# EC9A0: Pre-sessional Advanced Mathematics Course

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1 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 1 of 55 EC9A0: Pre-sessional Advanced Mathematics Course Slides 1: Matrix Algebra Peter J. Hammond & Pablo F. Beker Department of Economics University of Warwick Autumn 2013

2 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 2 of 55 Slides Outline 1 Solving Equations Substitution Gaussian Elimination 2 Matrices Matrices and Their Transposes Matrix Multiplication Special Kind of Matrices Linear versus Affine Functions Quadratic Forms 3 Determinants Definition Rules for Determinants Inverse Matrix Cramer s Rule

3 For a system like the one above, we are interested in: 1 Does a solution exists? 2 How many solutions are there? There are essentially three ways of solving such systems: 1 substitution University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 3 of 55 System of Equations Solving Equations We are interested in solving system of equations like: a 11 x 1 +a 12 x a 1n x n = b 1 a 21 x 1 +a 22 x a 2n x n = b 2... a m1 x 1 +a m2 x a mn x n = b m where each a ij is called the coefficient of the unknown x j in the i-th equation. Definition A solution to the system of equations is an n tuple of real numbers [x 1 x 2... x n ] which satisfies each of the equations.

4 Solving Equations Substitution University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 4 of 55 Substitution: Two Equations in Two Unknowns Consider the system of equations: x + y = 10 x y = 6 Use the second equation to write y in terms of x, y = x 6. (1) Substitute this expression for y into the 1st equation: x + (x 6) = 10 = x = 8 Use (1) to obtain, y = 2

5 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 5 of 55 Solving Equations Substitution Gaussian Elimination: Two Equations in Two Unknowns More generally, one can: 1 add the first equation to the second, to eliminate y; 2 obtain y from the first equation. This leads to the following transformation } x + y = 10 = x y = 6 { x + y = 10 2x = = x + y = 10 x = Obviously the solution is x = 1 2 (10 + 6) = 8, y = (10 + 6) = 2

6 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 6 of 55 Solving Equations Using Matrix Notation, I Gaussian Elimination Matrix notation allows the two equations 1x + 1y = 10 1x 1y = 6 to be expressed as ( ) ( ) 1 1 x = 1 1 y or as Az = b, where ( ) 1 1 A =, z = 1 1 ( ) 10 6 ( ) x, and b = y ( ) A is the coefficient matrix; 2 z is the vector of unknowns; 3 b is the vector of right-hand sides.

7 Solving Equations Using Matrix Notation, II Gaussian Elimination ( ) ( ) 1 1 x = 1 1 y ( ) ( ) ( ) x = y ( ) ( ) 1 1 x = 0 2 y ( ) 10 6 ( ( ) 10 4 ) ( ) 10 6 ( ) ( ) ( ) ( ) ( ) x = y ( ) ( ) ( ) 1 1 x 10 = 0 1 y 2 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 7 of 55

8 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 8 of 55 Solving Equations Using Matrix Notation, III Gaussian Elimination Also, the solution x = 1 2 (10 + 6), y = 1 2 (10 6) can be expressed as x = or as z = ( ) x y = ( y = ) ( ) 10 6 = Cb, where C = ( ).

9 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 9 of 55 Solving Equations Gaussian Elimination Gaussian Elimination: Two General Equations Consider the general system ax + by = u = 1u + 0v cx + dy = v = 0u + 1v of two equations in two unknowns, filled in with 1s and 0s. In matrix form, these equations can be written as ( ) ( ) ( ) ( ) a b x 1 0 u =. c d y 0 1 v In case a 0, we can eliminate x from the second equation by adding c/a times the first row to the second. After defining the scalar D := a[d + ( c/a)b] = ad bc, ( ) ( ) ( ) ( ) a b x 1 0 u = 0 D/a y c/a 1 v

