Matrices and Determinants

Chapter1 Matrices and Determinants 11 INTRODUCTION Matrix means an arrangement or array Matrices (plural of matrix) were introduced by Cayley in 1860 A matrix A is rectangular array of m n numbers (or functions) arranged in m horizontal lines (known as rows ) and in n vertical lines (known as columns ), denoted by A m n These m n numbers are known as the elements or entries of the matrix A and are enclosed in brackets [ ], or()or Theorder of the matrix is m n When m n, the matrix is said to be rectangular Row matrix (or row vector) B 1 n is a matrix having only one row (and several columns), column matrix (or column vector) C m 1 is a matrix having only one column (and several rows) A matrix is said to be a n-square matrix or simply square matrix if m n Thus the number of rows and number of columns in a square matrix are equal The elements of the matrix A are denoted by a ij and are located by the double subscript notation ij where the first subscript i denotes the row (position) and the second subscript j denotes the column position Thus capital letters are used to denote matrices, while the corresponding small letters with double subscript notation are used to denote the elements (or entries) Thus A [a ij ] Null or zero matrix denoted by 0 is a matrix with all its elements zero Equality: Two matrices A and B are said to be equal if they are of the same order and a ij b ij for every i and j Otherwise they are unequal, denoted by A B 12 MATRIX ALGEBRA Sum (or difference): C m n A m n ± B m n then C is said to be the sum (or difference) of A and B provided c ij a ij ± b ij for i, j ie the elements of C are obtained by adding (or subtracting) the corresponding elements of A and B Note that addition or subtraction of matrices A and B is possible only when both A and B are of the same order Submatrix of A is a matrix obtained from A after deleting some rows or columns or both Scalar multiplication: For any non zero scalar k, we have C ka when c ij ka ij ie every element of A is multiplied by k Thus B is considered as B multiplied by 1 Properties: 1 A + B B + A commutative 2 A + (B ± C) (A + B) ± C associative 3 k(a + B) ka + kb distributive 4 A B B A not commutative Transpose of a matrix A of order m n is denoted by A T or A is obtained from A by interchanging the rows and columns Thus B A T is of n m order and b ji a ij for any i, j, ie the i, j th element of A is placed in the j, i th location in A T Properties: 1 (A T ) T A 2 (ka) T ka T 3 (A + B) T A T + B T 11

12 MATHEMATICAL METHODS Matrix multiplication: Two matrices A and B are said to be conformable for multiplication if the number of columns in A is equal to the number of rows in B Then the product of two matrices A m p and B p n is a matrix c m n where c ij p a ik b kj k1 for i 1tom and j 1ton ie, i, j th element in the product matrix C is obtained by adding the p products obtained by multiplying each entry of the i th row of A by the corresponding entry of the j th column of B Thus matrix multiplication amounts to multiplication of rows (of the first matrix) into columns (of the second matrix) ie, row by column multiplication p n m A p B m C n ie mn scalar products of the m rows of A with n columns of B In the product C AB, the matrix B is premultiplied or multiplied from the left by A; while the matrix A is post multiplied or multiplied from the right by B Properties: 1 (ka)b k(ab) A(kB) kab 2 A(BC) (AB)C 3 (A + B)C AC + BC 4 A(B + C) AB + AC However 5 AB BA in general (not commutative) 6 AB 0 does not necessarily imply that A 0or B 0orBA 0 Also 7 AB AC does not necessarily imply that B C, even