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ma265 lecture notes - Chapter 4 DETERMINANTS a11 a12 , we...

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Unformatted text preview: Chapter 4 DETERMINANTS a11 a12 , we define the determinant of a21 a22 A, (also denoted by det A,) to be the scalar DEFINITION 4.0.1 If A = det A = a11 a22 − a12 a21 . The notation a11 a12 a21 a22 is also used for the determinant of A. If A is a real matrix, there is a geometrical interpretation of det A. If P = (x1 , y1 ) and Q = (x2 , y2 ) are points in the plane, forming a triangle 1 x1 y1 with the origin O = (0, 0), then apart from sign, 2 is the area x2 y2 of the triangle OP Q. For, using polar coordinates, let x1 = r1 cos θ1 and E y1 = r1 sin θ1 , where r1 = OP and θ1 is the angle made by the ray OP with 1 the positive x–axis. Then triangle OP Q has area 2 OP · OQ sin α, where α = ∠P OQ. If triangle OP Q has anti–clockwise orientation, then the ray E OQ makes angle θ2 = θ1 + α with the positive x–axis. (See Figure 4.1.) Also x2 = r2 cos θ2 and y2 = r2 sin θ2 . Hence Area OP Q = = = = 1 OP · OQ sin α 2 1 OP · OQ sin (θ2 − θ1 ) 2 1 OP · OQ(sin θ2 cos θ1 − cos θ2 sin θ1 ) 2 1 (OQ sin θ2 · OP cos θ1 − OQ cos θ2 · OP sin θ1 ) 2 71 72 CHAPTER 4. DETERMINANTS y T Q ¡ ¡ ¡d ¡ d ¡ d ¡ α ¨¨ ¡¨¨θ 1 ¨ ¡ ¨¨ d ¨P Ex O Figure 4.1: Area of triangle OP Q. = = 1 (y2 x1 − x2 y1 ) 2 1 x1 y1 . 2 x2 y2 Similarly, if triangle OP Q has clockwise orientation, then its area equals x1 y1 . −1 2x 2 y2 For a general triangle P1 P2 P3 , with Pi = (xi , yi ), i = 1, 2, 3, we can take P1 as the origin. Then the above formula gives 1 2 x2 − x1 y2 − y1 x3 − x1 y3 − y1 or − 1 2 x2 − x1 y2 − y1 x3 − x1 y3 − y1 , according as vertices P1 P2 P3 are anti–clockwise or clockwise oriented. We now give a recursive definition of the determinant of an n × n matrix A = [aij ], n ≥ 3. DEFINITION 4.0.2 (Minor) Let Mij (A) (or simply Mij if there is no ambiguity) denote the determinant of the (n − 1) × (n − 1) submatrix of A formed by deleting the i–th row and j –th column of A. (Mij (A) is called the (i, j ) minor of A.) Assume that the determinant function has been defined for matrices of size (n − 1) × (n − 1). Then det A is defined by the so–called first–row Laplace 73 expansion: det A = a11 M11 (A) − a12 M12 (A) + . . . + (−1)1+n M1n (A) n (−1)1+j a1j M1j (A). = j =1 For example, if A = [aij ] is a 3 × 3 matrix, the Laplace expansion gives det A = a11 M11 (A) − a12 M12 (A) + a13 M13 (A) a21 a22 a21 a23 a22 a23 + a13 − a12 = a11 a31 a32 a31 a33 a32 a33 = a11 (a22 a33 − a23 a32 ) − a12 (a21 a33 − a23 a31 ) + a13 (a21 a32 − a22 a31 ) = a11 a22 a33 − a11 a23 a32 − a12 a21 a33 + a12 a23 a31 + a13 a21 a32 − a13 a22 a31 . (The recursive definition also works for 2 × 2 determinants, if we define the determinant of a 1 × 1 matrix [t] to be the scalar t: det A = a11 M11 (A) − a12 M12 (A) = a11 a22 − a12 a21 .) EXAMPLE 4.0.1 If P1 P2 P3 is a triangle with Pi = (xi , yi ), i = 1, 2, 3, then the area of triangle P1 P2 P3 is 1 2 x1 y1 1 x2 y2 1 x3 y3 1 or 1 − 2 x1 y1 1 x2 y2 1 , x3 y3 1 according as the orientation of P1 P2 P3 is