Chapter 3 Determinants 3.1 The determinant of a matrix Homework: [Textbook, 3.1 Ex. 15, 17, 27, 33, 47, 55, 57; page 131]. The main point in this section is the following: 1. Define determinant of a matrix. 2. Compute determinant of a matrix, in various ways. 87
88 CHAPTER 3. DETERMINANTS Definition 3.1.1 We define determinant of square matrices. Here we give the definition only for square matrices of size 2 2 and 3 3. 1. Let [ ] a b A = c d be a matrix of size 2 2. Define determinant of A as det(a) = ad bc. 2. Let A = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 be a matrix of size 3 3. Now, write a new 3 5 matrix a 11 a 12 a 13 a 11 a 12 a 21 a 22 a 23 a 21 a 22 a 31 a 32 a 33 a 31 a 32 by augmenting that first two columns of A to itself. This matrix has 3 left to right diagonals and 3 right to left diagonals. The determinant of A is sum of product of all entries in each left to right diagonal minus sum of product of all entries in each right to left diagonal. More precisely, det(a) = (a 11 a 22 a 33 + a 12 a 23 a 31 + a 13 a 21 a 32 ) (a 13 a 22 a 31 + a 11 a 23 a 32 + a 12 a 21 a 33 ). 3. Determinant of square matrices of all sizes will be defined soon, in an unified way. Before we do that, let us analyze the definition in 3 3 case. (a) First, there are 6 terms and note 6 = 1 2 3. (If you know it, 6 is the number of possible permutaitions of 3 objects.)
3.1. THE DETERMINANT OF A MATRIX 89 (b) Each term has EXACTLY one factor of each row and each column. (c) There is a way to determine the sign for each term, which we will not discuss. To define determinant of square matrices of all sizes, we will have to define Minors and cofactors of a square matrices. Definition 3.1.2 Suppose A is a square matrix of size n n. 1. The minor M ij of the element a ij (or (ij) th minor) is the determinant of the matrix obtained by deleting the i th row and j th column from the matrix A. 2. The (ij) th cofactor is defined to be C ij = ( 1) i+j M ij. 3. Remark. We used the word determinant in this definition, without definiting determinant. This is not mathematically proper. (a) But we did define determinant of matrices of size 3 3. Note that the minors of matrices of size 4 4 are determinants of matrices of size 3 3. So, there is nothing improper in the definition of minors for matrices of size 4 4. (b) Subsesuently using the definition of cofactors of matrices of size 4 4, we will define determinant of matrices of size 4 4. (c) Then, using the definition of cofactors of matrices of size 5 5, we will define determinant of matrices of size 5 5.
90 CHAPTER 3. DETERMINANTS (d) This process continues and the definition of determinant of matrices of all sizes will be define. Such a process is called an inductive process, and in this case an inductive definition. Definition 3.1.3 Suppose A = [a ij ] is a matrix of order n (where n is atleast 2). Then, determinant of A is the sum of the entries in the first row of A multiplied by their cofactors. That means det(a) = n a 1j C 1j = a 11 C 11 + a 12 C 12 + + a 1n C 1n. j=1 1. We used the first row of A to give this definition. 2. There is nothing so special about the first row. This can be done using any row and there is a thoerem that says it will make no difference if we use the second or the third or any i th row. This means, we have det(a) = n a 1j C 1j = j=1 n a ij C ij = a i1 C i1 + a i2 C i2 + + a in C in. j=1 for any i with 1 i n. This will be called the i th row expansion of the determinant. 3. What is so special about rows and what is wrong with columns? Answer is none. There is a thoerem that says, we could use the columns to define determinants, instead of rows. This means, we have n n det(a) = a 1j C 1j = a ij C ij = a 1j C 1j +a 2j C 2j + +a nj C nj. j=1 i=1 4. We also use the notation det(a) = A.