10 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 10 of 55 Solving Equations Gaussian Elimination Gaussian Elimination: Two General Equations-Subcase 1A In Subcase 1A when D := ad bc 0, multiply the second row by a to obtain ( ) ( ) ( ) ( ) a b x 1 0 u = 0 D y c a v Adding b/d times the second row to the first yields ( ) ( ) ( ) ( ) a 0 x 1 + (bc/d) ab/d u = 0 D y c 1 v Recognizing that 1 + (bc/d) = (D + bc)/d = ad/d, then dividing the two rows/equations by a and D respectively, we obtain ( ) ( ) ( ) x 1 0 x = = 1 ( ) ( ) d b u y 0 1 y D c a v which implies the unique solution x = (1/D)(du bv) and y = (1/D)(av cu)

11 Solving Equations Gaussian Elimination Gaussian Elimination: Two General Equations-Subcase 1B In Subcase 1B when D := ad bc = 0, the multiplier ab/d is undefined and the system ( ) ( ) ( ) ( ) a b x 1 0 u = 0 D/a y c/a 1 v collapses to ( ) ( ) a b x = 0 0 y ( ) u. v c/a This leaves us with two subsubcases : if c av, then the left-hand side of the second equation is 0, but the right-hand side is non-zero, so there is no solution; if c = av, then the second equation reduces to 0 = 0, and there is a continuum of solutions satisfying the one remaining equation ax + by = u, or x = (u by)/a where y is any real number. University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 11 of 55

12 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 12 of 55 Solving Equations Gaussian Elimination Gaussian Elimination: Two General Equations-Case 2 In the final case when a = 0, simply interchanging the two equations ( ) ( ) ( ) ( ) a b x 1 0 u =. c d y 0 1 v gives ( c d 0 b ) ( ) x = y ( ) ( v u ). Provided that b 0, one has y = u/b and, assuming that c 0, also x = (v dy)/c = (bv du)/bc. On the other hand, if b = 0, we are back with two possibilities like those of Subcase 1B.

13 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 13 of 55 Matrices The Transpose of a Matrix Matrices and Their Transposes The transpose of the m n matrix A = (a ij ) m n is the n m matrix a 11 a a m1 A = (aij a 12 a a m2 ) n m = (a ji ) n m =... a 1n a 2n... a mn which results from transforming each column m-vector a j = (a ij ) m i=1 (j = 1, 2,..., n) of A into the corresponding row m-vector a j = (a ji )m i=1 of A. Equivalently, for each i = 1, 2,..., m, the ith row n-vector a i = (a ij ) n j=1 of A is transformed into the ith column n-vector a i = (a ji ) n j=1 of A. Either way, one has a ij = a ji for all relevant pairs i, j.

14 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 14 of 55 Matrix Multiplication I Matrices Matrix Multiplication A scalar, usually denoted by a Greek letter, is a real number α R. The product of any m n matrix A = (a ij ) m n and any scalar α R is the new m n matrix denoted by αa = (αa ij ) m n, each of whose elements αa ij results from multiplying the corresponding element a ij of A by α.

15 Matrices Matrix Multiplication University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 15 of 55 Matrix Multiplication II The matrix product of two matrices A and B is defined (whenever possible) as the matrix C = AB = (c ij ) m n whose element c ij in row i and column j is the inner product c ij = a i b j of: the ith row vector a i of the first matrix A; the jth column vector b j of the second matrix B. Note that the resulting matrix product C must have: as many rows as the first matrix A; as many columns as the second matrix B.

16 Matrices Matrix Multiplication University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 16 of 55 Laws of Matrix Multiplication The following laws of matrix multiplication hold whenever the matrices are compatible for multiplication. associative: A(BC) = (AB)C; distributive: A(B + C) = AB + AC and (A + B)C = AC + BC; transpose: (AB) = B A. shifting scalars: α(ab) = (αa)b = A(αB) for all α R. Exercise Let X be any m n matrix, and z any column n-vector. 1 Show that the matrix product z X Xz is well-defined, and that its value is a scalar.