when A 0 8 (AB) T B T A T ie, transpose of a product is the product of the transposes 13 SPECIAL SQUARE MATRICES The elements a ii of an n-square matrix are known as diagonal elements Trace of matrix A trace of A n a ii sum of the diagonal elements i1 Result: trace (A + B) trace A + trace B, trace (ka) k trace A A is singular matrix if A 0 A is Non-singular matrix if A 0 Upper triangular matrix if a ij 0fori>jie, can have non zero entries only on and above the main diagonal while any entry below the diagonal is zero Lower triangular matrix if a ij 0fori<jA matrix is said to be triangular if it is either upper or lower triangular matrix Diagonal matrix if any entry above or below, the main diagonal is zero However zero entries may be present in the diagonal Thus a ij 0fori j (however a ii 0, not for all i) Scalar matrix is a diagonal matrix in which all the diagonal entries are equal to a constant k ie, a ii k for every i and a ij 0 for any i and j, (i j) Identity matrix denoted by I is a scalar matrix with k 1 Thus 1 0 0 I 3 I 3 3 0 1 0 0 0 1 Positive integral power of a matrix A, denoted by A n, is obtained by multiplying A by itself n times 14 DIFFERENCES BETWEEN DETERMINANTS AND MATRICES Determinant D Matrix A 1 It has a numerical value It has no value It is a symbol representing an array of many numbers on which algebra can be performed 2 It can only a be square It can be rectangular 3 It is zero when elements It is zero only when all the of any one row (or elements in the matrix are zero column) are zero 4 It is multiplied by k It is multiplied by k if all the if elements of any elements of the matrix are one row (or column) multiplied by k are multiplied by k 5 Its value remains unalt- It gets altered (giving rise to ered by the interchange a new matrix) when rows and of rows and columns columns are interchanged 6 Its value is D when It gets changed to a new matrix adjacent rows (or when adjacent rows (or columns) are interchanged columns) are interchanged

MATRICES AND DETERMINANTS 13 WORKED OUT EXAMPLES Example 1: Classify the following matrices Also find the order of the matrices 1 2 3 4 2 3 4 5 1 (a) 3 4 0 6 3 7 8 10 (b) 2 3 (c)[2 4 6 7 8] 2 4 5 6 7 9 2 3 1 0 (d) 0 5 3 7 5 0 0 0 0 0 10 (e) 7 2 0 9 8 12 0 0 0 6 k 0 0 0 5 0 0 (f) 0 6 0 (g) 0 k 0 0 0 0 k 0 0 0 0 0 0 0 k 0 0 0 (h) 0 0 0 (i) [13] 0 0 0 Solution: (a) Rectangular matrix of order 5 4 (b) Column matrix of order 4 1 (c) Row matrix of order 1 5 (d) Upper triangular matrix, 4 4 (square) (e) Lower triangular matrix, 3 3 (square) (f) Diagonal matrix, 3 3 (square) (g) Scalar matrix, 4 4 (square) (h) Null or zero matrix, 3 3 (square) (i) 1 1 matrix which identifies the single entry Example 2: Suppose matrix A has m rows and m + 6 columns and matrix B has n rows and 12 n columns If both AB and BA exists, determine the orders of the matrices A and B Solution: Since A m m + 6 B n 12 n exists, the number of columns in A must be equal to the number of rows in B ie m +6nor m n 6 Similarly B n 12 n A m m + 6 exists, 12 n m or m + n 12 solving m 3,n 9 The order of A is 3 9 and of B is 9 3 Example [ ] 3: Determine [ AB ] and BA if A B 2 3 5 7 and A + B IsAB BA 1 2 4 1 Solution: Adding A B [ ] 6 7 A + B D we get 5 2 [ ] 2 3 C 1 2 [ ] 8 10 so 6 4 [ ] [ ] 2 3 6 7 2A C + D + 1 2 5 2 [ ] 4 5 A Then 3 2 [ ] [ ] [ ] 6 7 4 5 2 2 B D A 5 2 3 2 2 0 [ ][ ] [ ] 4 5 2 2 18 8 So AB and 3 2 2 0 10 6 [ ][ ] [ ] 2 2 4 5 14 14 BA 2 0 3 2 8 10 Note that AB BA, in general [ ] 11 25 Example 4: If A show that A 4 9 n [ ] 1 + 10n 25n using mathematical induction 4n 1 10n Solution: Consider A 2 A A [ ][ ] 11 25 11 25 4 9 4 9 [ ][ ] 21 50 1 + 10 2 25 2 A 2 8 19 4 2 1 10 2 [ ][ ] 11 25 21 50 Now A 3 A A 2 4 9 8 19 [ ] 31 75 12 29 So [ ] A 3 1 + 10 3 25 3 4 3 1 10 3 [ ] 1 + 10 k 25 k Assume A k Then 4 k 1 10 k