3.1. THE DETERMINANT OF A MATRIX 91 3.1.1 Triangular matrices In some cases, it is easier to compute determiants. They include diagonal matrices and triangular matrices. Definition 3.1.4 For a square matrix A = [a ij ], wehave the folowing definitions: 1. We say A is upper triangular, if all the enties below the main diagonal is zero. An upper triangular matrix looks like: a 11 a 12 a 13 a 1n 0 a 22 a 23 a 2n A = 0 0 a 33 a 3n 0 0 0 a nn 2. We say A is lower triangular, if all the enties above the main diagonal is zero. A lower triangular matrix looks like: a 11 0 0 0 a 21 a 22 0 0 A = a 31 a 32 a 33 0 a n1 a n2 a n3 a nn 3. We say A is a diagonal matrix, if all the off diagonal entries are zero. A diagonal matrix looks like a 11 0 0 0 0 a 22 0 0 A = 0 0 a 33 0 0 0 0 a nn
92 CHAPTER 3. DETERMINANTS Note that a diagonal matrix is both anupper triangular and lower triangular matrix. Theorem 3.1.5 Let A be a triangular matrix. Then the determinant of A is the product of the entries on the main diagonal. That means, with notations as in the definition above, det(a) = a 11 a 22 a nn. Proof. We prove it by the method of mathematical inducton. Let A be a lower triangular matix: a 11 0 0 0 a 21 a 22 0 0 A = a 31 a 32 a 33 0 a n1 a n2 a n3 a nn If n = 1 then det(a) = a 11 and the theorem is valid. If n = 2, then [ ] a11 0 A = and det(a) = a a 21 a 11 a 22. 22 So, the theorem is valid for n = 2. Now assume that the theorem is valid for matrices of order n 1. (This assumption would be called the induction hypothesis. ) We need to prove the theorem for matrices of order n. Then the minor M 11 of a 11 is the determinant of a lower triangular matrix and by our induction hypothesis, we have M 11 = a 22 a 33 a nn ; the cofactor C 11 = ( 1) 1+1 M 11 = a 22 a 33 a nn. From definition 3.1.3 of determinant, det(a) = a 11 C 11 + a 12 C 12 + + a 1n C 1n = a 11 a 22 a 33 a nn + 0C 12 + + 0C 1n = a 11 a 22 a 33 a nn. Therefore, by the method of mathematical induction, the theorem is established for square matrices of all sizes n.
3.1. THE DETERMINANT OF A MATRIX 93 Reading assignament: Read Example 1-5 (page 123-). Exercise 3.1.6 (Ex. 16, p. 130) Compute all theminors and cofactors of the matrix A = 3 4 2 6 3 1 4 7 8 Solution: We have 3 1 M 11 = 7 8 = 17 M 6 1 12 = 4 8 = 52 M 6 3 13 = 4 7 = 54 4 2 M 21 = 7 8 = 18 M 3 2 22 = 4 8 = 16 M 3 4 23 = 4 7 = 5 4 2 M 31 = 3 1 = 2 M 3 2 32 = 6 1 = 15 M 3 4 33 = 6 3 = 33. Now the formuala for cofactor is C ij = ( 1) i+j M ij. So, the cofactors are given by: C 11 = ( 1) i+j M ij = ( 1) 1+1 M 11 = 17, C 12 = ( 1) 1+2 M 12 = ( 1)( 52) = 52. C 13 = ( 1) 1+3 M 13 = 54 C 21 = ( 1) 2+1 M 21 = 18, C 22 = ( 1) 2+2 M 22 = 16, C 23 = ( 1) 2+3 M 23 = 5, C 31 = ( 1) 3+1 M 31 = 2, C 32 = ( 1) 3+2 M 32 = 15,. C 33 = ( 1) 3+3 M 33 = 33. Exercise 3.1.7 (Ex. 18, p. 130) Find the determinant of the matrix in Exercise 3.1.6, using the method of expansion by cofactors, as follows:
