2.6 The Inverse of a Square Matrix

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1 200/2/6 page CHAPTER 2 Matrices and Systems of Linear Equations i i 2i 5 A = A = i i 4 + i i + i + 5i 26 The Inverse of a Square Matrix In this section we investigate the situation when, for a given n n matrix A, there exists a matrix B satisfying AB = I n and BA = I n (26) and derive an efficient method for determining B (when it does exist) As a possible application of the existence of such a matrix B, consider the n n linear system Ax = b (262) Premultiplying both sides of (262) by an n n matrix B yields Assuming that BA = I n, this reduces to (BA)x = Bb x = Bb (26) Thus, we have determined a solution to the system (262) by a matrix multiplication Of course, this depends on the existence of a matrix B satisfying (26), and even if such a matrix B does exist, it will turn out that using (26) to solve n n systems is not very efficient computationally Therefore it is generally not used in practice to solve n n systems However, from a theoretical point of view, a formula such as (26) is very useful We begin the investigation by establishing that there can be at most one matrix B satisfying (26) for a given n n matrix A Theorem 26 Let A be an n n matrix Suppose B and C are both n n matrices satisfying AB = BA = I n, (264) AC = CA = I n, (265) respectively Then B = C Proof From (264), it follows that C = CI n = C(AB) That is, C = (CA)B = I n B = B, where we have used (265) to replace CA by I n in the second step Since the identity matrix I n plays the role of the number in the multiplication of matrices, the properties given in (26) are the analogs for matrices of the properties xx =, x x =, which holds for all (nonzero) numbers x It is therefore natural to denote the matrix B in (26) by A and to call it the inverse of A The following definition introduces the appropriate terminology

2 200/2/6 page 6 26 The Inverse of a Square Matrix 6 DEFINITION 262 Let A be an n n matrix If there exists an n n matrix A satisfying AA = A A = I n, then we call A the matrix inverse to A, orjustthe inverse of A We say that A is invertible if A exists Invertible matrices are sometimes called nonsingular, while matrices that are not invertible are sometimes called singular Remark It is important to realize that A denotes the matrix that satisfies AA = A A = I n It does not mean /A, which has no meaning whatsoever 2 0 Example 26 If A = 2, verify that B = is the inverse of A 0 Solution: By direct multiplication, we find that AB = 2 = 00 = I 0 00 and BA = 2 = 00 = I 0 00 Consequently, (26) is satisfied, hence B is indeed the inverse of A We therefore write 0 A = 0 We now return to the n n system of Equations (262) Theorem 264 If A exists, then the n n system of linear equations Ax = b has the unique solution for every b in R n x = A b

3 200/2/6 page CHAPTER 2 Matrices and Systems of Linear Equations Proof We can verify by direct substitution that x = A b is indeed a solution to the linear system The uniqueness of this solution is contained in the calculation leading from (262) to (26) Our next theorem establishes when A exists, and it also uncovers an efficient method for computing A Theorem 265 An n n matrix A is invertible if and only if rank(a) = n Proof If A exists, then by Theorem 264, any n n linear system Ax = b has a unique solution Hence, Theorem 259 implies that rank(a) = n Conversely, suppose rank(a) = n We must establish that there exists an n n matrix X satisfying AX = I n = XA Let e, e 2,, e n denote the column vectors of the identity matrix I n Since rank(a) = n, Theorem 259 implies that each of the linear systems Ax i = e i, i =, 2,,n (266) has a unique solution x i Consequently, if we let X =[x, x 2,,x n ], where x, x 2,,x n are the unique solutions of the systems in (266), then that is, A[x, x 2,,x n ]=[Ax,Ax 2,,Ax n ]=[e, e 2,,e n ]; We must also show that, for the same matrix X, AX = I n (26) XA = I n Postmultiplying both sides of (26) by A yields That is, (AX)A = A A(XA I n ) = 0 n (268) Now let y, y 2,,y n denote the column vectors of the n n matrix XA I n Equating corresponding column vectors on either side of (268) implies that Ay i = 0, i =, 2,,n (269) But, by assumption, rank(a) = n, and so each system in (269) has a unique solution that, since the systems are homogeneous, must be the trivial solution Consequently, each y i is the zero vector, and thus XA I n = 0 n Therefore, XA = I n (260) Notice that for an n n system Ax = b, if rank(a) = n, then rank(a # ) = n

