Exercise or Activity For this exercise it will be necessary for you to access the computer package DERIVE. DERIVE can be used to carry out many operations of matrix algebra. Let A be the matrix: A = 2 −1 3 1 −2 1 0 3 −2 To obtain its inverse using DERIVE we would first key in the matrix using Author:Matrix. Then, choosing the correct number of rows and columns, for A imput the matrix. DERIVE will respond #1 : 2 −1 3 1 −2 1 0 3 −2 To obtain its determinant it is advisable to give this matrix a name. To do this, simply go into the Author:Expression menu screen and type A := #1. DERIVE will respond: #2 : A := 2 −1 3 1 −2 1 0 3 −2 Now to obtain the inverse simply key in Author:Expression (A)∧(−1) =. DERIVE will respond; #3 : A−1 = 1 9 7 9 5 9 2 9 −4 9 1 9 1 3 −2 3 −1 3 It would be a useful exercise to check all the inverses obtained in this block by using DERIVE. Also choose two 3 × 3 matrices A and B at random. Check first that they have non-zero determinants and then verify that the property (AB)−1 = (B)−1 (A)−1 is always satisfied. 9 Engineering Mathematics: Open Learning Unit Level 1 7.4: Matrices
- 10. End of Block 7.4 Engineering Mathematics: Open Learning Unit Level 1 7.4: Matrices 10
- 11. AB = a b 2a 2b , BA = a + 2b 0 c + 2d 0 Back to the theory 11 Engineering Mathematics: Open Learning Unit Level 1 7.4: Matrices
- 12. a = 1, b = 0, 2a = 0, 2b = 1. Hence a = 1, b = 0, a = 0, b = 1 2 . Back to the theory Engineering Mathematics: Open Learning Unit Level 1 7.4: Matrices 12
- 13. AB = ad − bc 0 0 ad − bc = (ad − bc)I BA = ad − bc 0 0 ad − bc = (ad − bc)I Back to the theory 13 Engineering Mathematics: Open Learning Unit Level 1 7.4: Matrices
- 14. |A| = 1 × 3 − 0 × 2 = 3; |B| = 1 + 1 = 2; |C| = 2 − 2 = 0; |D| = 1 − 0 = 1. Therefore, A, B and D each have an inverse. C does not because it has a zero determinant. Back to the theory Engineering Mathematics: Open Learning Unit Level 1 7.4: Matrices 14
- 15. A−1 = 1 3 3 0 −2 1 , B−1 = 1 2 1 −1 1 1 , D−1 = 1 0 0 1 = D Back to the theory 15 Engineering Mathematics: Open Learning Unit Level 1 7.4: Matrices
- 16. AB = cos θ sin θ − sin θ cos θ cos θ − sin θ sin θ cos θ = cos2 θ + sin2 θ − cos θ sin θ + sin θ cos θ − sin θ cos θ + cos θ sin θ sin2 θ + cos2 θ = 1 0 0 1 similarly, BA = I effectively: a rotation through an angle θ followed by a rotation through angle −θ is equivalent to zero rotation (θ = 0). Back to the theory Engineering Mathematics: Open Learning Unit Level 1 7.4: Matrices 16
- 17. 0 1 1 2 3 −1 −1 2 1 1 0 0 0 1 0 0 0 1 R1 ↔ R2 ⇒ 2 3 −1 0 1 1 −1 2 1 0 1 0 1 0 0 0 0 1 2 3 −1 0 1 1 −1 2 1 0 1 0 1 0 0 0 0 1 R3 + 1 2 R1 ⇒ 2 3 −1 0 1 1 0 7 2 1 2 0 1 0 1 0 0 0 1 2 1 . Back to the theory 17 Engineering Mathematics: Open Learning Unit Level 1 7.4: Matrices
- 18. 2 3 −1 0 1 1 0 7 2 1 2 0 1 0 1 0 0 0 1 2 1 R3 − 7 2 R2 ⇒ 2 3 −1 0 1 1 0 0 −3 0 1 0 1 0 0 −7 2 1 2 1 2 3 −1 0 1 1 0 0 −3 0 1 0 1 0 0 −7 2 1 2 1 R1 − 1 3 R3 R2 + 1 3 R3 ⇒ 2 3 0 0 1 0 0 0 −3 +7 6 +5 6 −1 3 −1 6 1 6 1 3 −7 2 1 2 1 Back to the theory Engineering Mathematics: Open Learning Unit Level 1 7.4: Matrices 18
- 19. 2 3 0 0 1 0 0 0 −3 7 6 +5 6 −1 3 −1 6 1 6 1 3 −7 2 1 2 1 R1 − 3R2 ⇒ 2 0 0 0 1 0 0 0 −3 10 6 2 6 −4 3 −1 6 1 6 1 3 −7 2 1 2 1 2 0 0 0 1 0 0 0 −3 10 6 2 6 −4 3 −1 6 1 6 1 3 −7 2 1 2 1 R1 ÷ 2 R3 ÷ (−3) ⇒ 1 0 0 0 1 0 0 0 1 5 6 1 6 −2 3 −1 6 1 6 1 3 7 6 −1 6 −1 3 Back to the theory 19 Engineering Mathematics: Open Learning Unit Level 1 7.4: Matrices
- 20. |A| = 0 × 5 + 1 × (−1) + 1 × 7 = 6 Back to the theory Engineering Mathematics: Open Learning Unit Level 1 7.4: Matrices 20
- 21. 3 −1 2 1 2 −1 −1 1 2 3 −1 2 1 1 2 1 0 1 −1 1 0 1 −1 2 1 1 3 −1 0 1 2 −1 0 1 2 3 = 5 1 7 −1 1 1 −4 −2 −2 Back to the theory 21 Engineering Mathematics: Open Learning Unit Level 1 7.4: Matrices
- 22. 5 −1 7 1 1 −1 −4 2 −2 and, transposing, adj(A) = 5 1 −4 −1 1 2 7 −1 −2 . Back to the theory Engineering Mathematics: Open Learning Unit Level 1 7.4: Matrices 22
- 23. A−1 = 1 det(A) adj(A) = 1 6 5 1 −4 −1 1 2 7 −1 −2 as before. Back to the theory 23 Engineering Mathematics: Open Learning Unit Level 1 7.4: Matrices
- 24. 1. (a) −1 2 4 −2 −3 1 (b) −1 0 0 1 4 (c) 1 2 1 −1 1 1 2. (a) A−1 = −1 2 0 −2 1 −2 4 2 0 0 −1 T = −1 2 0 −2 0 −2 4 0 1 2 −1 (b) B−1 = 1 0 0 −1 1 0 0 −1 1 T = 1 −1 0 0 1 −1 0 0 1 Back to the theory Engineering Mathematics: Open Learning Unit Level 1 7.4: Matrices 24