94 CHAPTER 3. DETERMINANTS 1. Use the first row: Solution: We have, by expanding by first row: A = a 11 C 11 +a 12 C 12 +a 13 C 13 = 3 ( 17)+4 52+2 ( 54) = 151. 2. Use third row. Solution: We have, by expanding by third row: A = a 31 C 31 +a 32 C 32 +a 33 C 33 == 4 ( 2)+( 7) 15+( 8) ( 33) = 151. 3. Use the first column. Solution: We have, by expanding by first column: A = a 13 C 13 +a 21 C 21 +a 31 C 31 = 4 ( 17)+6 18+4 ( 2) = 151. Exercise 3.1.8 (Ex. 28. p. 131) Use expansion by cofactors to compute the determinant of 1 4 3 2 5 6 2 1 0 0 0 0 3 2 1 5. Solution: Looking at the problem, it is clear that it will be best to try to expand by second row: So, A = a 31 C 31 +a 32 C 32 +a 33 C 33 +a 34 C 34 = 0C 31 +0C 32 +0C 33 +0C 34 = 0. Exercise 3.1.9 (Ex. 30. p. 131) Use expansion by cofactors to compute the determinant of 3 0 7 0 2 6 11 12 4 1 1 2. 1 5 2 10
3.1. THE DETERMINANT OF A MATRIX 95 Solution:Since there are two zero entries in the first row, we will expand by first row: A = a 11 C 11 + a 12 C 12 + a 13 C 13 + a 34 1 34 = 3C 11 + 0C 12 + 7C 13 + 0C 34. Which is = 3C 11 + 7C 13. So, we will compute the cofactors C 11,C 13. 6 11 12 C 11 = ( 1) 1+1 M 11 = 1 1 2 = 6 ( 14) 11 0 + 12 7 = 0. 5 2 10 And 2 6 12 C 13 = ( 1) 1+3 M 13 = 4 1 2 = (2 0 6 38 + 12 19) = 0. 1 5 10 Therefore, A = 3C 11 + 7C 13 = 3 0 + 7 0 = 0. Exercise 3.1.10 (Ex. 46. p. 131) Use expansion by cofactors to compute the determinant of A = 7 0 0 0 0 8 1 4 0 0 0 4 5 2 0 0 3 3 5 1 0 1 13 4 1 2 Solution: This is a lower triangular matrix. By theorem 3.1.5, the determiant will be given by the product of the entries in the mail diagonal. So,. A = 7 1 2 ( 1) ( 2) = 7. 4
96 CHAPTER 3. DETERMINANTS Problems for Extra Practice: Ex. 20, 21, 22, 44, 45, page 131.
3.2. DETERMINANTS AND ELEMENTARY OPERATIONS 97 3.2 Determinants and Elementary Operations Homework: [Textbook, 3.2, page 140 Ex. 11, 15, 17, 27, 29, 31, 39, 49, 51] The main point in this section is to understtand elementary properties of determinants.
98 CHAPTER 3. DETERMINANTS Following is the main theorem in this section. Theorem 3.2.1 Suppose A,B be two square matrices of same size. 1. If B is obtained from A by interchanging two rows of A, then det(a) = det(b). 2. If B is obtained from A by adding a constant multiple of a row of A to another row of A, then det(a) = det(b). 3. If B is obtained from A by multiplying a row of A by a scalar c, then det(b) = c det(a). 4. All the above statements remain valid, if we replace the word row by column. Theorem 3.2.2 If A is a square matrix and any one of the following conditions is true, then det(a) = 0. 1. An entire row (or column) consists of zeros. 2. Two rows (or columns) are equal. 3. One row (or column) is a multiple of another row (or column). Reading assignment: Read examples 1-6 (page 133).