17 Matrices Matrix Multiplication Matrix Multiplication Does Not Commute The two matrices A and B commute just in case AB = BA. Note that typical pairs of matrices DO NOT commute, meaning that AB BA i.e., the order of multiplication matters. Indeed, suppose that A is l m and B is m n, as is needed for AB to be defined. Then the reverse product BA is undefined except in the special case when n = l. Hence, for both AB and BA to be defined, where B is m n, the matrix A must be n m. But then AB is n n, whereas BA is m m. Evidently AB BA unless m = n. Thus all four matrices A, B, AB and BA are m m = n n. We must be in the special case when all four are square matrices of the same dimension. University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 17 of 55

18 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 18 of 55 Matrices Matrix Multiplication Matrix Multiplication Does Not Commute, II Even if both A and B are n n matrices, implying that both AB and BA are also n n, one can still have AB BA. Here is a 2 2 example: Example ( ) ( ) = ( ) ( ) 0 0 = 1 0 ( ) ( )

19 Matrices Matrix Multiplication University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 19 of 55 More Warnings Regarding Matrix Multiplication Exercise Let A, B, C denote three matrices. Give examples showing that 1 The matrix AB might be defined, even if BA is not. 2 One can have AB = 0 even though A 0 and B 0. 3 If AB = AC and A 0, it does not follow that B = C.

20 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 20 of 55 Square Matrices Matrices Special Kind of Matrices A square matrix has an equal number of rows and columns, this number being called its dimension. The (principal) diagonal of a square matrix A = (a ij ) n n of dimension n is the list (a ii ) n i=1 = (a 11, a 22,..., a nn ) of its diagonal elements a ii. The other elements a ij with i j are the off-diagonal elements. A square matrix is often expressed in the form a 11 a a 1n a 21 a a 2n A = a n1 a n2... a nn with some extra dots along the diagonal.

21 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 21 of 55 Symmetric Matrices Matrices Special Kind of Matrices A square matrix A is symmetric if it is equal to its transpose i.e., if A = A. Example The product of two symmetric matrices need not be symmetric. ( ) ( ) ( ) ( ) ( ) ( ) = but =

22 Matrices Special Kind of Matrices University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 22 of 55 Two Exercises with Symmetric Matrices Exercise Let x be a column n-vector. 1 Find the dimensions of x x and of xx. 2 Show that one is a non-negative number which is positive unless x = 0, and that the other is a symmetric matrix. Exercise Let A be an m n-matrix. 1 Find the dimensions of A A and of AA. 2 Show that both A A and of AA are symmetric matrices. 3 What is a necessary condition for A A = AA? 4 What is a sufficient condition for A A = AA?

23 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 23 of 55 Diagonal Matrices Matrices Special Kind of Matrices A square matrix A = (a ij ) n n is diagonal just in case all of its off diagonal elements a ij with i j are 0. A diagonal matrix of dimension n can be written in the form d d D = 0 0 d = diag(d 1, d 2, d 3,..., d n ) = diag d d n where the n-vector d = (d 1, d 2, d 3,..., d n ) = (d i ) n i=1 consists of the diagonal elements of D. Obviously, any diagonal matrix is symmetric.

24 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 24 of 55 Matrices Multiplying by Diagonal Matrices Special Kind of Matrices Example Let D be a diagonal matrix of dimension n. Suppose that A and B are m n and n m matrices, respectively. Then E := AD and F := DB are well defined m n and n m matrices, respectively. By the law of matrix multiplication, their elements are e ij = n k=1 a ikd kj = a ij d jj and f ij = n k=1 d ikb kj = d ii b ij Thus, the jth column e j of AD is the product d jj a j of the jth element of D and the jth column a j of A. Similarly, the ith row fi of DB is the product d ii b i of the ith element of D and the ith row b i of B.