14 MATHEMATICAL METHODS [ ][ ] A k+1 A A k 11 25 1 + 10k 25k 4 9 4k 1 10k [ ] 11 + 10k 25 25k 4 + 4k 9 10k [ ] 1 + 10(k + 1) 25(k + 1) 4(k + 1) 1 10(k + 1) By mathematical induction the result follows ( ) 8 4 Example 5: If A prove that 2 2 A 2 10 A + 24 I 0 ( )( ) 8 4 8 4 Solution: A 2 A A 2 2 2 2 ( ) 56 40 Then A 20 4 2 10A + 24I ( ) ( ) ( ) 56 40 8 4 1 0 10 + 24 20 4 2 2 0 1 ( ) 56 80 + 24 40 + 40 + 0 20 20 + 0 4 20 + 24 ( ) 0 0 0 0 0 Example 6: Show that AB AC does not necessarily imply that B C where A 2 1 3, 4 3 1 1 3 2 1 4 1 0 2 1 1 2 B 2 1 1 1, C 3 2 1 1 1 2 1 2 2 5 1 0 Solution: A 3 3 B 3 4 D 3 4 3 3 0 1 1 15 0 5 3 15 0 5 A 3 3 C 3 4 however B C Example 7: Verify that (a) AB BA 0, (b) AC A, (c) CA C, (d) ACB CBA, (e) (A ± B) 2 A 2 + B 2, (f) (A B)(A + B)A 2 B 2 2 3 5 where A 1 4 5, 1 3 4 1 3 5 2 2 4 B 1 3 5, C 1 3 4 1 3 5 1 2 3 Solution: 2 3 5 1 3 5 (a) AB 1 4 5 1 3 5 1 3 4 1 3 5 0 0 0 0 0 0 0 0 0 0 Similarly 1 3 5 2 3 5 BA 1 3 5 1 4 5 1 3 5 1 3 4 0 0 0 0 0 0 0 0 0 0 Thus AB BA 0 2 3 5 2 2 4 (b) AC 1 4 5 1 3 4 1 3 4 1 2 3 2 3 5 1 4 5 A 1 3 4 2 2 4 2 3 5 (c) CA 1 3 4 1 4 5 1 2 3 1 3 4 2 2 4 1 3 4 C 1 2 3 (d) ACB (AC)B (A)B AB 0 CBA C(BA)C 00 (e) (A ± B) 2 A 2 ± AB ± BA + B 2 A 2 + B 2 since AB BA 0 (f) (A B)(A + B) A 2 BA + AB B 2 A 2 B 2 since AB BA 0 Example 8: Prove that (a) trace (A + B) trace A + trace B (b) trace (ka) k trace A n Solution: (a) Trace (A + B) (a ii + b ii ) a ii + b ii trace A + trace B i1

(b) Trace (ka) n (ka ii )k n a ii k trace A i1 i1 5 2 1 Example 9: Express A 7 1 5 as the 3 7 4 product of LU where L and U are lower and upper triangular matrices (known as LU-decomposition or Factorization) Solution: Let the lower triangular matrix be l 11 0 0 u 11 u 12 u 13 L l 21 l 22 0 while U 0 u 22 u 23 l 31 l 32 l 33 0 0 u 33 be the upper triangular matrix Then 5 2 1 A 7 1 5 LU 3 7 4 l 11 0 0 u 11 u 12 u 13 l 21 l 22 0 0 u 22 u 23 l 31 l 32 l 33 0 0 u 33 Equating the corresponding component on both sides, we have l 11 u 11 5,l 11 u 12 2,l 11 u 13 1, l 21 u 11 7,l 21 u 12 + l 22 u 22 1,l 21 u 13 + l 22 u 23 5, l 31 u 11 3,l 31 u 12 + l 32 u 22 7,l 31 u 13 + l 32 u 23 +l 33 u 33 4 Since there are 12 unknowns and 9 equations only, to get a unique solution, assume that l 11 l 22 l 33 1 Now u 11 5,u 12 2,u 13 1 Then l 21 7 u 11 7 5, u 22 (1 l 21 u 12 )/l 22 [ 1 7 5 ( 2)]/ 1 19 5, u 23 ( 5 l 21 u 13 )/l 22 ( 5 7 5 1) 32 5,l 31 3 u 11 3 5 Similarly l 32 41 19, u 33 327 Thus A LU 19 1 0 0 5 2 1 7 1 0 19 32 5 0 5 5 3 41 1 5 19 0 0 327 19 15 MATRICES EXERCISE 2 2 3 3 4 2 1 If A 5 0 2, B 4 2 5, 3 1 4 2 0 3 MATRICES AND DETERMINANTS 15 4 6 2 C 0 3 2 7 2 3 find (a) A + B, (b)a B, (c) 3B,(d) verify A + (B C) (A + B) C, (e) Find D such that C + D B, (f) AB, (g)ba, (h)isab BA, (i) AC, (j) verify A(B + C) AB + AC (k) Is AB AC, (l) Is AC CA 5 2 1 Ans (a) A + B 9 2 7 5 1 7 1 6 5 (b) A B 1 2 3 1 1 1 9 12 6 (c) 3B 12 6 15 6 0 9 1 10 0 (e) Hint: B C +4 1 3 D 5 2 0 8 4 5 (f) AB 19 20 16 13 14 13 8 4 9 (g) BA 33 3 12 13 1 9 (h) No In general AB BA 13 24 1 (i) AC 34 26 16 40 12 16 7 2 4 (j) Hint: B + C 4 5 7 9 2 6 44 6 8 (l) No (j) Hint: CA 21 2 14, No 13 11 13 AC CA 2 Determine the orders of the matrices A having m rows and m + 5 columns and B having n rows and 11 n columns if both AB and BA exist

