# Chapter 8. Matrices II: inverses. 8.1 What is an inverse?

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1 Chapter 8 Matrices II: inverses We have learnt how to add subtract and multiply matrices but we have not defined division. The reason is that in general it cannot always be defined. In this chapter, we shall explore when it can be. All matrices will be square and I will always denote an identity matrix of the appropriate size. 8.1 What is an inverse? The simplest kind of linear equation is ax = b where a and b are scalars. If a 0 we can solve this by multiplying by a 1 on both sides to get a 1 (ax = a 1 b. We now use associativity to get (a 1 ax = a 1 b. Finally, a 1 a = 1 and so 1x = a 1 b and this gives x = a 1 b. The number a 1 is the multiplicative inverse of the non-zero number a. We now try to emulate this approach for the matrix equation Ax = b. We suppose that there is a matrix B such that BA = I. Multiply on the left both sides of our equation Ax = b to get B(Ax = Bb. Because order matters when you multiply matrices, which side you multiply on also matters. Use associativity of matrix mulitiplication to get (BAx = Bb. Now use our assumption that BA = I to get Ix = Bb. Finally, we use the properties of the identity matrix to get x = Bb. 209

2 210 CHAPTER 8. MATRICES II: INVERSES We appear to have solved our equation, but we need to check it. We calculate A(Bb. By associativity this is (ABb. At this point we also have to assume that AB = I. This gives Ib = b, as required. We conclude that in order to copy the method for solving a linear equation in one unknown, our coefficient matrix A must have the property that there is a matrix B such that AB = I = BA. We take this as the basis of the following definition. A matrix A is said to be invertible if we can find a matrix B such that AB = I = BA. The matrix B we call it an inverse of A, and we say that the matrix A is invertible. Observe that A has to be square. A matrix that is not invertible is said to be singular. Example A real number r regarded as a 1 1 matrix is invertible if and only if it is non-zero, in which case an inverse is its reciprocal. Thus our definition of matrix inverse directly generalizes what we mean by the inverse of a number. It s clear that if A is a zero matrix, then it can t be invertible just as in the case of real numbers. However, the next example shows that even if A is not a zero matrix, then it need not be invertible. Example Let A be the matrix ( We shall show that there is no matrix B such that AB = I = BA. Let B be the matrix ( a b c d From BA = I we get a = 1 and a = 0. It s impossible to meet both these conditions at the same time and so B doesn t exist. On the other hand here is an example of a matrix that is invertible.

3 8.1. WHAT IS AN INVERSE? 211 Example Let and A = B = Check that AB = I = BA. We deduce that A is invertible with inverse B. As always, in passing from numbers to matrices things become more complicated. Before going any further, I need to clarify one point which will at least make our lives a little simpler. Lemma Let A be invertible and suppose that B and C are matrices such that AB = I = BA and AC = I = CA. Then B = C. Proof. Multiply AB = I both sides on the left by C. Then C(AB = CI. Now CI = C, because I is the identity matrix, and C(AB = (CAB since matrix multiplication is associative. But CA = I thus (CAB = IB = B. It follows that C = B. The above result tells us that if a matrix A is invertible then there is only one matrix B such that AB = I = BA. We call the matrix B the inverse of A. It is usually denoted by A 1. It is important to remember that we can only write A 1 if we know that A is invertible. In the following, we describe some important properties of the inverse of a matrix. Lemma If A is invertible then A 1 is invertible and its inverse is A. 2. If A and B are both invertible and AB is defined then AB is invertible with inverse B 1 A If A 1,..., A n are all invertible and A 1... A n is defined then A 1... A n is invertible and its inverse is A 1 n... A 1 1.

4 212 CHAPTER 8. MATRICES II: INVERSES Proof. (1 This is immediate from the equations A 1 A = I = AA 1. (2 Show that AB(B 1 A 1 = I = (B 1 A 1 AB. (3 This follows from (2 above and induction. We shall deal with the practical computation of inverse later. Let me conclude this section by returning to my original motivation for introducing an inverse. Theorem (Matrix inverse method. A system of linear equations Ax = b in which A is invertible has the unique solution Proof. Observe that x = A 1 b. A(A 1 b = (AA 1 b = Ib = b. Thus A 1 b is a solution. It is unique because if x is any solution then Ax = b giving and so A 1 (Ax = A 1 b x = A 1 b. Example We shall solve the following system of equations using the matrix inverse method x + 2y = 1 3x + y = 2 Write the equations in matrix form. ( ( 1 2 x 3 1 y = ( 1 2

