Lecture notes on matrices and Determinants

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LINEAR ALGEBRA W W L CHEN c W W L Chen, 1982, 2008. 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 nancial 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 nd 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 I = ( 1 ). If a6= 0, then clearly the matrix A is invertible, with inverse matrix 1 1 1 A = (a ): On the other hand, if a = 0, then clearly no matrix B can satisfy AB =BA =I , so that the matrix A 1 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 =6 0. Let us then agree on the following de nition. 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 24c Linear Algebra W W L Chen, 1982, 2008 Next, let us turn to 2 2 matrices, of the form   a b A = : c d We shall use elementary row operations to nd out when the matrix A is invertible. So we consider the array   a b 1 0 (AjI ) = ; (1) 2 c d 0 1 and try to use elementary row operations to reduce the left hand half of the array to I . Suppose rst 2 of all that a =c = 0. Then the array becomes   0 b 1 0 ; 0 d 0 1 and so it is impossible to reduce the left hand half of the array by elementary row operations to the matrix I . Consider next the case a6= 0. Multiplying row 2 of the array (1) by a, we obtain 2   a b 1 0 : ac ad 0 a Addingc times row 1 to row 2, we obtain   a b 1 0 : (2) 0 adbc c a If D =adbc = 0, then this becomes   a b 1 0 ; 0 0 c a and so it is impossible to reduce the left hand half of the array by elementary row operations to the matrixI . On the other hand, ifD =adbc6= 0, then the array (2) can be reduced by elementary row 2 operations to   1 0 d=D b=D ; 0 1 c=D a=D so that   1 d b 1 A = : c a adbc Consider nally the case c =6 0. Interchanging rows 1 and 2 of the array (1), we obtain   c d 0 1 : a b 1 0 Multiplying row 2 of the array by c, we obtain   c d 0 1 : ac bc c 0 Addinga times row 1 to row 2, we obtain   c d 0 1 : 0 bcad c a Chapter 3 : Determinants page 2 of 24c Linear Algebra W W L Chen, 1982, 2008 Multiplying row 2 by1, we obtain   c d 0 1 : (3) 0 adbc c a Again, if D =adbc = 0, then this becomes   c d 0 1 ; 0 0 c a and so it is impossible to reduce the left hand half of the array by elementary row operations to the matrixI . On the other hand, ifD =adbc =6 0, then the array (3) can be reduced by elementary row 2 operations to   1 0 d=D b=D ; 0 1 c=D a=D so that   1 d b 1 A = : c a adbc Finally, note thata =c = 0 is a special case ofadbc = 0. We therefore conclude that the valueadbc is a good \determinant" to determine whether the 2 2 matrix A is invertible, since the matrix A is invertible if and only if adbc6= 0. Let us then agree on the following de nition. Definition. Suppose that   a b A = c d is a 2 2 matrix. We write det(A) =adbc; 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 rm conclusion. Try the argument on 4 4 matrices if you must. Those who have their feet rmly on the ground will try a di erent approach. Our approach is inductive in nature. In other words, we shall de ne the determinant of 22 matrices in terms of determinants of 11 matrices, de ne the determinant of 33 matrices in terms of determinants of 2 2 matrices, de ne the determinant of 4 4 matrices in terms of determinants of 3 3 matrices, and so on. Suppose now that we have de ned the determinant of (n 1) (n 1) matrices. Let 0 1 a ::: a 11 1n . . A . . A = (4) . . a ::: a n1 nn Chapter 3 : Determinants page 3 of 24c Linear Algebra W W L Chen, 1982, 2008 be an nn 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 0 1 a ::: a  a ::: a 11 1n 1(j1) 1(j+1) B C . . . . . . . . . . B C . . . . . B C B C a ::: a  a ::: a (i1)1 (i1)(j1) (i1)(j+1) (i1)n B C B C A =  :::    :::  : (5) ij B C B C a ::: a  a ::: a (i+1)1 (i+1)(j1) (i+1)(j+1) (i+1)n B C B C . . . . . . . . . . A . . . . . a ::: a  a ::: a n1 nn n(j1) n(j+1) Here denotes that the entry has been deleted. i+j Definition. The number C = (1) det(A ) is called the cofactor of the entry a of A. In other ij ij ij words, the cofactor of the entrya is obtained fromA by rst deleting the row and the column containing ij the entry a , then calculating the determinant of the resulting (n 1) (n 1) matrix, and nally ij i+j multiplying by a sign (1) . Note that the entries of A in row i are given by (a ::: a ): i1 in Definition. By the cofactor expansion of A by row i, we mean the expression n X a C =a C +::: +a C : (6) ij ij i1 i1 in in j=1 Note that the entries of A in column j are given by 0 1 a 1j B C . . : A . a nj Definition. By the cofactor expansion of A by column j, we mean the expression n X a C =a C +::: +a C : (7) ij ij 1j 1j nj nj 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 nn 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 nn 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 24c Linear Algebra W W L Chen, 1982, 2008 Let us check whether this agrees with our earlier de nition of the determinant of a 2 2 matrix. Writing   a a 11 12 A = ; a a 21 22 we have C =a ; C =a ; C =a ; C =a : 11 22 12 21 21 12 22 11 It follows that by row 1 : a C +a C =a a a a ; 11 11 12 12 11 22 12 21 by row 2 : a C +a C =a a +a a ; 21 21 22 22 21 12 22 11 by column 1 : a C +a C =a a a a ; 11 11 21 21 11 22 21 12 by column 2 : a C +a C =a a +a a : 12 12 22 22 12 21 22 11 The four values are clearly equal, and of the form adbc as before. Example 3.2.1. Consider the matrix 0 1 2 3 5 A A = 1 4 2 : 2 1 5 Let us use cofactor expansion by row 1. Then   4 2 1+1 2 C = (1) det = (1) (20 2) = 18; 11 1 5   1 2 1+2 3 C = (1) det = (1) (5 4) =1; 12 2 5   1 4 1+3 4 C = (1) det = (1) (1 8) =7; 13 2 1 so that det(A) =a C +a C +a C = 36 3 35 =2: 11 11 12 12 13 13 Alternatively, let us use cofactor expansion by column 2. Then   1 2 1+2 3 C = (1) det = (1) (5 4) =1; 12 2 5   2 5 2+2 4 C = (1) det = (1) (10 10) = 0; 22 2 5   2 5 3+2 5 C = (1) det = (1) (4 5) = 1; 32 1 2 so that det(A) =a C +a C +a C =3 + 0 + 1 =2: 12 12 22 22 32 32 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 24c Linear Algebra W W L Chen, 1982, 2008 Example 3.2.2. Consider the matrix 0 1 2 3 0 5 B 1 4 0 2C A = : A 5 4 8 5 2 1 0 5 Here it is convenient to use cofactor expansion by column 3, since then 0 1 2 3 5 3+3 A det(A) =a C +a C +a C +a C = 8C = 8(1) det 1 4 2 =16; 13 13 23 23 33 33 43 43 33 2 1 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 de nition of the determinant by cofactor expansion. PROPOSITION 3B. Suppose that a square matrix A has a zero row or has a zero column. Then det(A) = 0. Proof. We simply use cofactor expansion by the zero row or zero column. Definition. Consider an nn matrix 0 1 a ::: a 11 1n . . A . . A = : . . a ::: a n1 nn If a = 0 whenever ij, then A is called an upper triangular matrix. If a = 0 whenever ij, then ij ij 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 0 1 1 2 3 A 0 4 5 0 0 6 is upper triangular. Example 3.3.2. A diagonal matrix is both upper triangular and lower triangular. PROPOSITION 3C. Suppose that the nn matrix 0 1 a ::: a 11 1n . . A . . A = . . a ::: a n1 nn is triangular. Then det(A) =a a :::a , the product of the diagonal entries. 