3.2. DETERMINANTS AND ELEMENTARY OPERATIONS 99 Exercise 3.2.3 (Ex. 28, p. 141) Use elementary row or column operations to compute the determinant: 3 1 3 A = 1 4 2. 3 1 1 Solution: we will try to reduce the matrix to a triangular matrix and use theorem 3.1.5. Switch first and second row: 1 4 2 A = 3 1 3 3 1 1 Add 3 times the second column to the first: 1 4 2 A = 0 1 3 0 1 1 Subtract second row from third: 1 4 2 A = 0 1 3 0 0 2 This is a triangular matrix and by theorem 3.1.5, A = ( 13) ( 1)(2) = 26. Exercise 3.2.4 (Ex. 34, p. 141) Use elementary row or column operations to compute the determinant: 9 4 2 5 A = 2 7 6 5 4 1 2 0. 7 3 4 10
100 CHAPTER 3. DETERMINANTS Solution: Subtract second and fourth row from first 0 14 8 0 A = 2 7 6 5 4 1 2 0. 7 3 4 10 Add 2 times second row to fourth: 0 14 8 0 A = 2 7 6 5 4 1 2 0 11 17 16 0. Now, interchange second and fourth row: 0 14 8 0 A = 11 17 16 0 4 1 2 0 2 7 6 5. Expanding by cofactors along the last column, we get 0 14 8 A = 5 11 17 16. 4 1 2 Now expand by first row: ( A = 5 ( 14)( 1) 1+2 11 16 4 2 + ( 8)( 1)1+3 11 17 4 1 ). So, A = 5 (14 ( 86) 8 ( 57)) = 3740..
3.3. PROPERTIES OF DETERMINANTS 101 3.3 Properties of determinants We skip this section. In any case, let me state the following without proofs. Theorem 3.3.1 Suppose A, B are two n n matrices. Then, det(ab) = det(a)det(b). Theorem 3.3.2 Suppose A is a n n matrix. Then, A is invertible (nonsingular) if and only if det(a) 0. 3.4 Introduction to eigen values We skip this section. 3.5 Application of Determinants Homework: [Textbook, 3.5 Ex. 5, 7, 27, 31, 47, 49, 51; page 169]. The main points in this section are the following: 1. We define adjoint of a matrix. 2. Using adjoint, we give formula to compute inverse of a matrix. 3. We describle Cramer s rule to solve non-singular system of linear equations.
102 CHAPTER 3. DETERMINANTS 3.5.1 Cofactors and adjoint of a matrix Definition 3.5.1 Suppose A is an n n matrix and as before C ij denotes the (i,j) th cofactor. The matrix of cofactors of A is defined as C 11 C 12 C 13 C 1n C 21 C 22 C 23 C 2n C 31 C 32 C 33 C 3n C = [C ij ] =. C n1 C n2 C n3 C nn We also call it the cofactor matrix of A. The transpose of the cofactor matrix is called the adjoint of A, which is denoted by Adj(A). So, C 11 C 21 C 31 C n1 C 12 C 22 C 32 C n2 Adj(A) = C T C 13 C 23 C 33 C n3 =. C 1n C 2n C 3n C nn Theorem 3.5.2 Suppose A is an n n invertible matrix. Then, A[Adj(A)] = det(a)i n = [Adj(A)]A. and A 1 = 1 det(a) Adj(A). Proof. In fact, later statement follows from the former statemnt: A[Adj(A)] = det(a)i n = [Adj(A)]A.