5 8.2. DETERMINANTS 213 Determine the inverse of the coefficient matrix. In this case, you can check that this is the following A 1 = 1 ( Now we may solve the equations. From Ax = b we get that x = A 1 b. Thus in this case x = 1 ( ( ( = Thus x = 3 and y = 1. Finally, it is always a good idea to check the solutions. 5 5 There are two (equivalent ways of doing this. The first is to check by direct substitution x + 2y = = 1 and 3x + y = = 2 Alternatively, you can check by matrix mutiplication which gives ( ( 1 2 ( You can see that both calculations are, in fact, identical. 8.2 Determinants The obvious questions that arise from the previous section are how do we decide whether a matrix is invertible or not and, if it is invertible, how do we compute its inverse? The material in this section is key to answering both of these questions. I shall define a number, called the determinant, that can be calculated from any square matrix. Unfortunately, the definition is unmotivated but it will justify itself by being extremely useful as well as having interesting properties.

6 214 CHAPTER 8. MATRICES II: INVERSES Let A be a square matrix. We denote its determinant by det(a or by replacing the round brackets of the matrix A with straight brackets. It is defined inductively: this means that I define an n n determinant in terms of (n 1 (n 1 determinants. The determinant of the 1 1 matrix ( a is a. The determinant of the 2 2 matrix ( a b A = c d denoted a b c d is the number ad bc. The determinant of the 3 3 matrix a b c d e f g h i denoted a b c d e f g h i is the number a e h f i b d g f i + c d g e h We could in fact define the determinant of any square matrix of whatever size in much the same way. However, we shall limit ourselves to calculating the determinants of 3 3 matrices at most. It s important to pay attention to the signs in the definition. You multiply alternately by plus one and minus one More generally, the signs are found by computing ( 1 i+j where i is the row of the element and j is the column. In the above definition, we are taking

7 8.2. DETERMINANTS 215 i = 1 throughout. We have defined determinants by expansion along the first row but in fact you can expand them along any row and also along any column. The fact that the same answers arise however you expand the determinant is a first indication of their remarkable properties. Examples = = = = 7 For our subsequent work, I want to single out two properties of determinants. Theorem Let A and B be square matrices having the same size. Then det(ab = det(a det(b. Proof. The result is true in general, but I shall only prove it for 2 2 matrices. Let ( ( a b e f A = and B = c d g h We prove directly that det(ab = det(a det(b. First ( ae + bg af + bh AB = ce + dg cf + dh Thus det(ab = (ae + bg(cf + dh (af + bh(ce + dg. The first bracket multiplies out as acef + adeh + bcgf + bdgh and the second as acef + adfg + bceh + bdgh.

8 216 CHAPTER 8. MATRICES II: INVERSES Subtracting these two expressions we get adeh + bcgf adfg bceh. Now we calculate det(a det(b. This is just which multiplies out to give (ad bc(eh fg adeh + bcfg adfg bceh. Thus the two sides are equal, and we have proved the result. I shall mention one other property of determinants that we shall need when we come to study vectors and which will be useful in developing the theory of inverses. It can be proved in the 2 2 and 3 3 cases by direct verification. Theorem Let A be a square matrix and let B be obtained from A by interchanging any two columns. Then det(b = det(a. An important consequence of the above result is the following. Proposition If two columns of a determinant are equal then the determinant is zero. Proof. Let A be a matrix with two columns equal. Then if we swap those two columns the matrix remains unchanged. Thus by Theorem 8.2.3, we have that det A = det A. It follows that det A = 0. I have defined determinants algebraically in this section. In Chapter 9, I shall also describe determinants geometrically. Exercises Compute the following determinants. (a

9 8.2. DETERMINANTS 217 (b (c (d (e (f (g (h 2. Solve 1 x x = Calculate x cos x sin x 1 sin x cos x 0 cos x sin x

