la03-d - LINEAR ALGEBRA W W L CHEN c W W L Chen, 1982,...

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Unformatted text preview: LINEAR ALGEBRA W W L CHEN c W W L Chen, 1982, 2005. This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990. It is available free to all individuals, on the understanding that it is not to be used for financial gain, and may be downloaded and/or photocopied, with or without permission from the author. However, this document may not be kept on any information storage and retrieval system without permission from the author, unless such system is not accessible to any individuals other than its owners. Chapter 3 DETERMINANTS 3.1. Introduction In the last chapter, we have related the question of the invertibility of a square matrix to a question of solutions of systems of linear equations. In some sense, this is unsatisfactory, since it is not simple to find an answer to either of these questions without a lot of work. In this chapter, we shall relate these two questions to the question of the determinant of the matrix in question. As we shall see later, the task is reduced to checking whether this determinant is zero or non-zero. So what is the determinant? Let us start with 1 × 1 matrices, of the form A = (a). Note here that I1 = ( 1 ). If a = 0, then clearly the matrix A is invertible, with inverse matrix A−1 = ( a−1 ) . On the other hand, if a = 0, then clearly no matrix B can satisfy AB = BA = I1 , so that the matrix A is not invertible. We therefore conclude that the value a is a good “determinant” to determine whether the 1 × 1 matrix A is invertible, since the matrix A is invertible if and only if a = 0. Let us then agree on the following definition. Definition. Suppose that A = (a) is a 1 × 1 matrix. We write det(A) = a, and call this the determinant of the matrix A. Chapter 3 : Determinants page 1 of 23 c Linear Algebra W W L Chen, 1982, 2005 Next, let us turn to 2 × 2 matrices, of the form ab cd A= . We shall use elementary row operations to find out when the matrix A is invertible. So we consider the array ab cd (A|I2 ) = (1) 1 0 0 1 , and try to use elementary row operations to reduce the left hand half of the array to I2 . Suppose first of all that a = c = 0. Then the array becomes 0 0 b d 1 0 0 1 , and so it is impossible to reduce the left hand half of the array by elementary row operations to the matrix I2 . Consider next the case a = 0. Multiplying row 2 of the array (1) by a, we obtain a ac b1 ad 0 0 a . Adding −c times row 1 to row 2, we obtain a b 10 0 ad − bc −c a (2) . If D = ad − bc = 0, then this becomes ab 00 10 −c a , and so it is impossible to reduce the left hand half of the array by elementary row operations to the matrix I2 . On the other hand, if D = ad − bc = 0, then the array (2) can be reduced by elementary row operations to 10 01 −b/D a/D d/D −c/D , so that A−1 = d −b −c a 1 ad − bc . Consider finally the case c = 0. Interchanging rows 1 and 2 of the array (1), we obtain cd ab 0 1 1 0 . Multiplying row 2 of the array by c, we obtain c ac d0 bc c 1 0 . Adding −a times row 1 to row 2, we obtain c 0 Chapter 3 : Determinants d bc − ad 0 c 1 −a . page 2 of 23 c Linear Algebra W W L Chen, 1982, 2005 Multiplying row 2 by −1, we obtain (3) c 0 d 01 ad − bc −c a . Again, if D = ad − bc = 0, then this becomes c 0 d 0 0 −c 1 a , and so it is impossible to reduce the left hand half of the array by elementary row operations to the matrix I2 . On the other hand, if D = ad − bc = 0, then the array (3) can be reduced by elementary row operations to 10 01 −b/D a/D d/D −c/D , so that A−1 = 1 ad − bc d −b −c a . Finally, note that a = c = 0 is a special case of ad − bc = 0. We therefore conclude that the value ad − bc is a good “determinant” to determine whether the 2 × 2 matrix A is invertible, since the matrix A is invertible if and only if ad − bc = 0. Let us then agree on the following definition. Definition. Suppose that A= ab cd is a 2 × 2 matrix. We write det(A) = ad − bc, and call this the determinant of the matrix A. 3.2. Determinants for Square Matrices of Higher Order If we attempt to repeat the argument for 2 × 2 matrices to 3 × 3 matrices, then it is very likely that we shall end up in a mess with possibly no firm conclusion. Try the argument on 4 × 4 matrices if you must. Those who have their feet firmly on the ground will try a different approach. Our approach is inductive in nature. In other words, we shall define the determinant of 2× 2 matrices in terms of determinants of 1 × 1 matrices, define the determinant of 3 × 3 matrices in terms of determinants of 2 × 2 matrices, define the determinant of 4 × 4 matrices in terms of determinants of 3 × 3 matrices, and so on. Suppose