3.5. APPLICATION OF DETERMINANTS 103 So, we prove this statement. For simpicity, I will first give a proof for matrices A of size 3 3. So, a 11 a 12 a 13 C 11 C 21 C 31 A = a 21 a 22 a 23 and Adj(A) = C 12 C 22 C 32. a 31 a 32 a 33 C 13 C 23 C 33 So, A[Adj(A)] = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 Let use compute the first row: C 11 C 21 C 31 C 12 C 22 C 32 C 13 C 23 C 33 = p 11 p 12 p 13 p 21 p 22 p 23 p 31 p 32 p 33 (say). p 11 = a 11 C 11 + a 12 C 12 + a 13 C 13 = det(a) (by definition 3.1.3). Then p 12 = a 11 C 21 +a 12 C 22 +a 13 C 23 = Then, p 13 = a 11 C 31 +a 12 C 32 +a 13 C 33 = a 11 a 12 a 13 a 11 a 12 a 13 a 31 a 32 a 33 a 11 a 12 a 13 a 21 a 22 a 23 a 11 a 12 a 13 = 0 = 0 Similarly, we compute the second row: a 21 a 22 a 23 p 21 = a 21 C 11 +a 22 C 12 +a 23 C 13 = a 21 a 22 a 23 a 31 a 32 a 33 = 0 and (b/c two rows are same). (b/c two rows are same). (b/c two rows are same). p 22 = a 21 C 21 + a 22 C 22 + a 23 C 23 = det(a) (by definition 3.1.3). p 23 = a 21 C 31 +a 22 C 32 +a 23 C 33 = a 11 a 12 a 13 a 21 a 22 a 23 a 21 a 22 a 23 = 0 (b/c two rows are same).
104 CHAPTER 3. DETERMINANTS Now, we compute the third row: p 31 = a 31 C 11 +a 32 C 12 +a 33 C 13 = Also p 32 = a 31 C 21 +a 32 C 22 +a 33 C 23 = Finally, a 31 a 32 a 33 a 21 a 22 a 23 a 31 a 32 a 33 a 11 a 12 a 13 a 31 a 32 a 33 a 31 a 32 a 33 = 0 = 0 (b/c two rows are same). (b/c two rows are same). p 33 = a 31 C 31 + a 32 C 32 + a 33 C 33 = det(a) (by definition 3.1.3). Therefore, we have, A[Adj(A)] = det(a) 0 0 0 det(a) 0 0 0 det(a) Similarly, (using column expansion definition 3.1.3), we can prove det(a) 0 0 [Adj(A)]A = 0 det(a) 0. 0 0 det(a) So, Dividing by det(a) we have ( ) 1 A det(a) Adj(A) = So, A 1 = 1 matrix. det(a) A[Adj(A)] = [Adj(A)]A = det(a)i 3. ( ) 1 det(a) Adj(A) A = I 3. Adj(A). So, the proof is complete when A is a 3 3 Proof in the general case: This means, A is an n n matrix and we will prove A[Adj(A)] = [Adj(A)]A = det(a)i n.
3.5. APPLICATION OF DETERMINANTS 105 In fact, I was too verbose in the 3 3 case, and I computed all the 9 terms of A[Adj(A)], explicitly. This is not possible in the general case. As usual we denote A = [a ij ] and Adj(A) = [C ji ], as above. The (i,j) th term p ij of A[Adj(A)] is given by p ij = a i1 C j1 + a i2 C j2 + + a in C jn = n a ik C jk. k=1 For the diagonal terms, we have i = j and we have, p ii = a i1 C i1 + a i2 C i2 + + a in C in = det(a), which follows from definition 3.1.3, by expanding along the i th row. If i j, then (assume i < j), we have a 11 a 12 a 13 a 1n a 21 a 22 a 23 a 2n p ij = a i1 a i2 a i3 a in = 0 (b/c two rows are same). a i1 a i2 a i3 a in a n1 a n2 a n3 a nn (It is the determinant of the matrix obtained from A, by replacing j th row of A by the i th row.) So, we have proved that det(a) 0 0 0 0 det(a) 0 0 A[Adj(A)] = 0 0 det(a) 0 = det(a)i n. 0 0 0 det(a) Using column expansion of determinants, similarly, we prove So, the proof is complete. [Adj(A)]A = det(a)i n.