10 218 CHAPTER 8. MATRICES II: INVERSES 4. Prove that a b c d = 0 if, and only if, one column is a scalar multiple of the other. Hint: consider two cases: ad = bc 0 and ad = bc = 0 for this case you will need to consider various possibilities. 8.3 When is a matrix invertible? Recall from Theorem that det(ab = det(a det(b. I use this property below to get a necessary condition for a matrix to be invertible. Lemma If A is invertible then det(a 0. Proof. By assumption, there is a matrix B such that AB = I. Take determinants of both side of the equation to get AB = I det(ab = det(i. By the key property of determinants recalled above and so But det(i = 1 and so In particular, det(a 0. det(ab = det(a det(b det(a det(b = det(i. det(a det(b = 1. Are there any other properties that a matrix must satisfy in order to have an inverse? The answer is, surprisingly, no. Specifically, I shall prove that a square matrix A is invertible if, and only if, det A 0. This solves the

11 8.3. WHEN IS A MATRIX INVERTIBLE? 219 theoretical question concerning inverses whereas the practical one of actually computing inverses is dealt with in the next section. I start with a 2 2 matrix A where everything is easy to calculate. Let ( a b A = c d We construct a new matrix as follows. Replace each entry a ij of A by the element you get when you cross out the ith row and jth column. Thus we get ( d c b a We now use the following matrix of signs ( +, + where the entry in row i and column j is the sign of ( 1 i+j, to get ( d c b a We now take the transpose of this matrix to get the matrix we call the adjugate of A ( d b adj(a = c a The defining characteristic of the adjugate is that Aadj(A = det(ai = adj(aa which can easily be checked. We deduce from the defining characteristic of the adjugate that if det(a 0 then ( A 1 1 d b = det(a c a We have therefore proved the following. Proposition A 2 2 matrix is invertible if and only if its determinant is non-zero.

13 8.3. WHEN IS A MATRIX INVERTIBLE? 221 Proof. We have verified the above result in the case of 2 2 matrices. I shall now prove it in the case of 3 3 matrices by means of an argument that generalizes. Let A = (a ij and we write B = adj(a = We shall compute AB. We have that c 11 c 21 c 31 c 12 c 22 c 32 c 13 c 23 c 33 (AB 11 = a 11 c 11 + a 12 c 12 + a 13 c 13 = det A by expanding the determinant along the top row. The next element is (AB 12 = a 11 c 21 + a 12 c 22 + a 13 c 23. But this is the determinant of the matrix a 11 a 12 a 13 a 11 a 12 a 13 a 31 a 32 a 33 which, having two rows equal, must be zero by Proposition This pattern now continues with all the off-diagonal entries being zero for similar reasons and the diagonal entries all being the determinant. We may now prove the main theorem of this section. Theorem (Existence of inverses. Let A be a square matrix. Then A is invertible if and only if det(a 0. When A is invertible, its inverse is given by A 1 1 = det(a adj(a. Proof. Let A be invertible. By our lemma above, det(a 0 and so we can form the matrix 1 det(a adj(a. We now calculate A 1 det(a adj(a = 1 det(a A adj(a = I

14 222 CHAPTER 8. MATRICES II: INVERSES by our theorem above. Thus A has the advertized inverse. Conversely, suppose that det(a 0. Then again we can form the matrix 1 det(a adj(a and verify that this is the inverse of A and so A is invertible. Example Let A = We show that A is invertible and calculate its inverse. adjugate. The matrix of minors is Calculate first its The matrix of cofactors is The adjugate is the transpose of the matrix of cofactors We now calculate A times it adjugate. We get 5I. It follows that det A = 5. The inverse of A is the adjugate with each entry divided by the determinant of A A 1 =

15 8.3. WHEN IS A MATRIX INVERTIBLE? 223 The Moore-Penrose Inverse We have proved that a square matrix has an inverse if, and only if, it has a non-zero determinant. For rectangular matrices, the existence of an inverse doesn t even come up for discussion. However, in later applications of matrix theory it is very convenient if every matrix have an inverse. Let A be any matrix. We say that A + is its Moore-Penrose inverse if the following conditions hold: 1. A = AA + A. 2. A + = A + AA (A + A T = A + A. 4. (AA + T = AA +. It is not obvious, but every matrix A has a Moore-Penrose inverse A + and, in fact, such an inverse is uniquely determined by the above four conditions. In the case, where A is invertible in the vanilla-sense, its Moore-Penrose inverse is just its inverse. But even singular matrices have Moore-Penrose inverse. You can check that the matrix defined below satisfies the four conditions above ( = 3 6 ( The Moore-Penrose inverse can be used to find approximate solutuions to systems of linear equations that might otherwise have no solution. Exercises Use the adjugate method to compute the inverses of the following matrices. In each case, check that your solution works. ( 1 0 (a 0 2 ( 1 1 (b 1 2