25 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 25 of 55 Matrices The Identity Matrix: Definition Special Kind of Matrices The identity matrix of dimension n is the diagonal matrix I n = diag(1, 1,..., 1) whose n diagonal elements are all equal to 1. Equivalently, it is the n n-matrix A = (a ij ) n n whose elements are all given by a ij = δ ij for the Kronecker delta function (i, j) δ ij defined on {1, 2,..., n} 2. Exercise Given any m n matrix A, verify that I m A = AI n = A.

26 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 26 of 55 Matrices Special Kind of Matrices Matrices Representing Elementary Row Operations E i (α) = α..., E ij (α) = α E ij is the matrix obtained by interchanging the ith and jth rows of the identity matrix. 2 E i (α) is the matrix obtained by multiplying the ith row of the identity matrix by α R, 3 E ij (α) is the matrix obtained by adding α R times row i to row j in the identity matrix. Theorem Let E be an elementary n n matrix obtained by performing a particular row operation on the n n identity. For any n m matrix A, EA is the matrix obtained by performing the same row operation on A.

27 Matrices Special Kind of Matrices Orthogonal Matrices An n-dimensional square matrix Q is said to be orthogonal just in case its columns form an orthonormal set i.e., they must be pairwise orthogonal unit vectors. Theorem A square matrix Q is orthogonal if and only if it satisfies QQ = I. Proof. The elements of the matrix product QQ satisfy (QQ ) ij = n q ik q jk = q i q j k=1 where q i (resp. q j ) denotes the ith (resp. jth) column vector of Q. But the columns of Q are orthonormal iff q i q j = δ ij for all i, j = 1, 2,..., n, and so iff QQ = I. University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 27 of 55

28 Matrices Special Kind of Matrices University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 28 of 55 Triangular Matrices: Definition Definition A square matrix is upper (resp. lower) triangular if all its non-zero off diagonal elements are above and to the right (resp. below and to the left) of the diagonal i.e., in the upper (resp. lower) triangle bounded by the principal diagonal. The elements of an upper (resp. lower) triangular matrix satisfy (U) ij = 0 whenever i > j, and (L) ij = 0 whenever i < j

29 Matrices Special Kind of Matrices University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 29 of 55 Triangular Matrices: Exercises Exercise Prove that the transpose: 1 U of any upper triangular matrix U is lower triangular; 2 L of any lower triangular matrix U is upper triangular. Exercise Consider the matrix E ij (α) that represents the elementary row operation of adding a multiple of α times row j to row i. Under what conditions is E ij (α) (i) upper triangular? (ii) lower triangular?

30 Matrices Special Kind of Matrices University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 30 of 55 Upper Triangular Matrices: Multiplication Theorem The product W = UV of any two upper triangular matrices U, V is upper triangular, with diagonal elements w ii = u ii v ii (i = 1,..., n) equal to the product of the corresponding diagonal elements of U, V. Proof. Given any two upper triangular n n matrices U and V, the elements (w ij ) n n of their product W = UV satisfy w ij = { j k=i u ikv kj if i j 0 if i > j because u ik v kj = 0 unless both i k and k j. So W = UV is upper triangular. Finally, putting j = i implies that w ii = u ii v ii for i = 1,..., n.

31 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 31 of 55 Matrices Special Kind of Matrices Lower Triangular Matrices: Multiplication Theorem The product of any two lower triangular matrices is lower triangular. Proof. Given any two lower triangular matrices L, M, taking transposes shows that (LM) = M L = U, where the product U is upper triangular, as the product of upper triangular matrices. Hence LM = U is lower triangular, as the transpose of an upper triangular matrix.

32 Matrices Special Kind of Matrices University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 32 of 55 Row Echelon Matrices: Definition Definition For each i {1,..., m}, the leading non-zero element in any non-zero row i of the m n matrix A is a i,ji, where j i := arg min{j {1,..., n} a i,j 0}. The m n matrix R = (a ij ) m n is in row echelon form if; 1 each zero row comes after each row that is not zero; 2 if the leading non-zero element in each row i {1,..., p}, where p is the number of non-zero rows, is denoted by a i,ji, then 1 j 1 < j 2 <... j p.