4 200/2/6 page The Inverse of a Square Matrix 65 Equations (26) and (260) imply that X = A We now have the following converse to Theorem 264 Corollary 266 Let A be an n n matrix If Ax = b has a unique solution for some column n-vector b, then A exists Proof If Ax = b has a unique solution, then from Theorem 259, rank(a) = n, and so from the previous theorem, A exists Remark In particular, the above corollary tells us that if the homogeneous linear system Ax = 0 has only the trivial solution x = 0, then A exists Other criteria for deciding whether or not an n n matrix A has an inverse will be developed in the next three chapters, but our goal at present is to develop a method for finding A, should it exist Assuming that rank(a) = n, letx, x 2,,x n denote the column vectors of A Then, from (266), these column vectors can be obtained by solving each of the n n systems Ax i = e i, i =, 2,,n As we now show, some computation can be saved if we employ the Gauss-Jordan method in solving these systems We first illustrate the method when n = In this case, from (266), the column vectors of A are determined by solving the three linear systems Ax = e, Ax 2 = e 2, Ax = e The augmented matrices of these systems can be written as A 0, A 0, A 0 0, 0 0 respectively Furthermore, since rank(a) = by assumption, the reduced row-echelon form of A is I Consequently, using elementary row operations to reduce the augmented matrix of the first system to reduced row-echelon form will yield, schematically, A 0 ERO 0 which implies that the first column vector of A is a x = a 2 a Similarly, for the second system, the reduction A 0 ERO 0 00a 00a 2, 00a 00b 00b 2 00b

5 200/2/6 page CHAPTER 2 Matrices and Systems of Linear Equations implies that the second column vector of A is b x 2 = b 2 b Finally, for the third system, the reduction A 0 00c 0 ERO 00c 2 00c implies that the third column vector of A is x = Consequently, a b c A =[x, x 2, x ]= a 2 b 2 c 2 a b c The key point to notice is that in solving for x, x 2, x we use the same elementary row operations to reduce A to I We can therefore save a significant amount of work by combining the foregoing operations as follows: A 00 00a b c 00 ERO 00a 2 b 2 c a b c The generalization to the n n case is immediate We form the n 2n matrix [A I n ] and reduce A to I n using elementary row operations Schematically, c c 2 c [A I n ] ERO [I n A ] This method of finding A is called the Gauss-Jordan technique Remark Notice that if we are given an n n matrix A, we likely will not know from the outset whether rank(a) = n, hence we will not know whether A exists However, if at any stage in the row reduction of [A I n ] we find that rank(a) < n, then it will follow from Theorem 265 that A is not invertible Example 26 Find A if A = Solution: Using the Gauss-Jordan technique, we proceed as follows

6 200/2/6 page 6 26 The Inverse of a Square Matrix 6 Thus, A = We leave it as an exercise to confirm that AA = A A = I A ( ) 2 A 2 ( ), A 2 ( 2) M ( /4) 4 A ( ), A 2 ( 2) Example 268 Continuing the previous example, use A to solve the system x + x 2 + x = 2, x 2 + 2x =, x + 5x 2 x = 4 Solution: The system can be written as Ax = b, where A is the matrix in the previous example, and 2 b = 4 Since A is invertible, the system has a unique solution that can be written as x = A b Thus, from the previous example we have x = = 4 Consequently, x = 5, x 2 =, and x = 2, so that the solution to the system is ( 5, ), 2 We now return to more theoretical information pertaining to the inverse of a matrix Properties of the Inverse The inverse of an n n matrix satisfies the properties stated in the following theorem, which should be committed to memory: 5 2 Theorem 269 Let A and B be invertible n n matrices Then A is invertible and (A ) = A 2 AB is invertible and (AB) = B A A T is invertible and (A T ) = (A ) T