anti–clockwise or clockwise. For from the definition of 3 × 3 determinants, we have 1 2 x1 y1 1 x2 y2 1 x3 y3 1 = 1 2 x1 = 1 2 x2 − x1 y2 − y1 x3 − x1 y3 − y1 y2 1 y3 1 − y1 x2 1 x3 1 + x2 y2 x3 y3 . One property of determinants that follows immediately from the definition is the following: THEOREM 4.0.1 If a row of a matrix is zero, then the value of the determinant is zero. 74 CHAPTER 4. DETERMINANTS (The corresponding result for columns also holds, but here a simple proof by induction is needed.) One of the simplest determinants to evaluate is that of a lower triangular matrix. THEOREM 4.0.2 Let A = [aij ], where aij = 0 if i < j . Then det A = a11 a22 . . . ann . (4.1) An important special case is when A is a diagonal matrix. If A =diag (a1 , . . . , an ) then det A = a1 . . . an . In particular, for a scalar matrix tIn , we have det (tIn ) = tn . Proof. Use induction on the size n of the matrix. The result is true for n = 2. Now let n > 2 and assume the result true for matrices of size n − 1. If A is n × n, then expanding det A along row 1 gives det A = a11 a22 a32 . . . 0 a33 ... ... 0 0 an1 an2 . . . ann = a11 (a22 . . . ann ) by the induction hypothesis. If A is upper triangular, equation 4.1 remains true and the proof is again an exercise in induction, with the slight difference that the column version of theorem 4.0.1 is needed. REMARK 4.0.1 It can be shown that the expanded form of the determinant of an n × n matrix A consists of n! signed products ±a1i1 a2i2 . . . anin , where (i1 , i2 , . . . , in ) is a permutation of (1, 2, . . . , n), the sign being 1 or −1, according as the number of inversions of (i1 , i2 , . . . , in ) is even or odd. An inversion occurs when ir > is but r < s. (The proof is not easy and is omitted.) The definition of the determinant of an n × n matrix was given in terms of the first–row expansion. The next theorem says that we can expand the determinant along any row or column. (The proof is not easy and is omitted.) 75 THEOREM 4.0.3 n (−1)i+j aij Mij (A) det A = j =1 for i = 1, . . . , n (the so–called i–th row expansion) and n (−1)i+j aij Mij (A) det A = i=1 for j = 1, . . . , n (the so–called j –th column expansion). REMARK 4.0.2 The expression (−1)i+j obeys the chess–board pattern of signs: + − + ... − + − ... + − + ... . . . . The following theorems can be proved by straightforward inductions on the size of the matrix: THEOREM 4.0.4 A matrix and its transpose have equal determinants; that is det At = det A. THEOREM 4.0.5 If two rows of a matrix are equal, the determinant is zero. Similarly for columns. THEOREM 4.0.6 If two rows of a matrix are interchanged, the determinant changes sign. EXAMPLE 4.0.2 If P1 = (x1 , y1 ) and P2 = (x2 , y2 ) are distinct points, then the line through P1 and P2 has equation xy1 x1 y1 1 x2 y2 1 = 0. 