33 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 33 of 55 Matrices Reduced Row Echelon Matrices Special Kind of Matrices A matrix is in reduced row echelon form (also called row canonical form) if it satisfies the additional condition: Every leading non-zero coefficient a i,ji is 1 and it is the only nonzero entry in column j i. Here is an example: b b b 3 Theorem For any k n matrix A there exists elementary matrices E 1, E 2,..., E m such that the matrix product E m E m 1...E 1 A = U where U is in (reduced) row echelon form.

34 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 34 of 55 Matrices Partitioned Matrices: Definition Special Kind of Matrices A partitioned matrix is a rectangular array of different matrices. Example Consider the (m + l) (n + k) matrix ( ) A B C D where the four submatrices A, B, C, D are of dimension m n, m k, l n and l k respectively. For any scalar α R, the scalar multiple of a partitioned matrix is ( ) ( ) A B αa αb α = C D αc αd

35 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 35 of 55 Matrices Partitioned Matrices: Addition Special Kind of Matrices Suppose the two partitioned matrices ( ) A B and C D ( ) E F G H have the property that the following four pairs of corresponding matrices have equal dimensions: (i) A and E; (ii) B and F; (iii) C and G; (iv) D and H. Then the sum of the two matrices is ( ) ( ) ( ) A B E F A + E B + F + = C D G H C + G D + H

36 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 36 of 55 Matrices Partitioned Matrices: Multiplication Special Kind of Matrices Provided that the two partitioned matrices ( ) ( ) A B E F and C D G H along with their sub-matrices are all compatible for multiplication, the product is defined as ( ) ( ) ( ) A B E F AE + BG AF + BH = C D G H CE + DG CF + DH This adheres to the usual rule for multiplying rows by columns.

37 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 37 of 55 Matrices Partitioned Matrices: Transposes Special Kind of Matrices The rule for transposing a partitioned matrix is C ( A B ) ( A = C ) D B D So the original matrix is symmetric iff A = A, D = D, B = C, and C = B. It is diagonal iff A, D are both diagonal, while B = 0 and C = 0. The identity matrix is diagonal with A = I, D = I, possibly identity matrices of different dimensions.

38 Matrices Linear versus Affine Functions University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 38 of 55 Linear Functions: Definition Definition A linear combination of vectors is the weighted sum k h=1 λ hx h, where x h V and λ h F for h = 1, 2,..., k. Definition A function V : u f (u) F is linear provided that whenever u, v V and λ, µ V. f (λu + µv) = λf (u) + µf (v)

39 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 39 of 55 Matrices Key Properties of Linear Functions Linear versus Affine Functions Exercise By induction on k, show that if the function f : V F is linear, then ( k ) f λ hx h = k λ hf (x h ) h=1 h=1 for all linear combinations k h=1 λ hx h in V i.e., f preserves linear combinations. Exercise In case V = R n and F = R, show that any linear function is homogeneous of degree 1, meaning that f (λv) = λf (v) for all λ > 0 and all v R n.

40 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 40 of 55 Affine Functions Matrices Linear versus Affine Functions Definition A function g : V F is said to be affine if there is a scalar additive constant α F and a linear function f : V F such that g(v) α + f (v).

41 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 41 of 55 Quadratic Forms Matrices Quadratic Forms Definition A quadratic form on R K is a real-valued function of the form Q(x 1,..., x n ) i j a ij x i x j = x T A x Example Q(x 1, x 2 ) = a 11 x1 2 + a 12x 1 x 2 + a 22 x2 2 can be written as ( ) ( 1 a x1 x 11 2 a ) ( ) 12 x a 21 a 22 x 2

42 Matrices Quadratic Forms University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 42 of 55 Negative and Semi-Negative Definite Matrices Definition Let A be an n n symmetric matrix, then A is: 1 Positive definite if x T Ax > 0 for all x 0 in R K, 2 positive semidefinite if x T Ax 0 for all x 0 in R K, 3 negative definite if x T Ax < 0 for all x 0 in R K, 4 negative semidefinite if x T Ax 0 for all x 0 in R K, 5 indefinite if x T Ax > 0 for some x R K and x T Ax < 0 for some other x R K.