now that we have defined the determinant of (n − 1) × (n − 1) matrices. Let (4) a11 . . A= . an1 Chapter 3 : Determinants ... a1n . . . . . . ann page 3 of 23 c Linear Algebra W W L Chen, 1982, 2005 be an n × n matrix. For every i, j = 1, . . . , n, let us delete row i and column j of A to obtain the (n − 1) × (n − 1) matrix (5) a11 . . . a(i−1)1 Aij = • a(i+1)1 . . . an1 ... a1(j −1) . . . . . . a(i−1)(j −1) ... • . . . a(i+1)(j −1) . . . ... an(j −1) • . . . • • • . . . • a1(j +1) . . . ... a(i−1)(j +1) • a(i+1)(j +1) . . . an(j +1) ... ... ... ... a1n . . . a(i−1)n • . a(i+1)n . . . ann Here • denotes that the entry has been deleted. Definition. The number Cij = (−1)i+j det(Aij ) is called the cofactor of the entry aij of A. In other words, the cofactor of the entry aij is obtained from A by first deleting the row and the column containing the entry aij , then calculating the determinant of the resulting (n − 1) × (n − 1) matrix, and finally multiplying by a sign (−1)i+j . Note that the entries of A in row i are given by ( ai1 ... ain ) . Definition. By the cofactor expansion of A by row i, we mean the expression n (6) aij Cij = ai1 Ci1 + . . . + ain Cin . j =1 Note that the entries of A in column j are given by a1j . . . . anj Definition. By the cofactor expansion of A by column j , we mean the expression n (7) aij Cij = a1j C1j + . . . + anj Cnj . i=1 We shall state without proof the following important result. The interested reader is referred to Section 3.8 for further discussion. PROPOSITION 3A. Suppose that A is an n × n matrix given by (4). Then the expressions (6) and (7) are all equal and independent of the row or column chosen. Definition. Suppose that A is an n × n matrix given by (4). We call the common value in (6) and (7) the determinant of the matrix A, denoted by det(A). Chapter 3 : Determinants page 4 of 23 c Linear Algebra W W L Chen, 1982, 2005 Let us check whether this agrees with our earlier definition of the determinant of a 2 × 2 matrix. Writing a11 a21 A= a12 a22 , we have C11 = a22 , C12 = −a21 , C21 = −a12 , C22 = a11 . It follows that by row 2 : by column 1 : a11 C11 + a12 C12 = a11 a22 − a12 a21 , a21 C21 + a22 C22 = −a21 a12 + a22 a11 , a11 C11 + a21 C21 = a11 a22 − a21 a12 , by column 2 : a12 C12 + a22 C22 = −a12 a21 + a22 a11 . by row 1 : The four values are clearly equal, and of the form ad − bc as before. Example 3.2.1. Consider the matrix 2 A = 1 2 3 4 1 5 2. 5 Let us use cofactor expansion by row 1. Then C11 = (−1)1+1 det 4 1 2 5 = (−1)2 (20 − 2) = 18, C12 = (−1)1+2 det 1 2 2 5 = (−1)3 (5 − 4) = −1, C13 = (−1)1+3 det 1 2 4 1 = (−1)4 (1 − 8) = −7, so that det(A) = a11 C11 + a12 C12 + a13 C13 = 36 − 3 − 35 = −2. Alternatively, let us use cofactor expansion by column 2. Then C12 = (−1)1+2 det 1 2 2 5 = (−1)3 (5 − 4) = −1, C22 = (−1)2+2 det 2 2 5 5 = (−1)4 (10 − 10) = 0, C32 = (−1)3+2 det 2 1 5 2 = (−1)5 (4 − 5) = 1, so that det(A) = a12 C12 + a22 C22 + a32 C32 = −3 + 0 + 1 = −2. When using cofactor expansion, we should choose a row or column with as few non-zero entries as possible in order to minimize the calculations. Chapter 3 : Determinants page 5 of 23 c Linear Algebra W W L Chen, 1982, 2005 Example 3.2.2. Consider the matrix 2 1 A= 5 2 3 4 4 1 5 2 . 5 5 0 0 8 0 Here it is convenient to use cofactor expansion by column 3, since then det(A) = a13 C13 + a23 C23 + a33 C33 + a43 C43 = 8C33 2 = 8(−1)3+3 det 1 2 3 4 1 5 2 = −16, 5 in view of Example 3.2.1. 3.3. Some Simple Observations In this section, we shall describe two simple observations which follow immediately from the definition of the determinant by cofactor expansion. PROPOSITION 3B. Suppose that a square matrix A has a zero row or a zero column. Then det(A) = 0. Proof. We simply use cofactor expansion by the zero row or zero column. Definition. Consider an n × n matrix a11 . . A= . ... a1n . . . . an1 ... ann If aij = 0 whenever i > j , then A is called an upper triangular matrix. If aij = 0 whenever i < j , then A is called a lower triangular matrix. We also say that A is a triangular matrix if it is upper triangular or lower triangular. Example 3.3.1. The matrix 1 0 0 3 5 6 2 4 0 is upper triangular. Example 3.3.2. A diagonal matrix is both upper triangular and lower triangular. PROPOSITION 3C. Suppose that the n × n matrix a11 . A= . . ... a1n . . . an1 ... ann is triangular. Then det(A) = a11 a22 . . . ann , the product of the diagonal entries. Chapter 3 : Determinants page 6 of 23 c Linear Algebra W W L Chen, 1982, 2005 Proof. Let us assume that A is upper triangular – for the case when A is lower triangular, change the term “left-most column” to the term “top row” in the proof. Using cofactor expansion by the left-most column at each step, we see that a22 . det(A) = a11 det . . an2 ... a2n a33 . . . = a11 a22 det . . . an3 . . . ann ... a3n . . = . . . = a11 a22 . . . ann . . . . ann as required. 