7 200/2/6 page CHAPTER 2 Matrices and Systems of Linear Equations Proof The proof of each result consists of verifying that the appropriate matrix products yield the identity matrix We must verify that A A = I n and AA = I n Both of these follow directly from Definition We must verify that (AB)(B A ) = I n and (B A )(AB) = I n We establish the first equality, leaving the second equation as an exercise We have We must verify that (AB)(B )(A ) = A(BB )A = AI n A = AA = I n A T (A ) T = I n and (A ) T A T = I n Again, we prove the first part, leaving the second part as an exercise First recall from Theorem 222 that A T B T = (BA) T Using this property with B = A yields A T (A ) T = (A A) T = I T n = I n The proof of property 2 of Theorem 269 can easily be extended to a statement about invertibility of a product of an arbitrary finite number of matrices More precisely, we have the following Corollary 260 Let A,A 2,,A k be invertible n n matrices Then A A 2 A k is invertible, and (A A 2 A k ) = A k A k A Proof The proof is left as an exercise (Problem 28) Some Further Theoretical Results Finally, in this section, we establish two results that will be required in Section 2 and also in a proof that arises in Section 2 Theorem 26 Let A and B be n n matrices If AB = I n, then both A and B are invertible and B = A Proof Let b be an arbitrary column n-vector Then, since AB = I n,wehave A(Bb) = I n b = b Consequently, for every b, the system Ax = b has the solution x = Bb But this implies that rank(a) = n To see why, suppose that rank(a) < n, and let A denote a rowechelon form of A Note that the last row of A is zero Choose b to be any column

8 200/2/6 page The Inverse of a Square Matrix 69 n-vector whose last component is nonzero Then, since rank(a) < n, it follows that the system A x = b is inconsistent But, applying to the augmented matrix [A b ] the inverse row operations that reduced A to row-echelon form yields [A b] for some b Since Ax = b has the same solution set as A x = b, it follows that Ax = b is inconsistent We therefore have a contradiction, and so it must be the case that rank(a) = n, and therefore that A is invertible by Theorem 265 We now establish that 8 A = B Since AB = I n by assumption, we have A = A I n = A (AB) = (A A)B = I n B = B, as required It now follows directly from property of Theorem 269 that B is invertible with inverse A Corollary 262 Let A and B be n n matrices If AB is invertible, then both A and B are invertible Proof If we let C = B(AB) and D = AB, then AC = AB(AB) = DD = I n It follows from Theorem 26 that A is invertible Similarly, if we let C = (AB) A, then CB = (AB) AB = I n Once more we can apply Theorem 26 to conclude that B is invertible Exercises for 26 Key Terms Inverse, Invertible, Singular, Nonsingular, Gauss-Jordan technique Skills Be able to check directly whether or not two matrices A and B are inverses of each other Be able to find the inverse of an invertible matrix via the Gauss-Jordan technique Be able to use the inverse of a coefficient matrix of a linear system in order to solve the system Know the basic properties related to how the inverse operation behaves with respect to itself, multiplication, and transpose (Theorem 269) True-False Review For Questions 0, decide if the given statement is true or false, and give a brief justification for your answer If true, you can quote a relevant definition or theorem from the text If false, provide an example, illustration, or brief explanation of why the statement is false An invertible matrix is also known as a singular matrix 8 Note that it now makes sense to speak of A, whereas prior to proving in the preceding paragraph that A is invertible, it would not have been legal to use the notation A

9 200/2/6 page 0 0 CHAPTER 2 Matrices and Systems of Linear Equations 2 Every square matrix that does not contain a row of 5 zeros is invertible 9 A = 2 26 A linear system Ax = b with an n n invertible coefficient matrix A has a unique solution 00 4 If A is a matrix such that there exists a matrix B with 0 A = 00 AB = I n, then A is invertible 02 5 If A and B are invertible n n matrices, then so is 42 A + B A = If A and B are invertible n n matrices, then so is AB 2 If A is an invertible matrix such that A 2 = A, then A 2 A = 26 2 is the identity matrix 4 8 If A is an n n invertible matrix and B and C are n n i 2 matrices such that AB = AC, then B = C A = + i 2i 2 2i 5 9 If A isa5 5matrix of rank 4, then A is not invertible 0 If A isa6 6 matrix of rank 6, then A is invertible 2 4 A = 2 4 Problems For Problems verify by direct multiplication that the 2 given matrices are inverses of one another [ ] [ ] 5 A = A =,A = [ ] [ ] 49 2 A =,A = 4 6 A = A = 2,A = Let 2 4 A = 5 2 For Problems 4 6, determine A, if possible, using the Gauss-Jordan method If A exists, check your answer by verifying that AA = I n [ ] 2 4 A = [ + i 5 A = i [ i 6 A = + i 2 [ ] 00 A = A = ] ] Find the second column vector of A without determining the whole inverse For Problems 8 22, use A to find the solution to the given system x + x 2 =, 2x + 5x 2 = x + x 2 2x = 2, x 2 + x =, 2x + 4x 2 x = x 2ix 2 = 2, (2 i)x + 4ix 2 = i