43 Determinants Definition Determinants of Order 2: Definition Consider again the pair of linear equations a 11 x 1 + a 12 x 2 = b 1 a 21 x 1 + a 12 x 2 = b 2 with its associated coefficient matrix ( ) a11 a A = 12 a 21 a 22 Let us define D := a 11 a 22 a 21 a 12. Provided that D 0, there is a unique solution given by x 1 = 1 D (b 1a 22 b 2 a 12 ), x 2 = 1 D (b 2a 11 b 1 a 21 ) The number D is called the determinant of the matrix A, and denoted by either det(a) or more concisely, A. University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 43 of 55

44 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 44 of 55 Determinants Definition Determinants of Order 2: Simple Rule Thus, for any 2 2 matrix A, its determinant D is A = a 11 a 12 a 21 a 22 = a 11a 22 a 21 a 12 For this special case of order 2 determinants, a simple rule is: 1 multiply the diagonal elements together; 2 multiply the off-diagonal elements together; 3 subtract the product of the off-diagonal elements from the product of the diagonal elements. Note that 1 0 A = a 11 a a a

45 Determinants Cramer s Rule in the 2 2 Case Definition Using determinant notation, the solution to the equations a 11 x 1 + a 12 x 2 = b 1 a 21 x 1 + a 12 x 2 = b 2 can be written in the alternative form x 1 = 1 D b 1 a 12 b 2 a 22, x 2 = 1 D a 11 b 1 a 21 b 2 This accords with Cramer s rule for the solution to Ax = b, which is the vector x = (x i ) n i=1 each of whose components x i is the fraction with: 1 denominator equal to the determinant D of the coefficient matrix A (provided, of course, that D 0); 2 numerator equal to the determinant of the matrix (A i, b) formed from A by replacing its ith column with the b vector of right-hand side elements. University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 45 of 55

46 Determinants Definition Determinants of Order 3: Definition Determinants of order 3 can be calculated from those of order 2 according to the formula a A = a 22 a a 32 a 33 a 12 a 21 a 23 a 31 a 33 + a 13 a 21 a 22 a 31 a 32 = 3 j=1 ( 1)1+j a 1j A 1j where, for j = 1, 2, 3, the 2 2 matrix A 1j is the sub-matrix obtained by removing both row 1 and column j from A. The scalar C ij = ( 1) i+j A ij is called the(i, j)th-cofactor of A. The result is the following sum A = a 11 a 22 a 33 a 11 a 23 a 32 + a 12 a 23 a 31 a 12 a 21 a 33 + a 13 a 21 a 32 a 13 a 22 a 31 of 3! = 6 terms, each the product of 3 elements chosen so that each row and each column is represented just once. University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 46 of 55

47 Determinants Definition Determinants of Order 3: Cofactor Expansion The determinant expansion A = a 11 a 22 a 33 a 11 a 23 a 32 + a 12 a 23 a 31 a 12 a 21 a 33 + a 13 a 21 a 32 a 13 a 22 a 31 is very symmetric, suggesting (correctly) that the cofactor expansion along the first row (a 11, a 12, a 13 ) A = 3 j=1 ( 1)1+j a 1j A 1j gives the same answer as the two cofactor expansions A = 3 j=1 ( 1)r+j a rj A rj = 3 along, respectively: the rth row (a r1, a r2, a r3 ) the sth column (a 1s, a 2s, a 3s ) i=1 ( 1)i+s a is A is University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 47 of 55