3.4. Elementary Row Operations We now study the effect of elementary row operations on determinants. Recall that the elementary row operations that we consider are: (1) interchanging two rows; (2) adding a multiple of one row to another row; and (3) multiplying one row by a non-zero constant. PROPOSITION 3D. (ELEMENTARY ROW OPERATIONS) Suppose that A is an n × n matrix. (a) Suppose that the matrix B is obtained from the matrix A by interchanging two rows of A. Then det(B ) = − det(A). (b) Suppose that the matrix B is obtained from the matrix A by adding a multiple of one row of A to another row. Then det(B ) = det(A). (c) Suppose that the matrix B is obtained from the matrix A by multiplying one row of A by a non-zero constant c. Then det(B ) = c det(A). Sketch of Proof. (a) The proof is by induction on n. It is easily checked that the result holds when n = 2. When n > 2, we use cofactor expansion by a third row, say row i. Then n aij (−1)i+j det(Bij ). det(B ) = j =1 Note that the (n − 1) × (n − 1) matrices Bij are obtained from the matrices Aij by interchanging two rows of Aij , so that det(Bij ) = − det(Aij ). It follows that n det(B ) = − aij (−1)i+j det(Aij ) = − det(A) j =1 as required. (b) Again, the proof is by induction on n. It is easily checked that the result holds when n = 2. When n > 2, we use cofactor expansion by a third row, say row i. Then n aij (−1)i+j det(Bij ). det(B ) = j =1 Note that the (n − 1) × (n − 1) matrices Bij are obtained from the matrices Aij by adding a multiple of one row of Aij to another row, so that det(Bij ) = det(Aij ). It follows that n aij (−1)i+j det(Aij ) = det(A) det(B ) = j =1 as required. Chapter 3 : Determinants page 7 of 23 c Linear Algebra W W L Chen, 1982, 2005 (c) This is simpler. Suppose that the matrix B is obtained from the matrix A by multiplying row i of A by a non-zero constant c. Then n caij (−1)i+j det(Bij ). det(B ) = j =1 Note now that Bij = Aij , since row i has been removed respectively from B and A. It follows that n caij (−1)i+j det(Aij ) = c det(A) det(B ) = j =1 as required. In fact, the above operations can also be carried out on the columns of A. More precisely, we have the following result. PROPOSITION 3E. (ELEMENTARY COLUMN OPERATIONS) Suppose that A is an n × n matrix. (a) Suppose that the matrix B is obtained from the matrix A by interchanging two columns of A. Then det(B ) = − det(A). (b) Suppose that the matrix B is obtained from the matrix A by adding a multiple of one column of A to another column. Then det(B ) = det(A). (c) Suppose that the matrix B is obtained from the matrix A by multiplying one column of A by a non-zero constant c. Then det(B ) = c det(A). Elementary row and column operations can be combined with cofactor expansion to calculate the determinant of a given matrix. We shall illustrate this point by the following examples. Example 3.4.1. Consider the matrix 2 1 A= 5 2 3 4 4 2 2 1 4 0 5 2 . 5 4 Adding −1 times column 3 to column 1, we have 3 4 4 2 2 1 4 0 5 2 . 5 4 3 4 3 2 0 0 det(A) = det 1 2 2 1 4 0 5 2 . 3 4 Adding −1/2 times row 4 to row 3, we have 0 0 det(A) = det 0 2 Using cofactor expansion by column 1, we have 3 det(A) = 2(−1)4+1 det 4 3 Adding −1 times row 1 to row 3, we have 2 1 4 5 3 2 = −2 det 4 3 3 3 det(A) = −2 det 4 0 Chapter 3 : Determinants 2 1 2 2 1 4 5 2. 3 5 2 . −2 page 8 of 23 c Linear Algebra W W L Chen, 1982, 2005 Adding 1 times column 2 to column 3, we have 3 det(A) = −2 det 4 0 7 3. 0 2 1 2 Using cofactor expansion by row 3, we have 3 4 det(A) = −2 · 2(−1)3+2 det 7 3 = 4 det 3 4 7 3 . Using the formula for the determinant of 2 × 2 matrices, we conclude that det(A) = 4(9 − 28) = −76. Let us start again and try a different way. Dividing row 4 by 2, we have 3 4 4 1 2 1 4 0 5 2 . 5 2 3 3 4 1 2 1 det(A) = 2 det 5 1 2 1 4 0 5 0 . 5 2 Adding −1 times row 4 to row 2, we have 2 0 det(A) = 2 det 5 1 Adding −3 times column 3 to column 2, we have 2 0 det(A) = 2 det 5 1 −3 0 −8 1 2 1 4 0 5 0 . 5 2 Using cofactor expansion by row 2, we have 2 det(A) = 2 · 1(−1)2+3 det 5 1 −3 −8 1 5 2 5 = −2 det 5 2 1 −3 −8 1 5 5. 2 Adding −2 times row 3 to row 1, we have −5 −8 1 0 det(A) = −2 det 5 1 1 5. 2 Adding −5 times row 3 to row 2, we have 0 det(A) = −2 det 0 1 −5 −13 1 1 −5 . 2 Using cofactor expansion by column 1, we have det(A) = −2 · 1(−1)3+1 det −5 −13 1 −5 = −2 det −5 −13 1 −5 . Using the formula for the determinant of 2 × 2 matrices, we conclude that det(A) = −2(25 + 13) = −76. Chapter 3 : Determinants page 9 of 23 c Linear Algebra W W L Chen, 1982, 2005 Example 3.4.2. Consider the matrix 2 2 A = 4 1 2 1 3 7 0 1 0 1 2 1 0 3 5 7. 