11 22 nn Chapter 3 : Determinants page 6 of 24c Linear Algebra W W L Chen, 1982, 2008 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 0 1 0 1 a ::: a a ::: a 22 2n 33 3n . . . . A A . . . . det(A) =a det =a a det =::: =a a :::a 11 11 22 11 22 nn . . . . a ::: a a ::: a n2 nn n3 nn as required. 3.4. Elementary Row Operations We now study the e ect 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 nn 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 X i+j det(B) = a (1) det(B ): ij ij j=1 Note that the (n 1) (n 1) matrices B are obtained from the matrices A by interchanging two ij ij rows of A , so that det(B ) = det(A ). It follows that ij ij ij n X i+j det(B) = a (1) det(A ) = det(A) ij ij j=1 as required. (b) Again, the proof is by induction onn. It is easily checked that the result holds whenn = 2. When n 2, we use cofactor expansion by a third row, say row i. Then n X i+j det(B) = a (1) det(B ): ij ij j=1 Note that the (n 1) (n 1) matricesB are obtained from the matricesA by adding a multiple of ij ij one row of A to another row, so that det(B ) = det(A ). It follows that ij ij ij n X i+j det(B) = a (1) det(A ) = det(A) ij ij j=1 as required. Chapter 3 : Determinants page 7 of 24c Linear Algebra W W L Chen, 1982, 2008 (c) This is simpler. Suppose that the matrix B is obtained from the matrixA by multiplying rowi of A by a non-zero constant c. Then n X i+j det(B) = ca (1) det(B ): ij ij j=1 Note now that B =A , since row i has been removed respectively from B and A. It follows that ij ij n X i+j det(B) = ca (1) det(A ) =c det(A) ij ij 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. PROPOSITION3E. (ELEMENTARY COLUMN OPERATIONS) Suppose thatA is annn 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 0 1 2 3 2 5 1 4 1 2 B C A = : A 5 4 4 5 2 2 0 4 Adding1 times column 3 to column 1, we have 0 1 0 3 2 5 B 0 4 1 2C det(A) = det A: 1 4 4 5 2 2 0 4 Adding1=2 times row 4 to row 3, we have 0 1 0 3 2 5 0 4 1 2 B C det(A) = det : A 0 3 4 3 2 2 0 4 Using cofactor expansion by column 1, we have 0 1 0 1 3 2 5 3 2 5 4+1 A A det(A) = 2(1) det 4 1 2 =2 det 4 1 2 : 3 4 3 3 4 3 Adding1 times row 1 to row 3, we have 0 1 3 2 5 A det(A) =2 det 4 1 2 : 0 2 2 Chapter 3 : Determinants page 8 of 24c Linear Algebra W W L Chen, 1982, 2008 Adding 1 times column 2 to column 3, we have 0 1 3 2 7 A det(A) =2 det 4 1 3 : 0 2 0 Using cofactor expansion by row 3, we have     3 7 3 7 3+2 det(A) =2 2(1) det = 4 det : 4 3 4 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 di erent way. Dividing row 4 by 2, we have 0 1 2 3 2 5 B 1 4 1 2C det(A) = 2 det : A 5 4 4 5 1 1 0 2 Adding1 times row 4 to row 2, we have 0 1 2 3 2 5 B 0 3 1 0C det(A) = 2 det : A 5 4 4 5 1 1 0 2 Adding3 times column 3 to column 2, we have 0 1 2 3 2 5 B 0 0 1 0C det(A) = 2 det A: 5 8 4 5 1 1 0 2 Using cofactor expansion by row 2, we have 0 1 0 1 2 3 5 2 3 5 2+3 A A det(A) = 2 1(1) det 5 8 5 =2 det 5 8 5 : 1 1 2 1 1 2 Adding2 times row 3 to row 1, we have 0 1 0 5 1 A det(A) =2 det 5 8 5 : 1 1 2 Adding5 times row 3 to row 2, we have 0 1 0 5 1 A det(A) =2 det 0 13 5 : 1 1 2 Using cofactor expansion by column 1, we have     5 1 5 1 3+1 det(A) =2 1(1) det =2 det : 13 5 13 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 24c Linear Algebra W W L Chen, 1982, 2008 Example 3.4.2. Consider the matrix 0 1 2 1 0 1 3 2 3 1 2 5 B C B C A =B 4 7 2 3 7 : C A 1 0 1 1 3 2 1 0 2 0 Here we have the least number of non-zero entries in column 3, so let us work to get more zeros into this column. Adding1 times row 4 to row 2, we have 0 1 2 1 0 1 3 B 1 3 0 1 2C B C det(A) = detB 4 7 2 3 7C: A 1 0 1 1 3 2 1 0 2 0 Adding2 times row 4 to row 3, we have 0 1 2 1 0 1 3 1 3 0 1 2 B C B C det(A) = det 