48 Determinants Rules for Determinants University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 48 of 55 Eight Basic Rules (Rules A H of EMEA, Section 16.4) Let A denote the determinant of any n n matrix A. 1 A = 0 if all the elements in a row (or column) of A are 0. 2 A = A, where A is the transpose of A. 3 If all the elements in a single row (or column) of A are multiplied by a scalar α, so is its determinant. 4 If two rows (or two columns) of A are interchanged, the determinant changes sign, but not its absolute value. 5 If two of the rows (or columns) of A are proportional, then A = 0. 6 The value of the determinant of A is unchanged if any multiple of one row (or one column) is added to a different row (or column) of A. 7 The determinant of the product AB of two n n matrices equals the product A B of their determinants. 8 If α is any scalar, then αa = α n A.

49 Determinants Rules for Determinants University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 49 of 55 The Adjugate Matrix Definition The adjugate (or (classical) adjoint ) adj A of an order n square matrix A has elements given by (adj A) ij = C ji. It is therefore the transpose of the cofactor matrix C whose elements are the respective cofactors of A. See Example 9.3 in page 195 of Simon and Blume for a detailed example.

50 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 50 of 55 Determinants Definition of Inverse Matrix Inverse Matrix Definition The n n matrix X is the inverse of the invertible n n matrix A provided that AX = XA = I n. In this case we write X = A 1, so A 1 denotes the (unique) inverse. Big question: does the inverse exist? is it unique?

51 Determinants Inverse Matrix University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 51 of 55 Existence Conditions Theorem An n n matrix A has an inverse if and only if A 0, which holds if and only if at least one of the equations AX = I n and XA = I n has a solution. Proof. Provided A 0, the identity (adj A)A = A(adj A) = A I n shows that the matrix X := (1/ A ) adj A is well defined and satisfies XA = AX = I n, so X is the inverse A 1. Conversely, if either XA = I n or AX = I n has a solution, then the product rule for determinants implies that 1 = I n = AX = XA = A X, and so A 0. The rest follows from the paragraph above.

52 Determinants Inverse Matrix University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 52 of 55 Singularity So A 1 exists if and only if A 0. Definition 1 In case A = 0, the matrix A is said to be singular; 2 In case A 0, the matrix A is said to be non-singular or invertible.

53 Determinants Inverse Matrix Example and Application to Simultaneous Equations Exercise Verify that A = ( ) = A 1 = C := ( by using direct multiplication to show that AC = CA = I 2. Example Suppose that a system of n simultaneous equations in n unknowns is expressed in matrix notation as Ax = b. Of course, A must be an n n matrix. Suppose A has an inverse A 1. Premultiplying both sides of the equation Ax = b by this inverse gives A 1 Ax = A 1 b, which simplifies to Ix = A 1 b. Hence the unique solution of the equation is x = A 1 b. University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 53 of 55 )

54 Determinants Cramer s Rule University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 54 of 55 Cramer s Rule: Statement Notation Given any m n matrix A, let [A j, b] denote the new m n matrix in which column j has been replaced by the column vector b. Evidently [A j, a j ] = A. Theorem Provided that the n n matrix A is invertible, the simultaneous equation system Ax = b has a unique solution x = A 1 b whose ith component is given by the ratio x i = [A i, b] / A. This result is known as Cramer s rule.

55 University of Warwick, EC9A0: Pre-sessional Advanced Mathematics Course Peter J. Hammond & Pablo F. Beker 55 of 55 Determinants Rule for Inverting Products Cramer s Rule Theorem Suppose that A and B are two invertible n n matrices. Then the inverse of the matrix product AB exists, and is the reverse product B 1 A 1 of the inverses. Proof. Using the associative law for matrix multiplication repeatedly gives: and (B 1 A 1 )(AB) = B 1 (A 1 A)B = B 1 (I)B = B 1 (IB) = B 1 B = I (AB)(B 1 A 1 ) = A(BB 1 )A 1 = A(I)A 1 = (AI)A 1 = AA 1 = I. These equations confirm that X := B 1 A 1 is the unique matrix satisfying the double equality (AB)X = X(AB) = I.

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