3 0 1 2 3 1 2 Here we have the least number of non-zero entries in column 3, so let us work to get more zeros into this column. Adding −1 times row 4 to row 2, we have 1 3 7 0 1 0 0 2 1 0 1 1 3 1 2 3 2 7. 3 0 1 3 7 0 1 2 1 det(A) = det 4 1 2 0 0 0 1 0 1 1 1 1 2 3 2 1. 3 0 Adding −2 times row 4 to row 3, we have 2 1 det(A) = det 2 1 2 Using cofactor expansion by column 3, we have 2 1 4+3 det(A) = 1(−1) det 2 2 1 3 7 1 3 2 2 1 = − det 1 2 0 2 1 1 1 2 1 3 7 1 1 1 1 2 3 2 . 1 0 Adding −1 times column 3 to column 1, we have 1 0 det(A) = − det 1 0 1 3 7 1 1 1 1 2 3 2 . 1 0 Adding −1 times row 1 to row 3, we have 1 0 det(A) = − det 0 0 1 3 6 1 1 1 0 2 3 2 . −2 0 Using cofactor expansion by column 1, we have 3 det(A) = −1(−1)1+1 det 6 1 1 0 2 2 3 −2 = − det 6 0 1 1 0 2 2 −2 . 0 Adding 1 times row 1 to row 2, we have 3 det(A) = − det 9 1 Chapter 3 : Determinants 1 1 2 2 0. 0 page 10 of 23 c Linear Algebra W W L Chen, 1982, 2005 Using cofactor expansion by column 3, we have 9 1 det(A) = −2(−1)1+3 det 1 2 = −2 det 9 1 1 2 . Using the formula for the determinant of 2 × 2 matrices, we conclude that det(A) = −2(18 − 1) = −34. Example 3.4.3. Consider the matrix 1 2 4 A= 3 2 1 0 4 6 5 4 0 2 5 1 0 5 2 4 7 9 1 3 5 1 6 2 2 6 1 Here note that rows 1 and 6 are almost identical. Adding 102 2 4 5 4 6 1 det(A) = det 3 5 0 245 000 Adding −1 times row 5 to row 2, we have 1 0 4 det(A) = det 3 2 0 Adding −4 times row 6 to row 2, we have 1 0 4 det(A) = det 3 2 0 0 2 1 . 5 2 0 −1 times row 1 to row 6, we have 410 7 6 2 9 2 1 . 1 2 5 362 100 0 0 6 5 4 0 2 0 1 0 5 0 4 4 9 1 3 1 1 0 2 2 6 0 0 0 1 . 5 2 0 0 0 6 5 4 0 2 0 1 0 5 0 4 0 9 1 3 1 1 0 2 2 6 0 0 0 1 . 5 2 0 It follows from Proposition 3B that det(A) = 0. 3.5. Further Properties of Determinants Definition. Consider the n × n matrix a11 . . A= . ... a1n . . . . an1 ... ann By the transpose At of A, we mean the matrix a11 t . . A= . a1n ... an1 . . . ... ann obtained from A by transposing rows and columns. Chapter 3 : Determinants page 11 of 23 c Linear Algebra Example 3.5.1. Consider the matrix 1 A = 4 7 2 5 8 3 6. 9 4 5 6 W W L Chen, 1982, 2005 7 8. 9 Then 1 At = 2 3 Recall that determinants of 2 × 2 matrices depend on determinants of 1 × 1 matrices; in turn, determinants of 3 × 3 matrices depend on determinants of 2 × 2 matrices, and so on. It follows that determinants of n × n matrices ultimately depend on determinants of 1 × 1 matrices. Note now that transposing a 1 × 1 matrix does not affect its determinant (why?). The result below follows in view of Proposition 3A. PROPOSITION 3F. For every n × n matrix A, we have det(At ) = det(A). Example 3.5.2. We have 2 1 det 0 1 3 2 3 1 2 5 4 7 2 3 7 1 0 1 1 3 2 2 1 2 0 = det 4 2 1 0 2 1 3 7 0 1 0 1 2 1 0 1 2 3 1 2 3 5 7 = −34. 3 0 Next, we shall study the determinant of a product. In Section 3.8, we shall sketch a proof of the following important result. PROPOSITION 3G. For every n × n matrices A and B , we have det(AB ) = det(A) det(B ). PROPOSITION 3H. Suppose that the n × n matrix A is invertible. Then det(A−1 ) = 1 . det(A) Proof. In view of Propositions 3G and 3C, we have det(A) det(A−1 ) = det(In ) = 1. The result follows immediately. Finally, the main reason for studying determinants, as outlined in the introduction, is summarized by the following result. PROPOSITION 3J. Suppose that A is an n × n matrix. Then A is invertible if and only if det(A) = 0. Proof. Suppose that A is invertible. Then det(A) = 0 follows immediately from Proposition 3H. Suppose now that det(A) = 0. Let us now reduce A by elementary row operations to reduced row echelon form B . Then there exist a finite sequence E1 , . . . , Ek of elementary n × n matrices such that B = Ek . . . E1 A. It follows from Proposition 3G that det(B ) = det(Ek ) . . . det(E1 ) det(A). Chapter 3 : Determinants page 12 of 23 c Linear Algebra W W L Chen, 1982, 2005 Recall that all elementary matrices are invertible and so have non-zero determinants. It follows that det(B ) = 0, so that B has no zero rows by Proposition 3B. Since B is an n × n matrix in reduced row echelon form, it must be In . We therefore conclude that A is row equivalent to In . It now follows from Proposition 2N(c) that A is invertible. Combining Propositions 2Q and 3J, we have the following result. PROPOSITION 3K. In the notation of Proposition 2N, the following statements are equivalent: (a) The matrix A is invertible. (b) The system Ax = 0 of linear equations has only the trivial solution. (c) The matrices A and In are row equivalent. (d) The system Ax = b of linear equations is soluble for every n × 1 matrix b. (e) The determinant det(A) = 0. 