16 MATHEMATICAL METHODS Ans A 3 8 ; B 8 3 3 Compute [ the ] product AB, [ BA given ] that A + B 1 1 3 1 and A B IsBA AB 3 0 1 4 [ ] [ ] 2 2 4 2 Ans AB, BA,No 0 6 2 4 [ ] [ ] 2 0 1 1 Hint: A, B 2 2 1 2 4 State why in general (a) (A ± B) 2 A 2 ± 2AB + B 2 (b) A 2 B 2 (A B)(A + B) (c) Verify the results (a) and (b) for A and B in the above problem 3 Ans Since AB BA in general 1 2 2 5 If A 0 2 1, verify that A 3 5A 2 + 1 2 2 8A 4I 0 [ ] cos θ sin θ 6 If A show that A sin θ cos θ [ ] cos nθ sin nθ by using mathematical sin nθ cos nθ induction 7 Is AB BA given that 1 3 0 2 3 4 A 1 2 1, B 0 0 2 1 2 3 1 1 2 8 Determine the values of x for which the matrix A 1 2x 3x is nonsingular where A 1 1 2 x 3 0 Ans A is nonsingular for any x other than 3 and 2 3 Hint: A 3x 2 11x+6 (x 3) ( x 2 3) 0 [ ] [ ] 3 1 1 5 9 If A, B then find (a) A 0 1 7 2 T (b) B T (c) (A + B) T (d) (A B) T (e) A T + B T (f) A T B T (g) Verify (A + B) T A T + B T (h)is(a B) T A T B T (i) (AB) T (j) B T A T (k) Verify that (AB) T B T A T [ ] [ ] 3 0 1 7 Ans (a) A T (b) B 1 1 T 5 2 [ ] 4 7 (c) (A + B) T 6 3 [ ] 2 7 (d) (A B) T 4 1 [ ] 4 7 (e) A T + B T (g) True 6 3 (h) (A B) T A T B T [ ] 10 7 (i) (AB) T 17 2 [ ] 10 7 (j) B T A T 17 2 (k) True 3 5 2 10 Express A 0 8 2 as product LU where L 6 2 8 and U are lower and upper triangular matrices 1 0 0 3 5 2 Ans L 0 1 0, U 0 8 2 2 1 1 0 0 6 16 DETERMINANTS Although determinants are inefficient in practical computations, they are useful in vector algebra, differential equations and eigenvalue problems A determinant is a scalar (numerical value) associated with only square matrix A [a ij ] and is denoted as determinant of A or det A or A Thus a determinant is a scalar-valued function whose domain is a set of square matrices A determinant of an n n square matrix A is a scalar given by a 11 a 12 a 1n a 21 a 22 a 2n D det A (1) a n1 a n2 a nn The determinant D issaidtobeofordern and contains n 2 elements or quantities (which may be numbers or functions), arranged in n rows and n columns

MATRICES AND DETERMINANTS 17 The principal diagonal of the determinant is the sloping line of elements from left top corner a 11 to a nn Note that in the matrix representation the elements a ij are enclosed between brackets []or()or, whereas in the determinant the elements are enclosed between vertical lines or bars For n 2, the second order determinant is defined by D deta a 11 a 12 a 21 a 2 a 11 a 22 a 21 a 12 (2) ie second order determinant difference between the product of elements of principal diagonal and the product of the elements of the other diagonal For n 1, D det[a 11 ] [a 11 ] a 11 Note: Here vertical bars does not denote absolute value Thus det [ 5] 5 5 Minor of an element a ij of a matrix A, denoted by M ij,isan(n 1) order determinant of the submatrix of A obtained by omitting the ith row and jth column in A Cofactor of an element a ij of a matrix A, denoted by C ij, is a signed minor of a ij ie C ij ( 1) i+j M ij LaplaceExpansion Laplace Expansion is the expansion of determinant in terms of the cofactors For n 2 n D a ij C ij a i1 C i1 + a i2 C i2 + +a in C in j1 (row wise for any i 1, 2, or n) (3) or D n a ij C ij a 1j C 1j +a 2j C 2j + +a nj C nj (4) i1 (column wise for any j 1, 2,or n) Thus the value of a determinant is the sum of the products of elements of any row (or column) and their respective cofactors However sum of products formed by multiplying the elements of a row (or column) of A by the corresponding cofactors of another n row (or column) of A is zero ie a ik C jk δ ij A k1 n or a kj C ki δ ij A Here δ ij is the Kronecker k1 delta So it is convenient to choose the row or column in the determinant with zeros in it, since these terms in the expansion will vanish The expansion of D by (3) or (4) involves n! determinants since D is defined in terms of n determinants of order (n 1), each of which is