2 7 0 1 1 : B C A 1 0 1 1 3 2 1 0 2 0 Using cofactor expansion by column 3, we have 0 1 0 1 2 1 1 3 2 1 1 3 1 3 1 2 1 3 1 2 B C B C 4+3 det(A) = 1(1) det = det : A A 2 7 1 1 2 7 1 1 2 1 2 0 2 1 2 0 Adding1 times column 3 to column 1, we have 0 1 1 1 1 3 B 0 3 1 2C det(A) = det : A 1 7 1 1 0 1 2 0 Adding1 times row 1 to row 3, we have 0 1 1 1 1 3 B 0 3 1 2 C det(A) = det A: 0 6 0 2 0 1 2 0 Using cofactor expansion by column 1, we have 0 1 0 1 3 1 2 3 1 2 1+1 A A det(A) =1(1) det 6 0 2 = det 6 0 2 : 1 2 0 1 2 0 Adding 1 times row 1 to row 2, we have 0 1 3 1 2 A det(A) = det 9 1 0 : 1 2 0 Chapter 3 : Determinants page 10 of 24c Linear Algebra W W L Chen, 1982, 2008 Using cofactor expansion by column 3, we have     9 1 9 1 1+3 det(A) =2(1) det =2 det : 1 2 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 0 1 1 0 2 4 1 0 2 4 5 7 6 2 B C B C 4 6 1 9 2 1 B C A = : B C 3 5 0 1 2 5 B C A 2 4 5 3 6 2 1 0 2 5 1 0 Here note that rows 1 and 6 are almost identical. Adding1 times row 1 to row 6, we have 0 1 1 0 2 4 1 0 B 2 4 5 7 6 2C B C B 4 6 1 9 2 1C det(A) = detB C: B 3 5 0 1 2 5C A 2 4 5 3 6 2 0 0 0 1 0 0 Adding1 times row 5 to row 2, we have 0 1 1 0 2 4 1 0 0 0 0 4 0 0 B C B C 4 6 1 9 2 1 B C det(A) = det : B C 3 5 0 1 2 5 B C A 2 4 5 3 6 2 0 0 0 1 0 0 Adding4 times row 6 to row 2, we have 0 1 1 0 2 4 1 0 B 0 0 0 0 0 0C B C B 4 6 1 9 2 1C det(A) = detB C: B 3 5 0 1 2 5C A 2 4 5 3 6 2 0 0 0 1 0 0 It follows from Proposition 3B that det(A) = 0. 3.5. Further Properties of Determinants Definition. Consider the nn matrix 0 1 a ::: a 11 1n . . A . . A = : . . a ::: a n1 nn t By the transpose A of A, we mean the matrix 0 1 a ::: a 11 n1 . . t A . . A = . . a ::: a 1n nn Chapter 3 : Determinants page 11 of 24c Linear Algebra W W L Chen, 1982, 2008 obtained from A by transposing rows and columns. Example 3.5.1. Consider the matrix 0 1 1 2 3 A A = 4 5 6 : 7 8 9 Then 0 1 1 4 7 t A A = 2 5 8 : 3 6 9 Recall that determinants of 22 matrices depend on determinants of 11 matrices; in turn, determi- nants of 33 matrices depend on determinants of 22 matrices, and so on. It follows that determinants of nn matrices ultimately depend on determinants of 1 1 matrices. Note now that transposing a 1 1 matrix does not a ect its determinant (why?). The result below follows in view of Proposition 3A. t PROPOSITION 3F. For every nn matrix A, we have det(A ) = det(A). Example 3.5.2. We have 0 1 0 1 2 2 4 1 2 2 1 0 1 3 B 1 3 7 0 1C B 2 3 1 2 5C B C B C detB 0 1 2 1 0C = detB 4 7 2 3 7C =34: A A 1 2 3 1 2 1 0 1 1 3 3 5 7 3 0 2 1 0 2 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 nn matrices A and B, we have det(AB) = det(A) det(B). PROPOSITION 3H. Suppose that the nn matrix A is invertible. Then 1 1 det(A ) = : det(A) 1 Proof. In view of Propositions 3G and 3C, we have det(A) det(A ) = det(I ) = 1. The result follows n immediately. Finally, the main reason for studying determinants, as outlined in the introduction, is summarized by the following result. PROPOSITION3J. Suppose thatA is annn matrix. ThenA is invertible if and only if det(A) =6 0. Proof. Suppose that A is invertible. Then det(A) 6= 0 follows immediately from Proposition 3H. Suppose now that det(A)6= 0. Let us now reduce A by elementary row operations to reduced row echelon form B. Then there exist a nite sequence E ;:::;E of elementary nn matrices such that 1 k B =E :::E A: k 1 It follows from Proposition 3G that det(B) = det(E )::: det(E ) det(A): k 1 Chapter 3 : Determinants page 12 of 24c Linear Algebra W W L Chen, 1982, 2008 Recall that all elementary matrices are invertible and so have non-zero