3.6. Application to Curves and Surfaces A special case of Proposition 3K states that a homogeneous system of n linear equations in n variables has a non-trivial solution if and only if the determinant if the coefficient matrix is equal to zero. In this section, we shall use this to solve some problems in geometry. We illustrate our ideas by a few simple examples. Example 3.6.1. Suppose that we wish to determine the equation of the unique line on the xy -plane that passes through two distinct given points (x1 , y1 ) and (x2 , y2 ). The equation of a line on the xy -plane is of the form ax + by + c = 0. Since the two points lie on the line, we must have ax1 + by1 + c = 0 and ax2 + by2 + c = 0. Hence xa + yb + c = 0, x1 a + y1 b + c = 0, x2 a + y2 b + c = 0. Written in matrix notation, we have x x1 x2 y y1 y2 1 a 0 1b = 0. 1 c 0 Clearly there is a non-trivial solution (a, b, c) to this system of linear equations, and so we must have xy1 det x1 y1 1 = 0, x2 y2 1 the equation of the line required. Example 3.6.2. Suppose that we wish to determine the equation of the unique circle on the xy -plane that passes through three distinct given points (x1 , y1 ), (x2 , y2 ) and (x3 , y3 ), not all lying on a straight line. The equation of a circle on the xy -plane is of the form a(x2 + y 2 ) + bx + cy + d = 0. Since the three 2 2 points lie on the circle, we must have a(x2 + y1 ) + bx1 + cy1 + d = 0, a(x2 + y2 ) + bx2 + cy2 + d = 0, and 1 2 2 2 a(x3 + y3 ) + bx3 + cy3 + d = 0. Hence (x2 + y 2 )a + xb + yc + d = 0, 2 (x2 + y1 )a + x1 b + y1 c + d = 0, 1 2 (x2 + y2 )a + x2 b + y2 c + d = 0, 2 2 (x2 + y3 )a + x3 b + y3 c + d = 0. 3 Chapter 3 : Determinants page 13 of 23 c Linear Algebra W W L Chen, 1982, 2005 Written in matrix notation, we have x2 + y 2 2 x2 + y1 1 2 2 x2 + y2 2 2 x3 + y3 x x1 x2 x3 y y1 y2 y3 1 a 0 1b 0 = . c 0 1 1 d 0 Clearly there is a non-trivial solution (a, b, c, d) to this system of linear equations, and so we must have x2 + y 2 2 x2 + y1 det 1 2 2 x2 + y2 2 2 x3 + y3 x x1 x2 x3 y y1 y2 y3 1 1 = 0, 1 1 the equation of the circle required. Example 3.6.3. Suppose that we wish to determine the equation of the unique plane in 3-space that passes through three distinct given points (x1 , y1 , z1 ), (x2 , y2 , z2 ) and (x3 , y3 , z3 ), not all lying on a straight line. The equation of a plane in 3-space is of the form ax + by + cz + d = 0. Since the three points lie on the plane, we must have ax1 + by1 + cz1 + d = 0, ax2 + by2 + cz2 + d = 0, and ax3 + by3 + cz3 + d = 0. Hence xa + yb + zc + d = 0, x1 a + y1 b + z1 c + d = 0, x2 a + y2 b + z2 c + d = 0, x3 a + y3 b + z3 c + d = 0. Written in matrix notation, we have x x1 x2 x3 y y1 y2 y3 z z1 z2 z3 1 a 0 1b 0 = . 1 c 0 1 d 0 Clearly there is a non-trivial solution (a, b, c, d) to this system of linear equations, and so we must have x x1 det x2 x3 y y1 y2 y3 z z1 z2 z3 1 1 = 0, 1 1 the equation of the plane required. Example 3.6.4. Suppose that we wish to determine the equation of the unique sphere in 3-space that passes through four distinct given points (x1 , y1 , z1 ), (x2 , y2 , z2 ), (x3 , y3 , z3 ) and (x4 , y4 , z4 ), not all lying on a plane. The equation of a sphere in 3-space is of the form a(x2 + y 2 + z 2 ) + bx + cy + dz + e = 0. Since the four points lie on the sphere, we must have 2 2 a(x2 + y1 + z1 ) + bx1 + cy1 + dz1 + e = 0, 1 2 2 a(x2 + y2 + z2 ) + bx2 + cy2 + dz2 + e = 0, 2 2 2 a(x2 + y3 + z3 ) + bx3 + cy3 + dz3 + e = 0, 3 and 2 2 a(x2 + y4 + z4 ) + bx4 + cy4 + dz4 + e = 0. 4 Chapter 3 : Determinants page 14 of 23 c Linear Algebra W W L Chen, 1982, 2005 Hence (x2 + y 2 + z 2 )a + xb + yc + zd + e = 0, 2 2 (x2 + y1 + z1 )a + x1 b + y1 c + z1 d + e = 0, 1 2 2 (x2 + y2 + z2 )a + x2 b + y2 c + z2 d + e = 0, 2 2 2 (x2 + y3 + z3 )a + x3 b + y3 c + z3 d + e = 0, 3 2 2 (x2 + y4 + z4 )a + x4 b + y4 c + z4 d + e = 0. 4 Written in matrix notation, we have x2 + y 2 + z 2 2 2 x2 + y1 + z1 1 2 2 2 x2 + y2 + z2 2 2 2 x3 + y3 + z3 2 2 2 x4 + y4 + z4 x x1 x2 x3 x4 y y1 y2 y3 y4 z z1 z2 z3 z4 1 a 0 1b 0 1c = 0. d 0 1 1 e 0 Clearly there is a non-trivial solution (a, b, c, d, e) to this system of linear equations, and so we must have x2 + y 2 + z 2 2 2 x2 + y1 + z1 1 2 2 2 det x2 + y2 + z2 2 2 2 x3 + y3 + z3 2 2 2 x4 + y4 + z4 x x1 x2 x3 x4 y y1 y2 y3 y4 z z1 z2 z3 z4 1 1 1 = 0, 1 1 the equation of the sphere required. 