in turn defined in terms of (n 1) determinants of order (n 2) and then (n 2) determinants of order (n 3) and so on As the number of calculations of an nth order determinant is N(n) en!, even for n 25, computing time is 4 10 19 sec s 10 12 years However with the use of several properties of determinants which are listed below, the determinant can be triangularized In this case, the number of calculations N(n) 2n3 Forn 25, computation time 3 is 001 second (against 10 12 years which is incredible!) Properties of Determinants 1 If all the elements of any one row (or column) are zero, then det A 0 2 If any two rows (or columns) are proportional to each other, then det A 0 3 If any row (or column) is a linear combination of other rows (or columns), then det A 0 4 If all the elements of any row or column are multiplied by k, giving rise to a new matrix B, then det B k det A 5 If B ka then det B det (ka) k n det A 6 det (A T ) det (A) 7 In general, det (αa + βb) α det (A) + β det (B) iedeterminant()isnot linear 8 If any two rows (or columns) of A are interchanged, yielding a new matrix B then det B det A 9 If k times the elements of any row (or column) in A are added to the corresponding elements of any row (or column) in A, giving rise to a new matrix B, then det B det A This operation is written symbolically as

18 MATHEMATICAL METHODS r i r i + kr j ie, elements of jth row multiplied by k are added to the corresponding elements of theith row Here the ith row gets modified Similarly c i c i + kc j 10 If A is a (upper or lower) triangular matrix or diagonal matrix then det A a 11 a 22 a nn ie value of the determinant is the product of the diagonal elements 11 If each element of a row (or column) consists of m terms (two: binomials, three: trinomial etc), then the determinant is expressed as the sum of m determinants Suppose any row (or column) of A is a binomial say a b + c then det A a det A b + det A c Here A a is the original matrix, A b is the matrix obtained from the original matrix A a by replacing a by b and similarly A c by replacing a by c Extending this, if the elements of say three rows (or columns) consists of m, n, p terms respectively then the original determinants can be expressed as the sum of m n p determinants as stated above Product of Determinants 12 For any n n matrices A and B det (AB) det (BA) det A det B Note: If A is singular, then AB is also singular so det A 0, det AB 0Thus00 13 If C det A det B, then the i, jth element of C is obtained by multiplying the ith row (or column) of A with jth column (or row) of B Note that in matrix multiplication i, j th element is obtained by multiplying the ith row of A with jth column of B Whereas in determinant multiplication, ij th element is obtained either by multiplying (ith row of A with jth column of B)or(ith row of A with jth row of B)or(ith column of A with jth column of B)or(ith column of A with jth row of B) Thus, in determinant multiplication we can multiply row by row, row by column, column by row or column by column 14 Derivative of a Determinant If the elements a ij of a matrix A are differentiable functions of a parameter t, then the derivative of the determinant A equals to sum of n determinants obtained by replacing in all possible ways the elements of one row (or column) of A by their derivatives wrt t, ie da 11 da 1n dt dt a 11 a 1n d a 21 a 2n da 21 da 2n dt dt (det A) dt + + a n1 a nn a n1 a nn a 11 a 1n + + a n 1, 1 a n+1,n da n1 dt da nn dt 15 Factor Theorem Consider an nth order functional determinant A in which the elements a ij are functions of x Suppose for x x, any two rows (or columns) of A become equal, then det A 0 Then det A must contain a factor (x x ) Suppose for x x,k rows (or columns) become identical then det A 0 So consequently det A must contain a factor (x x ) k 1 16 Singular A matrix A is said to be singular if A 0, otherwise A is said to be nonsingular ie, determinant of A is non zero ( A 0) WORKED OUT EXAMPLES Example 1: Find all the cofactors and evaluate A 2 3 4 5 Solution: C 11 cofactor of 2 5, C 12 cofactor of 3 4, C 21 cofactor of 4 3,