16 224 CHAPTER 8. MATRICES II: INVERSES (c (d (e (f Computing inverses The practical way to compute inverses is to use elementary row operations. I shall first describe the method and then I shall prove that it works. Let A be a square n n matrix. We want to determine whether it is invertible and, if it is, we want to calculate its inverse. We shall do this at the same time and we shall not need to calculate a determinant. We write down a new kind of augmented matrix this time of the form B = (A I where I is the n n identity matrix. The first part of the algorithm is to carry out elementary row operations on B guided by A. Our goal is to convert A into a row echelon matrix. This will have zeros below the leading diagonal. We are interested in what entries lie on the leading diagonal. If one of them is zero we stop and say that A is not invertible. If all of them are 1 then the algorithm continues. We now use the 1 s that lie on the leading diagonal to remove all element above each 1. Our original matrix B now has the following form (I A. I claim that A = A 1. I shall illustrate this method by means of an example. Example Let A = We shall show that A is invertible and calculate its inverse. We first write

17 8.4. COMPUTING INVERSES 225 down the augmented matrix We now carry out a sequence of elementary row operations to get the following where the matrix to the left of the partition is an echelon matrix The leading diagonal contains only 1 s and so our original matrix is invertible. We now use these 1 s to insert zeros above using elementary row operations It follows that the inverse of A is A 1 = At this point, it is always advisable to check that A 1 A = I in fact rather than just in theory. We now need to explain why this method works. An n n matrix E is called an elementary matrix if it is obtained from the n n identity matrix by means of a single elementary row operation. Example Let s find all the 2 2 elementary matrices. The first one is obtained by interchanging two rows and so is ( Next we obtain two matrices by multiplying each row by a non-zero scalar λ ( λ 0 ( λ

18 226 CHAPTER 8. MATRICES II: INVERSES Finally, we obtain two matrices by adding a scalar multiple of one row to another row ( 1 λ ( λ 1 There are now two key results we shall need. Lemma Let B be obtained from A by means of a single elementary row operation ρ. Thus B = ρ(a. Let E = ρ(i. Then B = EA. 2. Each elementary row matrix is invertible. Proof. (1 This has to be verified for each of the three types of elementary row operation. I shall deal with the third class of such operations: R j R j +λr i. Apply this elementary row operation to the n n identity matrix I to get the matrix E. This agrees with the identity matrix everywhere except the jth row. There it has a λ in the ith column and, of course, a 1 in the jth column. We now calculate the effect of E on any suitable matrix A. Then EA will be the same as A except in the jth row. This will consist of the jth row of A to which λ times the ith row of A has been added. (2 Let E be the elementary matrix that arises from the elementary row operation ρ. Thus E = ρ(i. Let ρ be the elementary row operation that undoes the effect of ρ. Thus ρρ and ρ ρ are both identity functions. Let E = ρ (I. Then E E = ρ (E = ρ (ρ(i = I. Similarly, EE = I. It follows that E is invertible with inverse E. Example We give an example of 2 2 elementary matrices. Consider the elementary matrix ( 1 0 E = λ 1 which is obtained from the 2 2 identity matrix by carrying out the elementary row operation R 2 R 2 + λr 1. We now calculate the effect of this matrix when we multiply it into the following matrix ( a b c A = d e f and we get ( EA = a b c λa + d λb + e λc + f

19 8.4. COMPUTING INVERSES 227 But this matrix is what we would get if we applied the elementary row operation directly to the matrix A. We may now prove that our the elementary row operation method for calculating inverses which we described above really works. Proposition If (I B can be obtained from (A I by means of elementary row operations then A is invertible with inverse B. Proof. Let the elementary row operations that transform (A I to (I B be ρ 1,..., ρ n in this order. Thus (ρ n... ρ 1 (A = I and (ρ n... ρ 1 (I = B. Let E i be the elementary matrix corresponding to the elementary row operation ρ i. Then (E n... E 1 A = I and (E n... E 1 I = B. Now the matrices E i are invertible and so A = (E n... E 1 1 and B = E n... E 1. Thus B is the inverse of A as claimed. Exercises Use elementary row operations to compute the inverses of the following matrices. In each case, check that your solution works. ( 1 0 (a 0 2 ( 1 1 (b (c