3.7. Some Useful Formulas In this section, we shall discuss two very useful formulas which involve determinants only. The first one enables us to find the inverse of a matrix, while the second one enables us to solve a system of linear equations. The interested reader is referred to Section 3.8 for proofs. Recall first of all that for any n × n matrix a11 . . A= . ... a1n . . , . an1 ... ann the number Cij = (−1)i+j det(Aij ) is called the cofactor of the entry aij , and the (n − 1) × (n − 1) matrix ... a1(j −1) • a1(j +1) ... a1n a11 . . . . . . . . . . . . . . . a(i−1)1 . . . a(i−1)(j −1) • a(i−1)(j +1) . . . a(i−1)n Aij = • ... • • • ... • a(i+1)1 . . . a(i+1)(j −1) • a(i+1)(j +1) . . . a(i+1)n . . . . . . . . . . . . . . . an1 ... an(j −1) • an(j +1) ... ann is obtained from A by deleting row i and column j ; here • denotes that the entry has been deleted. Chapter 3 : Determinants page 15 of 23 c Linear Algebra Definition. The n × n matrix Cn1 . . . ... C11 . adj(A) = . . C1n W W L Chen, 1982, 2005 . . . Cnn is called the adjoint of the matrix A. Remark. Note that adj(A) is obtained from the matrix A first by replacing each entry of A by its cofactor and then by transposing the resulting matrix. PROPOSITION 3L. Suppose that the n × n matrix A is invertible. Then 1 adj(A). det(A) A−1 = Example 3.7.1. Consider the matrix 1 A = 0 2 0 2. 3 −1 1 0 Then 1 det 0 0 adj(A) = − det 2 0 det 2 2 3 2 3 1 0 On the other hand, adding 1 times we have 1 det(A) = det 0 2 −1 0 03 10 det 23 1 −1 − det 20 −1 0 12 10 − det 02 1 −1 det 01 − det det 3 = 4 −2 3 3 −2 −2 −2 . 1 column 1 to column 2 and then using cofactor expansion on row 1, −1 0 1 1 2 = det 0 03 2 0 1 2 0 2 = det 3 1 2 2 3 = −1. It follows that A−1 −3 = −4 2 −3 −3 2 2 2 . −1 Next, we turn our attention to systems of n linear equations in n unknowns, of the form a11 x1 + . . . + a1n xn = b1 , . . . an1 x1 + . . . + ann xn = bn , represented in matrix notation in the form Ax = b, Chapter 3 : Determinants page 16 of 23 c Linear Algebra W W L Chen, 1982, 2005 where a11 . . A= . (8) an1 a1n . . . ... b1 . b= . . and bn . . . ann represent the coefficients and x1 . x= . . (9) xn represents the variables. For every j = 1, . . . , k , write a11 . Aj (b) = . . an1 (10) ... a1(j −1) . . . an(j −1) ... b1 . . . bn a1(j +1) . . . an(j +1) ... ... a1n . ; . . ann in other words, we replace column j of the matrix A by the column b. PROPOSITION 3M. (CRAMER’S RULE) Suppose that the matrix A is invertible. Then the unique solution of the system Ax = b, where A, x and b are given by (8) and (9), is given by x1 = det(A1 (b)) , det(A) xn = ..., det(An (b)) , det(A) where the matrices A1 (b), . . . , A1 (b) are defined by (10). Example 3.7.2. Consider the system Ax = b, where 1 A = 0 2 0 2 3 −1 1 0 1 b = 2. 3 and Recall that det(A) = −1. By Cramer’s rule, we have 1 −1 0 det 2 1 2 303 x1 = = −3, det(A) 110 det 0 2 2 233 x2 = = −4, det(A) 1 −1 1 det 0 1 2 203 x3 = = 3. det(A) Let us check our calculations. Recall from Example 3.7.1 that A−1 −3 = −4 2 −3 −3 2 2 2 . −1 We therefore have x1 −3 x2 = −4 2 x3 Chapter 3 : Determinants −3 −3 2 2 1 −3 2 2 = −4 . −1 3 3 page 17 of 23 c Linear Algebra W W L Chen, 1982, 2005 3.8. Further Discussion In this section, we shall first discuss a definition of the determinant in terms of permutations. In order to do so, we need to make a digression and discuss first the rudiments of permutations on non-empty finite sets. Definition. Let X be a non-empty finite set. A permutation φ on X is a function φ : X → X which is one-to-one and onto. If x ∈ X , we denote by xφ the image of x under the permutation φ. It is not difficult to see that if φ : X → X and ψ : X → X are both permutations on X , then φψ : X → X , defined by xφψ = (xφ)ψ for every x ∈ X so that φ is followed by ψ , is also a permutation on X . Remark. Note that we use the notation xφ instead of our usual notation φ(x) to denote the image of x under φ. Note also that we write φψ to denote the composition ψ ◦ φ. We shall do this only for permutations. The reasons will become a little clearer later in the discussion. Since the set X is non-empty and finite, we may assume, without loss of generality, that it is {1, 2, . . . , n}, where n ∈ N. We now let Sn denote the set of all permutations on the set {1, 2, . . . , n}. In other words, Sn denotes the collection of all functions from {1, 2, . . . , n} to {1, 2, . . . , n} that are both one-to-one and onto. PROPOSITION 3N. For every n ∈ N, the set Sn has n! elements. Proof. There are n choices for 1φ. For each such choice, there are (n − 1) choices left for 2φ. And so on. To represent particular elements of Sn , there are various notations. For example, we can use the notation 1 2 ... n 1φ 2φ . . . nφ to denote the permutation φ. Example 3.8.1. In S4 , 1 2 2 4 3 1 4 3 denotes the permutation φ, where 1φ = 2, 2φ = 4, 3φ = 1 and 4φ = 3. On the other hand, the reader can easily check that 1 2 2 4 3 1 4 3 1 3 2 2 3 4 4 1 1 2 = 2 1 3 3 4 4 . A more convenient way is to use the cycle notation. The permutations 1 2 2 4 3 1 4 3 1 3 and 2 2 3 4 4 1 can be represented respectively by the cycles (1 2 4 3) and (1 3 4). Here the