determinants. It follows that det(B) =6 0, so that B has no zero rows by Proposition 3B. Since B is an nn matrix in reduced row echelon form, it must be I . We therefore conclude that A is row equivalent to I . It now follows from n n 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 I are row equivalent. n (d) The system Ax =b of linear equations is soluble for every n 1 matrix b. (e) The determinant det(A)6= 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 coecient 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. Example3.6.1. Suppose that we wish to determine the equation of the unique line on thexy-plane that passes through two distinct given points (x ;y ) and (x ;y ). The equation of a line on the xy-plane is 1 1 2 2 of the form ax +by +c = 0. Since the two points lie on the line, we must have ax +by +c = 0 and 1 1 ax +by +c = 0. Hence 2 2 xa + yb +c = 0; x a +y b +c = 0; 1 1 x a +y b +c = 0: 2 2 Written in matrix notation, we have 0 10 1 0 1 x y 1 a 0 A A A x y 1 b = 0 : 1 1 x y 1 c 0 2 2 Clearly there is a non-trivial solution (a;b;c) to this system of linear equations, and so we must have 0 1 x y 1 A det x y 1 = 0; 1 1 x y 1 2 2 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 (x ;y ), (x ;y ) and (x ;y ), not all lying on a straight 1 1 2 2 3 3 2 2 line. The equation of a circle on thexy-plane is of the forma(x +y ) +bx +cy +d = 0. Since the three 2 2 2 2 points lie on the circle, we must have a(x +y ) +bx +cy +d = 0,a(x +y ) +bx +cy +d = 0, and 1 1 2 2 1 1 2 2 2 2 a(x +y ) +bx +cy +d = 0. Hence 3 3 3 3 2 2 (x +y )a + xb + yc +d = 0; 2 2 (x +y )a +x b +y c +d = 0; 1 1 1 1 2 2 (x +y )a +x b +y c +d = 0; 2 2 2 2 2 2 (x +y )a +x b +y c +d = 0: 3 3 3 3 Chapter 3 : Determinants page 13 of 24c Linear Algebra W W L Chen, 1982, 2008 Written in matrix notation, we have 0 10 1 0 1 2 2 x +y x y 1 a 0 2 2 Bx +y x y 1CBbC B 0C 1 1 1 1 = : A A A 2 2 x +y x y 1 c 0 2 2 2 2 2 2 x +y x y 1 d 0 3 3 3 3 Clearly there is a non-trivial solution (a;b;c;d) to this system of linear equations, and so we must have 0 1 2 2 x +y x y 1 2 2 Bx +y x y 1C 1 1 1 1 det = 0; A 2 2 x +y x y 1 2 2 2 2 2 2 x +y x y 1 3 3 3 3 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 (x ;y ;z ), (x ;y ;z ) and (x ;y ;z ), not all lying on a 1 1 1 2 2 2 3 3 3 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 ax +by +cz +d = 0, ax +by +cz +d = 0, and 1 1 1 2 2 2 ax +by +cz +d = 0. Hence 3 3 3 xa + yb + zc +d = 0; x a +y b +z c +d = 0; 1 1 1 x a +y b +z c +d = 0; 2 2 2 x a +y b +z c +d = 0: 3 3 3 Written in matrix notation, we have 0 10 1 0 1 x y z 1 a 0 Bx y z 1CBbC B 0C 1 1 1 = : A A A x y z 1 c 0 2 2 2 x y z 1 d 0 3 3 3 Clearly there is a non-trivial solution (a;b;c;d) to this system of linear equations, and so we must have 0 1 x y z 1 x y z 1 B C 1 1 1 det = 0; A x y z 1 2 2 2 x y z 1 3 3 3 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 (x ;y ;z ), (x ;y ;z ), (x ;y ;z ) and (x ;y ;z ), not all lying 1 1 1 2 2 2 3 3 3 4 4 4 2 2 2 on a plane. The equation of a sphere in 3-space is of the form a(x +y +z ) +bx +cy +dz +e = 0. Since the four points lie on the sphere, we must have 2 2 2 a(x +y +z ) +bx +cy +dz +e = 0; 1 1 1 1 1 1 2 2 2 a(x +y +z ) +bx +cy +dz +e = 0; 2 2 2 2 2 2 2 2 2 a(x +y +z ) +bx +cy +dz +e = 0; 3 3 3 3 3 3 2 2 2 a(x +y +z ) +bx +cy +dz +e = 0: 4 4 4 4 4 4 Hence 2 2 2 (x +y +z )a + xb + yc + zd +e = 0; 2 2 2 (x +y +z )a +x b +y c +z d +e = 0; 1 1 1 1 1 1 2 2 2 (x +y +z )a +x b +y c +z d +e = 0; 2 2 2 2 2 2 2 2 2 (x +y +z )a +x b +y c +z d +e = 0; 3 3 3 3 3 3 2 2 2 (x +y +z )a +x b +y c +z d +e = 0: 4 4 4 4 4 4 Chapter 3 : Determinants page 14 of 24c Linear Algebra W W L Chen, 1982, 2008 Written in matrix notation, we have 0 10 1 0 1 2 2 2 x +y +z x y z 1 a 0 2 2 2 Bx +y +z x y z 1CBbC B 0C 1 1 1 1 1 1 B CB C B C 2 2 2 Bx +y +z x y z 1CBcC =B 0C: 2 2 2 2 2 2 A A A 2 2 2 x +y +z x y z 1 d 0 3 3 3 3 3 3 2 2 2 x +y +z x y z 1 e 0 4 4 4 4 4 4 Clearly there is a non-trivial solution (a;b;c;d;e) to this system of linear equations, and so we must have 0 1 2 2 2 x +y +z x y z 1 2 2 2 Bx +y +z x y z 1C 1 1 1 1 1 1 B C 2 2 2 detBx +y +z x y z 1C = 0; 2 2 2 2 2 2 A 2 2 2 x +y +z x y z 1 3 3 3 3 3 3 2 2 2 x +y +z x y z 1 4 4 4 4 4 4 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 rst one enables us to nd 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 rst of all that for any nn matrix 0 1 a ::: a 11 1n . . A . . A = ; . . a ::: a n1 nn i+j the numberC = (1) det(A ) is called the cofactor of the entrya , and the (n1)(n1) matrix ij ij ij 0 1 a ::: a  a ::: a 11 1(j1) 1(j+1) 1n B C . . . . . . . . . . B C . . . . . B C B C a ::: a  a ::: a (i1)1 (i1)(j1) (i1)(j+1) (i1)n B C B C A =  :::    :::  ij B C B C a ::: a  a ::: a (i+1)1 (i+1)(j1) (i+1)(j+1) (i+1)n B C B C . . . . . . . . . . A . . . . . a ::: a  a ::: a n1 nn n(j1) n(j+1) is obtained from A by deleting row i and column j; here denotes that the entry has been deleted. Definition. The nn matrix 0 1 C ::: C 11 n1 . . A . . adj(A) = . . C ::: C 1n nn is called the adjoint of the matrix A. Remark. Note that adj(A) is obtained from the matrix A rst by replacing each entry of A by its cofactor and then by transposing the resulting matrix. Chapter 3 : Determinants page 15 of 24c Linear Algebra W W L Chen, 1982, 2008 PROPOSITION 3L. Suppose that the nn matrix A is invertible. Then 1 1 A = adj(A): det(A) Example 3.7.1. Consider the matrix 0 1 1 1 0 A A = 0 1 2 : 2 0 3 Then       0 1 1 2 1 0 1 0 det det det B C 0 1 0 3 0 3 1 2 B      C 3 3 2 B C 0 2 1 0 1 0 B C A adj(A) = det det det = 4 3 2 : B C 2 3 2 3 0 2 B C       2 2 1 A 0 1 1 1 1 1 det det det 2 0 2 0 0 1 On the other hand, adding 1 times column 1 to column 2 and then using cofactor expansion on row 1, we have 0 1 0 1   1 1 0 1 0 0 1 2 A A det(A) = det 0 1 2 = det 0 1 2 = det =1: 2 3 2 0 3 2 2 3 It follows that 0 1 3 3 2 1 A A = 4 3 2 : 2 2 1 Next, we turn our attention to systems of n linear equations in n unknowns, of the form a x +::: +a x = b ; 11 1 1n n 1 . . . a x +::: +a x =b ; n1 1 nn n n represented in matrix notation in the form Ax =b; where 0 1 0 1 a ::: a b 11 1n 1 . . . A A . . . A = and b = (8) . . . a ::: a b n1 nn n represent the coecients and 0 1 x 1 . A . x = (9) . x n represents the variables. Chapter 3 : Determinants page 16 of 24c Linear Algebra W W L Chen, 1982, 2008 For every j = 1;:::;k, write 0 1 a ::: a b a ::: a 11 1(j1) 1 1(j+1) 1n B C . . . . . A (b) = . . . . . ; (10) A j . . . . . a ::: a b a ::: a n1 n nn n(j1) n(j+1) 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 det(A (b)) det(A (b)) 1 n x = ; :::; x = ; 1 n det(A) det(A) where the matrices A (b);:::;A (b) are de ned by (10). 1 1 Example 3.7.2. Consider the system Ax =b, where 0 1 0 1 1 1 0 1 A A A = 0 1 2 and b = 2 : 2 0 3 3 Recall that det(A) =1. By Cramer's rule, we have 0 1 0 1 0 1 1 1 0 1 1 0 1 1 1 A A A det 2 1 2 det 0 2 2 det 0 1 2 3 0 3 2 3 3 2 0 3 x = =3; x = =4; x = = 3: 1 2 3 det(A) det(A) det(A) Let us check our calculations. Recall from Example 3.7.1 that 0 1 3 3 2 1 A A = 4 3 2 : 2 2 1 We therefore have 0 1 0 10 1 0 1 x 3 3 2 1 3 1 A A A A x = 4 3 2 2 = 4 : 2 x 2 2 1 3 3 3 3.8. Further Discussion In this section, we shall rst discuss a de nition of the