20 228 CHAPTER 8. MATRICES II: INVERSES (d (e (f The Cayley-Hamilton theorem The goal of this section is to prove a major theorem about square matrices. It is true in general, although I shall only prove it for 2 2 matrices. It provides a first indication of the importance of certain polynomials in studying matrices. Let A be a square matrix. We can therefore form the product AA which we write as A 2. When it comes to multiplying A by itself three times there are apparently two possibilities: A(AA and (AAA. However, matrix multiplication is associative and so these two products are equal. We write this as A 3. In general A n+1 = AA n = A n A. We define A 0 = I, the identity matrix the same size as A. The usual properties of exponents hold A m A n = A m+n and (A m n = A mn. One important consequence is that powers of A commute so that A m A n = A n A m. We can form powers of matrices, multiply them by scalars and add them together. We can therefore form sums like A 3 + 3A 2 + A + 4I. In other words, we can substitute A in the polynomial x 3 + 3x 2 + x + 4 remembering that 4 = 4x 0 and so has to be replaced by 4I.

21 8.5. THE CAYLEY-HAMILTON THEOREM 229 Example Let f(x = x 2 + x + 2 and let A = ( We calculate f(a. Remember that x 2 + x + 2 is really x 2 + x + 2x 0. Replace x by A and so x 0 is replaced by A 0 which is I. We therefore get A 2 + A + 2I and calculating gives ( It is important to remember that when a square matrix A is substituted into a polynomial, you must replace the constant term of the polynomial by the constant term times the identity matrix. The identity matrix you use will have the same size as A. We now come to an important extension of what we mean by a root. If f(x is a polynomial and A is a square matrix, we say that A is a matrix root of f(x if f(a is the zero matrix. Let A be a square n n matrix. Define χ A (x = det(a xi. Observe that x is essentially a complex variable and so cannot be replaced by a matrix. Then χ A (x is a polynomial of degree n called the characteristic polynomial of A. It is worth observing that when x = 0 we get that χ A (0 = det(a, which is therefore the value of the constant term of the characteristic polynomial. Theorem (Cayley-Hamilton. Every square matrix is a root of its characteristic polynomial. Proof. I shall only prove this theorem in the 2 2 case, though it is true in general. Let ( a b A = c d Then from the definition the characteristic polynomial is a x b c d x

22 230 CHAPTER 8. MATRICES II: INVERSES Thus We now calculate χ A (A which is just ( a b c d χ A (x = x 2 (a + dx + (ad bc. 2 ( a b (a + d c d This simplifies to ( ( ad bc 0 0 ad bc which proves the theorem in this case. The general proof uses the adjugate matrix of A xi. There is one very nice application of this theorem. Proposition Let A be an invertible n n matrix. Then the inverse of A can be written as a polynomial in A of degree n 1. Proof. We may write χ A (x = f(x + det(a where f(x is a polynomial with constant term zero. Thus f(x = xg(x for some polynomial g(x of degree n 1. By the Cayley-Hamilton theorem, 0 = Ag(A + det(ai Thus Ag(A = det(ai. Put B = 1 g(a. Then AB = I. But A and B det(a commute since B is a polynomial in A. Thus BA = I. We have therefore proved that A 1 = B. The characteristic polynomial of a matrix plays a fundamental rôle in more advanced matrix theory. The roots of the characteristic polynomial of the matrix A are called the eigenvalues of A. From our point of view this is further evidence for the interconnectedness of mathematics. 8.6 Complex numbers via matrices Consider all matrices that have the following shape ( a b b a where a and b are arbitrary real numbers. You should show first that the sum, difference and product of any two matrices having this shape is also a matrix

23 8.6. COMPLEX NUMBERS VIA MATRICES 231 of this shape. Rather remarkably matrix multiplication is commutative for matrices of this shape. Observe that the determinant of our matrix above is a 2 +b 2. It follows that every non-zero matrix of the above shape is invertible. The inverse of the above matrix in the non-zero case is 1 ( a b a 2 + b 2 b a and again has the same form. It follows that the set of all these matrices satisfies the axioms of high-school algebra. Define ( = 0 1 and i = ( We may therefore write our matrices in the form Observe that a1 + bi. i 2 = 1. It follows that our set of matrices can be regarded as the complex numbers in disguise.

24 232 CHAPTER 8. MATRICES II: INVERSES

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