76 CHAPTER 4. DETERMINANTS For, expanding the determinant along row 1, the equation becomes ax + by + c = 0, where a= y1 1 y2 1 = y1 − y2 and b = − x1 1 x2 1 = x2 − x1 . This represents a line, as not both a and b can be zero. Also this line passes through Pi , i = 1, 2. For the determinant has its first and i–th rows equal if x = xi and y = yi and is consequently zero. There is a corresponding formula in three–dimensional geometry. If P1 , P2 , P3 are non–collinear points in three–dimensional space, with Pi = (xi , yi , zi ), i = 1, 2, 3, then the equation xyz x1 y1 z1 x2 y2 z2 x3 y3 z3 1 1 1 1 =0 represents the plane through P1 , P2 , P3 . For, expanding the determinant along row 1, the equation becomes ax + by + cz + d = 0, where a= x1 z1 1 y1 z 1 1 y2 z2 1 , b = − x2 z2 1 , c = x3 z3 1 y3 z 3 1 x1 y1 1 x2 y2 1 . x3 y3 1 As we shall see in chapter 6, this represents a plane if at least one of a, b, c 1 is non–zero. However, apart from sign and a factor 2 , the determinant expressions for a, b, c give the values of the areas of projections of triangle P1 P2 P3 on the (y, z ), (x, z ) and (x, y ) planes, respectively. Geometrically, it is then clear that at least one of a, b, c is non–zero. It is also possible to give an algebraic proof of this fact. Finally, the plane passes through Pi , i = 1, 2, 3 as the determinant has its first and i–th rows equal if x = xi , y = yi , z = zi and is consequently zero. We now work towards proving that a matrix is non–singular if its determinant is non–zero. DEFINITION 4.0.3 (Cofactor) The (i, j ) cofactor of A, denoted by Cij (A) (or Cij if there is no ambiguity) is defined by Cij (A) = (−1)i+j Mij (A). 77 REMARK 4.0.3 It is important to notice that Cij (A), like Mij (A), does not depend on aij . Use will be made of this observation presently. In terms of the cofactor notation, Theorem 4.0.3 takes the form THEOREM 4.0.7 n aij Cij (A) det A = j =1 for i = 1, . . . , n and n aij Cij (A) det A = i=1 for j = 1, . . . , n. Another result involving cofactors is THEOREM 4.0.8 Let A be an n × n matrix. Then n aij Ckj (A) = 0 if i = k. aij Cik (A) = 0 (a) if j = k. j =1 Also n (b) i=1 Proof. If A is n × n and i = k , let B be the matrix obtained from A by replacing row k by row i. Then det B = 0 as B has two identical rows. Now expand det B along row k . We get n bkj Ckj (B ) 0 = det B = j =1 n aij Ckj (A), = j =1 in view of Remark 4.0.3. 78 CHAPTER 4. DETERMINANTS DEFINITION 4.0.4 (Adjoint) If A = [aij ] is an n × n matrix, the adjoint of A, denoted by adj A, is the transpose of the matrix of cofactors. Hence C11 C21 · · · Cn1 C12 C22 · · · Cn2 adj A = . . . . . . . C1n C2n · · · Cnn Theorems 4.0.7 and 4.0.8 may be combined to give THEOREM 4.0.9 Let A be an n × n matrix. Then A(adj A) = (det A)In = (adj A)A. Proof. n aij (adj A)jk (A adj A)ik = j =1 n aij Ckj (A) = j =1 = δik det A = ((det A)In )ik . Hence A(adj A) = (det A)In . The other equation is proved similarly. COROLLARY 4.0.1 (Formula for the inverse) If det A = 0, then A is non–singular and 1 A−1 = adj A. det A EXAMPLE 4.0.3 The matrix 123 A= 4 5 6 889 is non–singular. For 46 56 −2 89 89 = −3 + 24 − 24 det A = = −3 = 0. +3 45 88 79 Also A−1 C C21 1 11 C12 C22 = −3 C13 C23 56 89 1 46 = − − 89 3 45 88 −3 6 1 = − 12 −15 3 −8 8 C31 C32 C33 23 − 89 13 89 − 12 88 23 56 − 13 46 12 45 −3 6 . −3 The following theorem is useful for simplifying and numerically evaluating a determinant. Proofs are