determinant in terms of permutations. In order to do so, we need to make a digression and discuss rst the rudiments of permutations on non-empty nite sets. Definition. LetX be a non-empty nite set. A permutation  onX is a function :XX which is one-to-one and onto. If x2X, we denote by x the image of x under the permutation . It is not dicult to see that if  : X X and : X X are both permutations on X, then  :XX, de ned byx = (x) for everyx2X so that is followed by , is also a permutation on X. Chapter 3 : Determinants page 17 of 24c Linear Algebra W W L Chen, 1982, 2008 Remark. Note that we use the notation x instead of our usual notation (x) to denote the image of x under . Note also that we write  to denote the composition . 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 nite, we may assume, without loss of generality, that it is f1; 2;:::;ng, wheren2N. We now letS denote the set of all permutations on the setf1; 2;:::;ng. In n other words, S denotes the collection of all functions fromf1; 2;:::;ng tof1; 2;:::;ng that are both n one-to-one and onto. PROPOSITION 3N. For every n2N, the set S has n elements. n Proof. There are n choices for 1. For each such choice, there are (n 1) choices left for 2. And so on. To represent particular elements of S , there are various notations. For example, we can use the n notation   1 2 ::: n 1 2 ::: n to denote the permutation . Example 3.8.1. In S , 4   1 2 3 4 2 4 1 3 denotes the permutation , where 1 = 2, 2 = 4, 3 = 1 and 4 = 3. On the other hand, the reader can easily check that      1 2 3 4 1 2 3 4 1 2 3 4 = : 2 4 1 3 3 2 4 1 2 1 3 4 A more convenient way is to use the cycle notation. The permutations     1 2 3 4 1 2 3 4 and 2 4 1 3 3 2 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 1 = 2, 2 = 4, 4 = 3 and 3 = 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 3 4 1 2 3 4 1 2 3 4 = 2 4 1 3 3 2 4 1 2 1 3 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. Example 3.8.2. In S , the permutation 6   1 2 3 4 5 6 2 4 1 3 6 5 can be represented in cycle notation as (1 2 4 3)(5 6). Example 3.8.3. In S or S , we have (1 2 4 3) = (1 2)(1 4)(1 3). 4 6 Chapter 3 : Determinants page 18 of 24c Linear Algebra W W L Chen, 1982, 2008 The last example motivates the following important idea. Definition. Suppose that n2 N. A permutation in S that interchanges two numbers among the n elements off1; 2;:::;ng 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 (x x ::: x ) and (y y ::: y ) in S are said to be disjoint if the elements 1 2 k 1 2 l n x ;:::;x ;y ;:::;y are all di erent. 1 k 1 l The interested reader may try to prove the following result. PROPOSITION 3P. Suppose that n2N. (a) Every permutation in S can be written as a product of disjoint cycles. n (b) For every subsetfx ;x ;:::;xg of the setf1; 2;:::;ng, where the elements x ;x ;:::;x are dis- 1 2 k 1 2 k tinct, the cycle (x x ::: x ) satis es 1 2 k (x x ::: x ) = (x x )(x x )::: (x x ); 1 2 k 1 2 1 3 1 k in other words, every cycle can be written as a product of transpositions. (c) Consequently, every permutation in S can be written as a product of transpositions. n Example 3.8.4. In S , the permutation 9   1 2 3 4 5 6 7 8 9 3 2 5 1 7 8 4 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 n2N. Then a permutation in S is said to be even if it is representable as n 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  +1 if  is even, () = 1 if  is odd. Remark. It can be shown that no permutation can be simultaneously odd and even. We are now in a position to de ne the determinant of a matrix. Suppose that 0 1 a ::: a 11 1n . . A . . A = (11) . . a ::: a n1 nn is an nn 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 a a :::a ; 1(1) 2(2) n(n) where  is a permutation in S . n Chapter 3 : Determinants page 19 of 24c Linear Algebra WWL Chen, 1982, 2006 We are now in a position to define the determinant of a matrix. Suppose that   a ... a 11 1n . .   . . (11) A = . . a ... a n1 nn is an n×n matrix. Definition. By an elementary product from the matrix A,we mean the product of n entries of A, no c Line twoo arfAwhic lgebrah are from the same row or same column. W W L Chen, 1982, 2008 It follows that any such elementary product must be of the form a a ...a , 1(1φ) 2(2φ) n(nφ) Definition. By the determinant of an nn matrix A of the form (11), we mean the sum where φ is a permutation in S . n X det(A) = ()a a :::a ; (12) 1(1) 2(2) n(n) Definition. By the determinant of an n×n matrix A of the form (11), we mean the sum 2Sn % (12) det(A)= "(φ)a a ...a , 1(1φ) 2(2φ) n(nφ) where the summation is over all the n permutations  in S . n φ∈S n where It isthe be sho summation wn that the is ov determinan er all the n tp de ned ermutations in thisφ win ayS is .the same as that de ned earlier by row or n column expansions. Indeed, one can use (12) to establish Proposition 3A. The very interested reader It is be shown that the determinant defined in this way is the same as that defined earlier by row or may wish to make an attempt. Here we con ne our study to the special cases when n = 2 and n = 3. column expansions. Indeed, one can use (12) to establish Proposition 3A. The very interested reader In the two examples below, we use e to denote the identity permutation. 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: Example 3.8.5. Suppose that n=2.Wehave the following: elementary product permutation sign contribution elementary product permutation sign contribution a a e +1 +a a 11 22 11 22 a a (1 2) 1 a a 12 21 12 21 a a e +1 +a a 11 22 11 22 a a (1 2) −1 −a a 12 21 12 21 Hence det(A) =a a a a as shown before. 11 22 12 21 Hence det(A)= a a −a a as shown before. 11 22 12 21 Example 3.8.6. Suppose that n = 3. We have the following: Example 3.8.6. Suppose that n=3.Wehave the following: elementary product permutation sign contribution elementary product permutation sign contribution a a a e +1 +a a a 11 22 33 11 22 33 a a a e +1 +a a a 11 22 33 11 22 33 a a a (1 2 3) +1 +a a a 12 23 31 12 23 31 a a a (1 2 3) +1 +a a a 12 23 31 12 23 31 a a a (1 3 2) +1 +a a a 13 21 32 13 21 32 a a a (1 3 2) +1 +a a a 13 21 32 13 21 32 a a a (1 3) 1 a a a 13 22 31 13 22 31 a a a (1 3) −1 −a a a 13 22 31 13 22 31 a a a (2 3) 1 a a a 11 23 32 11 23 32 a a a (2 3) −1 −a a a 11 23 32 11 23 32 a a a (1 2) 1 a a a 12 21 33 12 21 33 a a a (1 2) −1 −a a a 12 21 33 12 21 33 Hence det(A) = a a a +a a a +a a a a a a a a a a a a . We have the Hence det(A)= a a a + a a a + a a a − a a a − a a a − a a a . We have the 11 11 22 22 33 33 12 12 23 23 31 31 13 13 21 21 32 32 13 13 22 22 31 31 11 11 23 23 32 32 12 12 21 21 33 33 picture below: picture below: + + + − − − a a a a a 11 12 13 11 12 a a a a a 21 22 23 21 22 a a a a a 31 32 33 31 32 Chapter 3 : Determinants page 20 of 24 Next, we discuss brie y 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 I , then det(E) =1. n (b) If E arises from adding one row of I to another row, then det(E) = 1. n (c) If E arises from multiplying one row of I by a non-zero constant c, then det(E) =c. n Combining Propositions 3D and 3Q, we can establish the following intermediate result. Chapter 3 : Determinants page 20 of 24

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