10 200/2/6 page 26 The Inverse of a Square Matrix 2 22 x + 4x 2 + 5x =, 2x + 0x 2 + x =, 4x + x 2 + 8x = x + x 2 + 2x = 2, x + 2x 2 x = 24, 2x x 2 + x = 6 An n n matrix A is called orthogonal if A T = A For Problems 2 26, show that the given matrices are orthogonal [ ] 0 2 A = 0 [ ] /2 /2 24 A = /2 /2 [ cos α sin α 25 A = sin α cos α 26 A = + 2x 2 ] 2x 2x 2 2x 2x 2 2x 2x 2 2x 2 Complete the proof of Theorem 269 by verifying the remaining properties in parts 2 and 28 Prove Corollary 260 For Problems 29 0, use properties of the inverse to prove the given statement 29 If A is an n n invertible symmetric matrix, then A is symmetric 0 If A is an n n invertible skew-symmetric matrix, then A is skew-symmetric Let A be an n n matrix with A 4 = 0 Prove that I n A is invertible with (I n A) = I n + A + A 2 + A 2 Prove that if A, B, C are n n matrices satisfying BA = I n and AC = I n, then B = C If A, B, C are n n matrices satisfying BA = I n and CA = I n, does it follow that B = C? Justify your answer 4 Consider the general 2 2 matrix [ ] a a A = 2 a 2 a 22 and let = a a 22 a 2 a 2 with a 0 Show that if 0, A = [ ] a22 a 2 a 2 a The quantity defined above is referred to as the determinant of A We will investigate determinants in more detail in the next chapter 5 Let A be an n n matrix, and suppose that we have to solve the p linear systems Ax i = b i, i =, 2,,p where the b i are given Devise an efficient method for solving these systems 6 Use your method from the previous problem to solve the three linear systems if Ax i = b i, i =, 2, A = 2 4, b =, 6 2 b 2 = 2, b = 5 2 Let A be an m n matrix with m n (a) If rank(a) = m, prove that there exists a matrix B satisfying AB = I m Such a matrix is called a right inverse of A (b) If A = [ ], 24 determine all right inverses of A For Problems 8 9, reduce the matrix [A I n ] to reduced row-echelon form and thereby determine, if possible, the inverse of A A = A is a randomly generated 4 4 matrix For Problems 40 42, use built-in functions of some form of technology to determine rank(a) and, if possible,a 5 40 A =

11 200/2/6 page 2 2 CHAPTER 2 Matrices and Systems of Linear Equations Consider the n n Hilbert matrix 4 A = [ ] H n =, i, j n i + j 42 A is a randomly generated 5 5 matrix 4 For the system in Problem 2, determine A and use it to solve the system (a) Determine H 4 and show that it is invertible (b) Find H 4 and use it to solve H 4 x = b if b = [2,,, 5] T 2 Elementary Matrices and the LU Factorization We now introduce some matrices that can be used to perform elementary row operations on a matrix Although they are of limited computational use, they do play a significant role in linear algebra and its applications DEFINITION 2 Any matrix obtained by performing a single elementary row operation on the identity matrix is called an elementary matrix In particular, an elementary matrix is always a square matrix In general we will denote elementary matrices by E If we are describing a specific elementary matrix, then in keeping with the notation introduced previously for elementary row operations, we will use the following notation for the three types of elementary matrices: Type : P ij permute rows i and j in I n Type 2: M i (k) multiply row i of I n by the nonzero scalar k Type : A ij (k) add k times row i of I n to row j of I n Example 22 Write all 2 2 elementary matrices Solution: From Definition 2 and using the notation introduced above, we have Permutation matrix: P 2 = 2 Scaling matrices: M (k) = Row combinations: A 2 (k) = [ ] 0 0 [ ] k 0, M 0 2 (k) = [ ] 0, A k 2 (k) = [ ] 0 0 k [ ] k 0 We leave it as an exercise to verify that the n n elementary matrices have the following structure:

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