obtained by expanding along the corresponding row or column. THEOREM 4.0.10 The determinant is a linear function of each row and column. For example (a) ′ ′ a11 + a′ 11 a12 + a12 a13 + a13 a21 a22 a23 a31 a32 a33 (b) ta11 ta12 ta13 a21 a22 a23 a31 a32 a33 = ′ ′ a′ a11 a12 a13 11 a12 a13 a21 a22 a23 + a21 a22 a23 a31 a32 a33 a31 a32 a33 a11 a12 a13 = t a21 a22 a23 a31 a32 a33 . COROLLARY 4.0.2 If a multiple of a row is added to another row, the value of the determinant is unchanged. Similarly for columns. Proof. We illustrate with a 3 × 3 example, but the proof is really quite general. a11 + ta21 a12 + ta22 a13 + ta23 a21 a22 a23 a31 a32 a33 = a11 a12 a13 a21 a22 a23 a31 a32 a33 + ta21 ta22 ta23 a21 a22 a23 a31 a32 a33 80 CHAPTER 4. DETERMINANTS = a11 a12 a13 a21 a22 a23 a31 a32 a33 a21 a22 a23 + t a21 a22 a23 a31 a32 a33 = a11 a12 a13 a21 a22 a23 a31 a32 a33 +t×0 = a11 a12 a13 a21 a22 a23 a31 a32 a33 . To evaluate a determinant numerically, it is advisable to reduce the matrix to row–echelon form, recording any sign changes caused by row interchanges, together with any factors taken out of a row, as in the following examples. EXAMPLE 4.0.4 Evaluate the determinant 123 456. 889 Solution. Using row operations R2 → R2 − 4R1 and R3 → R3 − 8R1 and then expanding along the first column, gives 123 456 889 1 2 3 0 −3 −6 0 −8 −15 = 1 2 −8 −15 = −3 = −3 −6 −8 −15 = −3 12 01 EXAMPLE 4.0.5 Evaluate the determinant 1 3 7 1 1 1 6 1 2 4 1 3 1 5 . 2 4 Solution. 1 3 7 1 1 1 6 1 2 4 1 3 1 5 2 4 = 1 1 2 1 0 −2 −2 2 0 −1 −13 −5 0 0 1 3 = −3. 81 1 1 2 1 0 1 1 −1 = −2 0 −1 −13 −5 0 0 1 3 1 0 = −2 0 0 1 2 1 1 1 −1 0 −12 −6 0 1 3 1 0 =2 0 0 1 2 1 1 1 −1 0 1 3 0 −12 −6 1 0 =2 0 0 1 1 0 0 2 1 1 −1 1 3 0 30 = 60. EXAMPLE 4.0.6 (Vandermonde determinant) Prove that 111 abc a2 b2 c2 = (b − a)(c − a)(c − b). Solution. Subtracting column 1 from columns 2 and 3 , then expanding along row 1, gives 111 abc a2 b2 c2 = = 1 0 0 a b−a c−a a2 b2 − a2 c2 − a2 b−a c−a 2 c2 − a2 −a b2 = (b − a)(c − a) 1 1 b+a c+a = (b − a)(c − a)(c − b). REMARK 4.0.4 From theorems 4.0.6, 4.0.10 and corollary 4.0.2, we deduce (a) det (Eij A) = −det A, (b) det (Ei (t)A) = t det A, if t = 0, 82 CHAPTER 4. DETERMINANTS (c) det (Eij (t)A) =det A. It follows that if A is row–equivalent to B , then det B = c det A, where c = 0. Hence det B = 0 ⇔ det A = 0 and det B = 0 ⇔ det A = 0. Consequently from theorem 2.5.8 and remark 2.5.7, we have the following important result: THEOREM 4.0.11 Let A be an n × n matrix. Then (i) A is non–singular if and only if det A = 0; (ii) A is singular if and only if det A = 0; (iii) the homogeneous system AX = 0 has a non–trivial solution if and only if det A = 0. EXAMPLE 4.0.7 Find the rational numbers a for which the following homogeneous system has a non–trivial solution and solve the system for these values of a: x − 2y + 3z = 0 ax + 3y + 2z = 0 6x + y + az = 0. Solution. The coefficient determinant of the system is ∆= 1 −2 3 a 32 6 1a = 1 −2 3 0 3 + 2a 2 − 3a 0 13 a − 18 3 + 2a 2 − 3a 13 a − 18 = (3 + 2a)(a − 18) − 13(2 − 3a) = = 2a2 + 6a − 80 = 2(a + 8)(a − 5). So ∆ = 0 ⇔ a = −8 or a = 5 and these