cycle (1 2 4 3) gives the information 1φ = 2, 2φ = 4, 4φ = 3 and 3φ = 1. Note also that in the latter case, since the image of 2 is 2, it is not necessary to include this in the cycle. Furthermore, the information 1 2 2 4 3 1 4 3 1 3 2 2 3 4 4 1 = 1 2 2 1 3 3 4 4 can be represented in cycle notation by (1 2 4 3)(1 3 4) = (1 2). We also say that the cycles (1 2 4 3), (1 3 4) and (1 2) have lengths 4, 3 and 2 respectively. Chapter 3 : Determinants page 18 of 23 c Linear Algebra W W L Chen, 1982, 2005 Example 3.8.2. In S6 , the permutation 1 2 2 4 3 1 4 3 5 6 6 5 can be represented in cycle notation as (1 2 4 3)(5 6). Example 3.8.3. In S4 or S6 , we have (1 2 4 3) = (1 2)(1 4)(1 3). The last example motivates the following important idea. Definition. Suppose that n ∈ N. A permutation in Sn that interchanges two numbers among the elements of {1, 2, . . . , n} and leaves all the others unchanged is called a transposition. Remark. It is obvious that a transposition can be represented by a 2-cycle, and is its own inverse. Definition. Two cycles (x1 x2 . . . xk ) and (y1 y2 . . . yl ) in Sn are said to be disjoint if the elements x1 , . . . , xk , y1 , . . . , yl are all different. The interested reader may try to prove the following result. PROPOSITION 3P. Suppose that n ∈ N. (a) Every permutation in Sn can be written as a product of disjoint cycles. (b) For every subset {x1 , x2 , . . . , xk } of the set {1, 2, . . . , n}, where the elements x1 , x2 , . . . , xk are distinct, the cycle (x1 x2 . . . xk ) satisfies (x1 x2 . . . xk ) = (x1 x2 )(x1 x3 ) . . . (x1 xk ); in other words, every cycle can be written as a product of transpositions. (c) Consequently, every permutation in Sn can be written as a product of transpositions. Example 3.8.4. In S9 , the permutation 1 3 2 2 3 5 4 1 5 7 6 8 7 4 8 9 9 6 can be written in cycle notation as (1 3 5 7 4)(6 8 9). By Theorem 3P(b), we have (1 3 5 7 4) = (1 3)(1 5)(1 7)(1 4) and (6 8 9) = (6 8)(6 9). Hence the permutation can be represented by (1 3)(1 5)(1 7)(1 4)(6 8)(6 9). Definition. Suppose that n ∈ N. Then a permutation in Sn is said to be even if it is representable as the product of an even number of transpositions and odd if it is representable as the product of an odd number of transpositions. Furthermore, we write (φ) = +1 if φ is even, −1 if φ is odd. Remark. It can be shown that no permutation can be simultaneously odd and even. Chapter 3 : Determinants page 19 of 23 c Linear Algebra W W L Chen, 1982, 2005 We are now in a position to define the determinant of a matrix. Suppose that a11 . . . a1n . . . (11) A= . . . an1 . . . ann is an n × n matrix. Definition. By an elementary product from the matrix A, we mean the product of n entries of A, no two of which are from the same row or same column. It follows that any such elementary product must be of the form a1(1φ) a2(2φ) . . . an(nφ) , where φ is a permutation in Sn . Definition. By the determinant of an n × n matrix A of the form (11), we mean the sum (12) det(A) = (φ)a1(1φ) a2(2φ) . . . an(nφ) , φ∈Sn where the summation is over all the n! permutations φ in Sn . It is be shown that the determinant defined in this way is the same as that defined earlier by row or column expansions. Indeed, one can use (12) to establish Proposition 3A. The very interested reader may wish to make an attempt. Here we confine our study to the special cases when n = 2 and n = 3. In the two examples below, we use e to denote the identity permutation. Example 3.8.5. Suppose that n = 2. We have the following: elementary product permutation sign contribution a11 a22 a12 a21 e (1 2) +1 −1 +a11 a22 −a12 a21 Hence det(A) = a11 a22 − a12 a21 as shown before. Example 3.8.6. Suppose that n = 3. We have the following: elementary product permutation sign contribution a11 a22 a33 a12 a23 a31 a13 a21 a32 a13 a22 a31 a11 a23 a32 a12 a21 a33 e (1 2 3) (1 3 2) (1 3) (2 3) (1 2) +1 +1 +1 −1 −1 −1 +a11 a22 a33 +a12 a23 a31 +a13 a21 a32 −a13 a22 a31 −a11 a23 a32 −a12 a21 a33 Hence det(A) = a11 a22 a33 + a12 a23 a31 + a13 a21 a32 − a13 a22 a31 − a11 a23 a32 − a12 a21 a33 . We have the picture below: +D Chapter 3 : Determinants D +E D D a11D E +E E E a12 E E E Ey y a13 y EE y y − y y y − ay az y 11 z 12 Ey y z D E E D yy E yy E zz a21 a22 a23 aE a22 yD D yE E z 21 E y y z E y yD D zE E E y z y a31 a32 a33 a31 a32 D z z z − page 20 of 23 c Linear Algebra W W L Chen, 1982, 2005 Next, we discuss briefly how one may prove Proposition 3G concerning the determinant of the product of two matrices. The idea is to use elementary matrices. Corresponding to Proposition 3D, we can easily establish the following result. PROPOSITION 3Q. Suppose that E is an elementary matrix. (a) If E arises from interchanging