MATRICES AND DETERMINANTS 19 C 22 cofactor of 5 2 By Laplace expansion about the first row, A 2 C 11 + 3 C 12 2(5) + 3(4) 10 + 12 2 By Laplace expansion about the second row A 4 C 21 + 5 C 22 4( 3) + 5( 2) 12 10 2 Similarly about first column A ( 2)(5) 4( 3) 10 + 12 2 About 2nd column, A 3(+4) + 5( 2) 2 Example 2: Find the minors M 21,M 13, cofactors C 22,C 32 and evaluate the determinant 12 27 12 A 28 18 24 70 15 40 Solution: M 21 minor of 28 27 12 15 40 1080 180 900 M 13 Minor of 12 28 18 70 15 840 C 22 cofactor of 18 ( 1) 2+2 M 22 12 12 70 40 360 C 32 cofactor of 15 ( 1) 3+2 M 32 12 12 28 24 ( 48) 48 Expanding the determinant by element of first row, we have A 12(18 40 15 24) 27(28 40 70 24) +12(28 15 70 18) 12(360) 27( 560) + 12( 840) 4320 + 15120 11080 9360 Example 3: where Evaluate A by triangularization 1 2 3 4 A 2 1 4 3 3 4 2 1 4 3 1 2 Solution: R 2 R 2 2 R 1,R 3 R 3 3 R 1, R 4 R 4 4 R 1 1 2 3 4 A 0 3 2 5 0 2 7 11 0 5 11 14 Expanding by first column and taking minus from the three rows, 3 2 5 A ( 1) 2 7 11 5 11 14 R 2 R 2 R 1 3 2 5 ( 1) 1 5 6 5 11 14 R 1 R 1 + 3R 2 0 17 23 ( 1) R 3 R 3 + 5R 2 1 5 6 0 36 44 Expanding by first column A ( 1) ( 1)( 1)[17 44 23 36] A [748 828] 80 Example 4: Evaluate 0 0 0 0 0 1 0 0 0 0 2 1 A 0 0 0 3 2 1 0 0 4 3 2 1 0 5 4 3 2 1 6 5 4 3 2 1 Solution: Interchanging all the rows and columns we get 1 0 0 0 0 0 1 2 0 0 0 0 A B 1 2 3 0 0 0 1 2 3 4 0 0 (1 2 3 4 5 6) 1 2 3 4 5 0 720 1 2 3 4 5 6 since B is an lower triangular matrix Example 5: 1 15 14 4 Find the value of 12 6 7 9 8 10 11 5 13 3 2 16

110 MATHEMATICAL METHODS Solution: R 3 R 3 R 2,R 4 R 4 R 1 1 15 14 4 12 6 7 9 4 4 4 4 12 12 12 12 Taking 4 from 3rd row and 12 from 4th row, 1 15 14 4 4 12 12 6 7 9 1 1 1 1 ; 1 1 1 1 1 15 14 4 R 4 R 4 + R 1 48 12 6 7 9 1 1 1 1 0 0 0 0 48 0 0 since all the entries of the 4th row are zero Example 6: a + 2b a + 4b a + 6b Evaluate a + 3b a + 5b a + 7b a + 4b a + 6b a + 8b Solution: Performing R 3 R 3 R 2,R 1 R 1 R 2, a + 2b a + 4b a + 6b b b b 2b 2b 2b 0sincethelasttworow are proportional Example 7: Solution: column, Evaluate the nth order determinant a b b b b a b b b b a b b b b a Adding all the (n 1) columns to the first a + (n 1)b b b a + (n 1)b a b a + (n 1)b b b a + (n 1)b b a 1 b b b 1 a b b a + (n 1)b 1 b a b 1 b b a when a b, R 1 R 2 R 3 R n ie all the n rows are identical and so determinant vanish Thus by factor theorem, (a b) n 1 is factor of Thus (a b) n 1 [a + (n 1)b] a + b b+ c c+ a Example 8: Show that b + c c+ a a + b c + a a + b b+ c a b c 2 c a b Solution: Here the 3 rows contains binomials So the given determinant can be expressed as the sum of 2 2 2 8 determinants as follows a + b b+ c c+ a b + c c+ a a + b c + a a + b b+ c a b+ c c+ a b c+ a a + b c a + b b+ c + b b+ c c+ a + c c+ a a + b a a + b b+ c a b c+ a + b c a b+ c + a c c+ a b a a + b c b b+ c + b b c+ a c c a + b a a b+ c + b c c+ a + c a a + b a b b+ c In this the 3rd determinant is zero because the 1st and 2nd columns are identical Then a b c c a b + a b a b c b c a c + a c c b a a c b b + a c a + b a b c b c + b c c c a a a b b + c a b a b c a b c c a b + c a b a b c