values of a are the only values for which the given homogeneous system has a non–trivial solution. If a = −8, the coefficient matrix has reduced row–echelon form equal to 1 0 −1 0 1 −2 00 0 83 and so the complete solution is x = z, y = 2z , with z arbitrary. If a = 5, the coefficient matrix has reduced row–echelon form equal to 10 1 0 1 −1 00 0 and so the complete solution is x = −z, y = z , with z arbitrary. EXAMPLE 4.0.8 Find the values of t for which the following system is consistent and solve the system in each case: x+y = 1 tx + y = t (1 + t)x + 2y = 3. Solution. Suppose that the given system has a solution (x0 , y0 ). Then the following homogeneous system x+y+z = 0 tx + y + tz = 0 (1 + t)x + 2y + 3z = 0 will have a non–trivial solution x = x0 , y = y0 , z = −1. Hence the coefficient determinant ∆ is zero. However ∆= 1 11 t 1t 1+t 2 3 = 1 0 0 t 1−t 0 1+t 1−t 2−t = 1−t 0 1−t 2−t = (1−t)(2−t). Hence t = 1 or t = 2. If t = 1, the given system becomes x+y = 1 x+y = 1 2x + 2y = 3 which is clearly inconsistent. If t = 2, the given system becomes x+y = 1 2x + y = 2 3x + 2y = 3 84 CHAPTER 4. DETERMINANTS which has the unique solution x = 1, y = 0. To finish this section, we present an old (1750) method of solving a system of n equations in n unknowns called Cramer’s rule . The method is not used in practice. However it has a theoretical use as it reveals explicitly how the solution depends on the coefficients of the augmented matrix. THEOREM 4.0.12 (Cramer’s rule) The system of n linear equations in n unknowns x1 , . . . , xn a11 x1 + a12 x2 + · · · + a1n xn = b1 a21 x1 + a22 x2 + · · · + a2n xn = b2 . . . an1 x1 + an2 x2 + · · · + ann xn = bn has a unique solution if ∆ = det [aij ] = 0, namely ∆1 ∆2 ∆n , x2 = , . . . , xn = , ∆ ∆ ∆ where ∆i is the determinant of the matrix formed by replacing the i–th column of the coefficient matrix A by the entries b1 , b2 , . . . , bn . x1 = Proof. Suppose the coefficient determinant ∆ = 0. Then by corollary 4.0.1, 1 A−1 exists and is given by A−1 = ∆ adj A and the system has the unique solution b1 b1 x1 C11 C21 · · · Cn1 b2 x2 1 C12 C22 · · · Cn2 b2 . = A−1 . = . . . . . . . . ∆ . . . . . bn bn xn C1n C2n · · · Cnn b1 C11 + b2 C21 + . . . + bn Cn1 1 b2 C12 + b2 C22 + . . . + bn Cn2 = . . . ∆ . bn C1n + b2 C2n + . . . + bn Cnn However the i–th component of the last vector is the column i. Hence ∆1 /∆ ∆1 x1 ∆2 ∆2 /∆ x2 1 . = . = . . . ∆ . . . . xn ∆n ∆n /∆ expansion of ∆i along . 85 4.1. PROBLEMS 4.1 PROBLEMS . 1. If the points Pi = (xi , yi ), i = 1, 2, 3, 4 form a quadrilateral with vertices in anti–clockwise orientation, prove that the area of the quadrilateral equals 1 2 x1 x2 y1 y2 + x2 x3 y2 y3 + x3 x4 y3 y4 + x4 x1 y4 y1 . (This formula generalizes to a simple polygon and is known as the Surveyor’s formula.) 2. Prove that the following identity holds by expressing the left–hand side as the sum of 8 determinants: a+x b+y c+z x+u y+v z+w u+a v+b w+c 3. Prove that abc =2 x y z . uvw n2 (n + 1)2 (n + 2)2 2 (n + 2)2 (n + 3)2 (n + 1) (n + 2)2 (n + 3)2 (n + 4)2 = −8. 4. Evaluate the following determinants: (a) 246 427 327 1014 543 443 −342 721 621 (b) 1 2 3 4 −2 1 −4 3 . 