two rows of In , then det(E ) = −1. (b) If E arises from adding one row of In to another row, then det(E ) = 1. (c) If E arises from multiplying one row of In by a non-zero constant c, then det(E ) = c. Combining Propositions 3D and 3Q, we can establish the following intermediate result. PROPOSITION 3R. Suppose that E is an n × n elementary matrix. Then for any n × n matrix B , we have det(EB ) = det(E ) det(B ). Proof of Proposition 3G. Let us reduce A by elementary row operations to reduced row echelon form A . Then there exist a finite sequence G1 , . . . , Gk of elementary matrices such that A = Gk . . . G1 A. Since elementary matrices are invertible with elementary inverse matrices,it follows that there exist a finite sequence E1 , . . . , Ek of elementary matrices such that (13) A = E1 . . . Ek A . Suppose first of all that det(A) = 0. Then it follows from (13) that the matrix A must have a zero row. Hence A B must have a zero row, and so det(A B ) = 0. But AB = E1 . . . Ek (A B ), so it follows from Proposition 3R that det(AB ) = 0. Suppose next that det(A) = 0. Then A = In , and so it follows from (13) that AB = E1 . . . Ek B . The result now follows on applying Proposition 3R. We complete this chapter by establishing the two formulas discussed in Section 3.7. Proof of Proposition 3L. It suffices to show that (14) A adj(A) = det(A)In , as this clearly implies A giving the result. To show (14), note that a11 . . (15) A adj(A) = . an1 1 adj(A) det(A) = In , ... C11 a1n . . . . . . ... ann Suppose that the right hand side of (15) is equal to b11 . . . . . B= . bn 1 . . . C1n ... Cn1 . . . . . . . Cnn b1n . . . . bnn Then for every i, j = 1, . . . , n, we have (16) bij = ai1 Cj 1 + . . . + ain Cjn . It follows that when i = j , we have bii = ai1 Ci1 + . . . + ain Cin = det(A). On the other hand, if i = j , then (16) is equal to the determinant of the matrix obtained from A by replacing row j by row i. This matrix has therefore two identical rows, and so the determinant is 0 (why?). The identity (14) follows immediately. Chapter 3 : Determinants page 21 of 23 c Linear Algebra W W L Chen, 1982, 2005 Proof of Proposition 3M. Since A is invertible, it follows from Proposition 3L that A−1 = 1 adj(A). det(A) By Proposition 2P, the unique solution of the system Ax = b is given by x = A−1 b = Written in full, this becomes x1 C11 1 . . = . . . . det(A) xn C1n ... 1 adj(A)b. det(A) b1 Cn1 . . . . = . . bn . . . Cnn b1 C11 + . . . + bn Cn1 . . . . 1 det(A) b1 C1n + . . . + bn Cnn Hence, for every j = 1, . . . , n, we have xj = b1 C1j + . . . + bn Cnj . det(A) To complete the proof, it remains to show that b1 C1j + . . . + bn Cnj = det(Aj (b)). Note, on using cofactor expansion by column j , a11 . . . a(i−1)1 n i+j det(Aj (b)) = bi (−1) det • a(i+1)1 i=1 . . . that an1 n a1(j −1) . . . . . . a(i−1)(j −1) ... • . . . a(i+1)(j −1) . . . ... an(j −1) • . . . • • • . . . • a1(j +1) . . . ... a(i−1)(j +1) • a(i+1)(j +1) . . . an(j +1) ... ... ... ... a1n . . . a(i−1)n • a(i+1)n . . . ann n bi (−1)i+j det(Aij ) = = ... i=1 bi Cij i=1 as required. Problems for Chapter 3 1. Compute the determinant of each of the matrices in Problem 2.6. 2. Find the determinant 1 P = 8 2 of each of the following matrices: 32 1 1 −1 4 0, Q = 1 −1 1 , 12 −1 1 1 a R =b c a2 b2 c2 a3 b3 . c3 3. Find the determinant of the matrix 3 1 2 7 Chapter 3 : Determinants 4 0 3 2 5 1 6 9 2 0 . 3 4 page 22 of 23 c Linear Algebra W W L Chen, 1982, 2005 4. By using suitable elementary row and column operations as well as row and column expansions, show that 2 2 det 2 4 5 3 3 3 6 8 7 7 6 2 7 1 1 1 3 4 3 5 9 = 2. 4 5 [Remark: Note that rows 1 and 2 of the matrix are almost identical.] 5. By using suitable elementary row and column operations as well as row and column expansions, show that 2 2 det 4 4 2 1 1 3 3 1 5 5 2 2 6 1 1 1 0 π 3 2 1 = 2. 1 7 [Remark: The entry π is not a misprint!] 6. If A and B are square matrices of the same size and det A = 2 and det B = 3, find det(A2 B −1 ). 7. a) Compute the Vandermonde determinants 1 a a2 det 1 b b2 and 1 c c2 1a 1 b det 1c 1d a2 b2 c2 d2 a3 b3 . c3 d3 b) Establish a formula for the Vandermonde determinant 1 a a2 . . . an−1 1 1 1 n−1 1 a2 a2 . . . a2 2 det . . . . . . . . . . . . . 1 an a2 n ... an−1 n 8. Compute the determinant a b c det a + x b + x c + x . a+y b+y c+y Harder Problems for Chapter 3 9. For each of the matrices below, compute its adjoint and use Proposition 3L to calculate its inverse: 113 354 a) 2 −2 1 b) 2 1 1 010 101 10. Use Cramer’s rule to solve the system of linear equations 2x1 + x2 + x3 = 4, −x1 + 2x3 = 2, 3x1 + x2 + 3x3 = −2. Chapter 3 : Determinants page 23 of 23 ...
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