MATRICES AND DETERMINANTS 111 Since in these 2nd, 3rd, 4th, 5th determinants are zero because of identical columns a b c c a b + ( 1)( 1) a b c c a b a b c 2 c a b Example 9: Prove that for the nth order determinant where 1 + a 1 a 2 a 3 a n a 1 1 + a 2 a 3 a n a 1 a 2 1 + a 3 a n a 1 a 2 a 3 1 + a n 1 + a 1 + a 2 + +a n Solution: Performing R 2 R 2 R 1,R 3 R 3 R 1,,R n R n R 1 (ie subtracting the first row from the remaining (n 1) rows) we get 1 + a 1 a 2 a 3 a 4 a n 1 1 0 0 0 1 0 1 0 0 1 0 0 1 0 1 0 0 0 0 1 Adding C 2,C 3,,C n columns to the first column C 1 ie C 1 C 1 + C 2 + C 3 + +C n,wehave 1 + a 1 + a 2 + +a n a 2 a 3 a 4 a n 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 Since this is an upper triangular matrix, product of the diagonal elements (1 + a 1 + a 2 + +a n ) 1 1 1 Example 10: Solve the following equation a x c b A c b x a b a c x 0 or for what value of x, is zero (ie, matrix A is singular) Solution: Adding C 2,C 3 columns to the first column C 1, ie, C 1 C 1 + C 2 + C 3 we have a x + c + b c b c + b x + a b x a b + a + c x a c x 1 c b (a + b + c x) 1 b x a 1 a c x performing R 2 R 2 R 1,R 3 R 3 R 1 we get 1 c b (a + b + c x) 0 b x c a b 0 a c c x b (a + b + c x) 1 [(b x c)(c x b) (a c)(a b)] (a + b + c x) (x 2 a 2 b 2 c 2 + + ab + bc + ca) Thus 0 when x a + b + c or x ± a 2 + b 2 + c 2 ab bc ca Example 11: Compute the product directly (a) Row by column (b) Row by row (c) Column by column (d) Column by row (e) By individually, calculating the determinants where 1 1 1 A 2 1 3 1 0 1, B 2 1 1 3 1 0 1 2 4 Solution: Product of the determinants (a) Row by column 1 1 1 2 1 1 AB 2 1 3 3 1 0 1 0 1 1 2 4 2 0 3 4 9 14 3 3 5 39 (b) Row by row 1 1 1 2 1 1 AB 2 1 3 3 1 0 1 0 1 1 2 4 2 4 3 0 7 12 1 3 3 39 (c) Column by column 1 1 1 2 1 1 AB 2 1 3 3 1 0 1 0 1 1 2 4 3 5 5 1 2 1 10 4 3 39

112 MATHEMATICAL METHODS (d) Column by row 1 1 1 2 1 1 AB 2 1 3 3 1 0 1 0 1 1 2 4 1 5 7 1 4 1 6 0 11 39 (e) A 3, B 13, so AB (3)( 13) 39 Example 12: Find the derivative of the determinant wrt x (a) using formula (b) by differentiating the value of the (expanded) determinant wrt x d x 1 2 dx x 2 2x + 1 x 3 0 3x 2 x 2 + 1 Solution: (a) By formula, the derivative sum of 3 determinants where the 1st, 2nd, 3rd rows are differentiated respectively wrt x Thus d x 1 2 dx x 2 2x + 1 x 3 0 3x 2 x 2 + 1 1 0 0 x 2 2x + 1 x 3 0 3x 2 x 2 + 1 + x 1 2 + 2x 2 3x 2 0 3x 2 x 2 + 1 + x 1 2 x 2 2x + 1 x 3 0 3 2x [(2x + 1)(x 2 + 1) x 3 (3x 2)] + [x 2 (x 2 + 1) 3x 2 (3x 2) 1(2x(x 2 + 1) 0)+ + 2(2x(3x 2))] + [x((2x + 1)2x 3x 3 ) x 2 (2x 6)] 1 6x + 21x 2 + 12x 3 15x 4 (b) Expanding the given determinant x[(2x + 1)(x 2 + 1) x 3 (3x 2)] 1[x 2 (x 2 + 1) 0] + 2[x 2 (3x 2) 0] 3x 5 + 3x 4 + 7x 3 3x 2 + x So d d of determinant dx dx ( 3x5 + 3x 4 + 7x 3 3x 2 + x) 15x 4 + 12x 3 + 21x 2 6x + 1 Example 13: Determine the values of x for which matrix A is non singular given 3 x 2 2 A 2 4 x 1 2 4 1 x Solution: 3 x 2 2 A det A 2 4 x 1 2 4 1 x Expanding the determinant A (3 x)[(4 x)( 1 x) + 4] 2[2( 1 x) + 2) + 2[ 8 + 2(4 x)] x(x 2 6x 9) x(x 3) 2 Then A 0 when x 0 or 3 Thus matrix A is non singular ie A 0 for any x other than zero and 3 EXERCISE 1 Evaluate the following determinants (a) 4 8 1 2 (b) cos θ sin θ sin θ cos θ 1 3 4 (c) 3 4 1 4 1 3 (d) 1 p p 2 1 q q 2 1 r r 2 0 1 2 3 1 1 1 1 (e) 1 0 1 2 2 1 0 3 (f) 1 2 3 4 1 3 6 10 3 2 3 0 1 4 10 20 Ans (a)16(b)1(c) 126 (d) qr(r q) + rp(p r) + pq(q p) (e)4 (f) 1 2 Evaluate the determinants using triangularization 1 2 3 2 2 2 1 1 3 2 1 2 3 4 (a) 1 1 2 1 1 (b) 2 1 4 3 1 4 3 2 5 2 3 4 5 3 2 2 2 2 3 4 5 6 21 17 7 10 3 2 2 2 (c) 24 22 6 10 6 8 2 3 (d) 2 3 2 2 2 2 3 2 6 7 1 2 2 2 2 3 Ans (a) 118 (b) 304 (c) 0 (d) 9 3 Determine the values of x for which the determinant is zero x + 2 2x + 3 3x + 4 (a) 2x + 3 3x + 4 4x + 5 3x + 5 5x + 8 10x + 17