3 −4 −1 2 4 3 −2 −1 [Answers: (a) −29400000; (b) 900.] 5. Compute the inverse of the matrix 1 0 −2 4 A= 3 1 5 2 −3 by first computing the adjoint matrix. −11 −4 2 7 −10 .] [Answer: A−1 = −1 29 13 1 −2 1 86 CHAPTER 4. DETERMINANTS 6. Prove that the following identities hold: (i) 2a 2b b−c 2b 2a a + c a+b a+b b = −2(a − b)2 (a + b), (ii) b+c b c c c+a a b a a+b = 2a(b2 + c2 ). 7. Let Pi = (xi , yi ), i = 1, 2, 3. If x1 , x2 , x3 are distinct, prove that there is precisely one curve of the form y = ax2 + bx + c passing through P1 , P2 and P3 . 8. Let 1 1 −1 k . A= 2 3 1k 3 Find the values of k for which det A = 0 and hence, or otherwise, determine the value of k for which the following system has more than one solution: x+y−z = 1 2x + 3y + kz = 3 x + ky + 3z = 2. Solve the system for this value of k and determine the solution for which x2 + y 2 + z 2 has least value. [Answer: k = 2; x = 10/21, y = 13/21, z = 2/21.] 9. By considering the coefficient determinant, find all rational numbers a and b for which the following system has (i) no solutions, (ii) exactly one solution, (iii) infinitely many solutions: x − 2y + bz = 3 ax + 5x + 2y 2z = 2 = 1. Solve the system in case (iii). [Answer: (i) ab = 12 and a = 3, no solution; ab = 12, unique solution; 7 a = 3, b = 4, infinitely many solutions; x = − 2 z + 2 , y = 5 z − 6 , with 3 3 3 z arbitrary.] 87 4.1. PROBLEMS 10. Express the determinant of the matrix 11 2 1 1 2 3 4 B= 2 4 7 2t + 6 2 2 6−t t as as polynomial in t and hence determine the rational values of t for which B −1 exists. [Answer: det B = (t − 2)(2t − 1); t = 2 and t = 1 .] 2 11. If A is a 3 × 3 matrix over a field and det A = 0, prove that det (adj A) = (det A)2 , 1 (ii) (adj A)−1 = A = adj (A−1 ). det A (i) 12. Suppose that A is a real 3 × 3 matrix such that At A = I3 . (i) Prove that At (A − I3 ) = −(A − I3 )t . (ii) Prove that det A = ±1. (iii) Use (i) to prove that if det A = 1, then det (A − I3 ) = 0. 13. If A is a square matrix such that one column is a linear combination of the remaining columns, prove that det A = 0. Prove that the converse also holds. 14. Use Cramer’s rule to solve the system −2x + 3y − z = 1 x + 2y − z = 4 −2x − y + z = −3. [Answer: x = 2, y = 3, z = 4.] 15. Use remark 4.0.4 to deduce that det Eij = −1, det Ei (t) = t, det Eij (t) = 1 and use theorem 2.5.8 and induction, to prove that det (BA) = det B det A, if B is non–singular. Also prove that the formula holds when B is singular. 88 CHAPTER 4. DETERMINANTS 16. Prove that a+b+c a+b a a a+b a+b+c a a a a a+b+c a+b a a a+b a+b+c = c2 (2b + c)(4a +2b + c). 17. Prove that 1 + u1 u1 u1 u1 u2 1 + u2 u2 u2 u3 u3 1 + u3 u3 u4 u4 u4 1 + u4 = 1 + u1 + u2 + u3 + u4 . 18. Let A ∈ Mn×n (F ). If At = −A, prove that det A = 0 if n is odd and 1 + 1 = 0 in F . 19. Prove that 1 r r r 1 1 r r 1 1 1 r 1 1 1 1 = (1 − r)3 . 20. Express the determinant 1 a2 − bc a4 1 b2 − ca b4 1 c2 − ab c4 as the product of one quadratic and four linear factors. [Answer: (b − a)(c − a)(c − b)(a + b + c)(b2 + bc + c2 + ac + ab + a2 ).] ...
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