1 + x 2 3 4 (b) 1 2+ x 3 4 1 2 3+ x 4 1 2 3 4+ x x + 1 2x + 1 3x + 1 (c) 2x 4x + 3 6x + 3 4x + 1 6x + 4 8x + 4 Ans (a) x 1, 1, 2 (b)x 0, 10 (c) x 0, 1 2 4 Find the value of the determinant 1 a b+ c a b c d (a) 1 b c+ a 1 c a + b (b) a b c d a b c d a b c d 1 a a 2 1 a a 2 a 3 (c) 1 b b 2 (d) 1 b b 2 b 3 1 c c 2 1 c c 2 c 3 1 d d 2 d 3 1 aa 2 a 3 1 x 1 x1 2 x n 1 1 (e) 1 bb 2 b 3 1 x 2 x2 2 x n 1 2 1 c c 2 c 3 (f) 1 dd 2 d 3 1 x n 1 xn 1 2 xn 1 n 1 1 x n xn 2 xn n 1 Ans (a) 0 Hint: C 3 C 3 + C 2,take(a + b + c) common) (b) 8 abcd (Hint: R 2 + R 1,R 3 + R 1,R 4 + R 1, diagonal, product of diagonal elements) (c) (a b)(b c)(c a) (Hint: a b, a c, b c, 0 Assume L(a b)(b c)(c a), determine constant L 1 by comparing the diagonal element) (d) (a b)(a c)(a d)(b c)(b d)(c d) (e) (a b)(a c)(a d)(b c)(b d) (c d)(a + b + c + d) Hint: By Factor Theorem L(a b)(a c)(a d)(b c)(b d)(c d) since at a b, a c, a d,b c, b d,c d, the determinant is zero Since principal diagonal is bc 2 d 4 is of 7th degree, introduce a linear factor (a + b + c + d), then determine L 1 MATRICES AND DETERMINANTS 113 (f) ( 1) n(n 1) 2 π where π product of factors (x i x j ) with i<j( n) Hint: At x 1 x 2,x 3,,x n,the 0so (x 1 x 2 )(x 1 x 3 ) (x 1 x n ) are factors of Similarly at x 2 x 3,x 4, x n,the 0 so (x 2 x 3 )(x 2 x 4 ) (x 2 x n ) are factors of and so on At x n 1 x n,the 0,so (x n 1 x n ) is a factor of Compare the principal diagonal term to get the value of the constant coefficient 1 + a 1 1 1 5 Prove that 1 1+ b 1 1 1 1 1+ c 1 1 1 1 1+ d abcd ( 1 + 1 a + 1 b + 1 c + 1 ) d Hint: Take a, b, c, d common from R 1,R 2,R 3,R 4 Add R 2,R 3,R 4 to R 1, take common 1 + 1 a + 1 b + 1 c + 1 d, subtract C 1 from C 2,C 3,C 4 a 2 + x ab ac ad 6 Evaluate ab b 2 + x bc bd ac bc c 2 + x cd ad bd cd d 2 + x Ans x 3 (a 2 + b 2 + c 2 + d 2 + x) Hint: Divide by abcd, multiply C 1,C 2,C 3,C 4 by a, b, c, d Take a, b, c, d common from R 1,R 2,R 3,R 4 Add C 2,C 3,C 4 to C 1, subtract R 1 from R 2,R 3,R 4 (b + c) 2 a 2 a 2 7 Find b 2 (c + a) 2 b 2 c 2 c 2 (a + b) 2 Ans 2abc(a + b + c) 3 Hint: a,b,c, are factors (ie 0 when a 0) (a + b + c) 2 is factor since 3 columns are equal, when a + b + c 0 Principal diagonal 6thdegreeso(a + b + c) isalsoafactor a + b c c 8 Show that a b+ c a b b c+ a 4abc Hint: Expand into 8 determinants

114 MATHEMATICAL METHODS 9 Show that a b a b b a b a c d c d 4(a 2 + b 2 )(c 2 + d 2 ) d c d c Hint: Add C 1 to C 3,C 2 to C 4 Take2 2 4 common Then subtract C 3 from C 1,C 4 from C 2 expand 2b 1 + c 1 c 1 + 3a 1 2a 1 + 3b 1 10 Show that 2b 2 + c 2 c 2 + 3a 2 2a 2 + 3b 2 2b 3 + c 3 c 3 + 3a 3 2a 3 + 3b 3 a 1 b 1 c 1 31 a 2 b 2 c 2 a 3 b 3 c 3 Hint: Expand into 8 determinants, 27D + 4D, remaining six determinants are zeros 11 Find the product of the determinants 2 1 1 3 1 0 1 2 4 and 1 3 4 2 1 0 0 1 3 Ans ( 13)( 13) 169 2 1 1 1 12 If A 1 2 1 0 1 0, B 0 1 2 1 0 1 1 2 3 Then compute 15A 2 2AB B 2 without calculating A and B independently Ans 15 1 2 1 1 12 2 1 1 2 1 1 Hint: A 2 1 2 1 1 2 1 0 1 0 0 1 0 6 5 1 5 6 2 1 2 1 1 0 1 7 1 Similarly AB 0 8 2 1, 0 0 2 5 3 5 B 2 1 4 3 2 2 5 2 14 1 13 Find the derivatives of x2 x 3 2x 3x + 1 Ans 2x + 9x 2 8x 3 x 2 x + 1 3 14 Find the derivative of 1 2x 1 x 3 0 x 2 Ans 5 + 4x 12x 2 6x 5 b 2 + ac bc c 2 15 Evaluate ab 2ac bc a 2 ab b 2 + ac Ans 4a 2 b 2 c 2 b c 0 2 Hint: A a 0 c 0 a b 16 Obtain all solutions of the following equations 1 3 x (a) 2x 3 1 x 3x + 1 9x 28 2 x 2 1 x x + 2 x 2 100 0 x x 2 x + 2 100 (b) 0 0 x + 2 x 2 100 0 0 0 0 x 2 x + 2 0 0 0 0 100 Ans (a) x 1, ±3i (b) x 0, ±2