106 CHAPTER 3. DETERMINANTS 3.5.2 Cramer s Rule Cramer s Rule is a method of solving systems of linear equations, when number of equations is same as the mumber is unknowns. Theorem 3.5.3 Suppose Ax = b is a system of n linear equations in n variables x 1,x 2,...,x n. Assume det(a) 0. Then, x 1 = det(a 1) det(a),x 2 = det(a 2) det(a),,x n = det(a n) det(a), where the metrix A i is obtained from A by replacing the i th column of A the column b of constants. Proof. Left-multiply the system Ax = b by Adj(A), we have By theorem 3.5.2, it follows So, det(a)i n x = Adj(A)b (Adj(A))Ax = Adj(A)b x 1 x 2 x n or det(a)x 1 det(a)x 2 det(a)x n = 1 det(a) Adj(A)b. = Adj(A)b. So, x i is the i th entry (row) of the column matrix 1 det(a) Adj(A)b. Therefore, x i = 1 det(a) = b 1C 1i + b 2 C 2i + + b n C ni = 1 det(a) det(a i), by expanding det(a i ) along the i th column of A i. This completes the proof. Reading assignment: Read Examples 1-4, page 158-.
3.5. APPLICATION OF DETERMINANTS 107 Exercise 3.5.4 (Ex. 4, p. 168) Let 1 2 3 A = 0 1 1. 2 2 2 Find the adjoint of A and use it to compute the inverse of A. Solution: We will use TI to compute determinants, whenever needed. First, we compute the cofactors: C 11 = ( 1) 1+1 1 1 2 2 = 4 C 12 = ( 1) 1+2 0 1 2 2 = 2 and C 13 = ( 1) 1+3 0 1 2 2 = 2 C 21 = ( 1) 2+1 2 3 2 2 = 2 and C 22 = ( 1) 2+2 1 3 2 2 = 4 C 23 = ( 1) 2+3 1 2 2 2 = 2 and C 31 = ( 1) 3+1 2 3 1 1 = 5 C 32 = ( 1) 3+2 1 3 0 1 = 1 and So, the cofactor matrix is 4 2 2 C = 2 4 2 5 1 1 C 33 = ( 1) 3+3 1 2 0 1 = 1 and Adj(A) = C t = 4 2 5 2 4 1 2 2 1.
108 CHAPTER 3. DETERMINANTS Using TI,wehave det(a) = 6. So, ( ) 1 A 1 = Adj(A) = 1 4 2 5 2 4 1 det(a) 6 2 2 1 Exercise 3.5.5 (Ex. 6, p. 168) Let 0 1 1 A = 1 2 3. 1 1 2 = 2 3 1 3 Find the adjoint of A and use it to compute the inverse of A. 1 3 5 6 2 3 1 6 1 3 1 3 1 6 Solution: We will use TI to compute determinants, whenever needed. Firsr, we compute the cofactors: C 11 = ( 1) 1+1 2 3 1 2 = 1 C 12 = ( 1) 1+2 1 3 1 2 = 1. and and C 13 = ( 1) 1+3 1 2 1 1 C 22 = ( 1) 2+2 0 1 1 2 = 1 C 21 = ( 1) 2+1 = 1 C 23 = ( 1) 2+3 1 1 1 2 0 1 1 1 = 1 = 1 and C 31 = ( 1) 3+1 1 1 2 3 = 1 C 32 = ( 1) 3+2 0 1 1 3 = 1 and C 33 = ( 1) 3+3 0 1 1 2 = 1
3.5. APPLICATION OF DETERMINANTS 109 So, the cofactor matrix is C = 1 1 1 1 1 1 1 1 1 and Adj(A) = C t = 1 1 1 1 1 1 1 1 1 Using TI,wehave det(a) = 0. So, A does not have an inverse..
110 CHAPTER 3. DETERMINANTS Problems on Cramer s Rule Exercise 3.5.6 (Ex. 28, p. 169) Use Cramer s Rule to solve the system of linear equations: 4x 1 2x 2 +3x 3 = 2 2x 1 +2x 2 +5x 3 = 16 8x 1 5x 2 2x 3 = 4 Solution: Here the coefficient matrix is 4 2 3 A = 2 2 5 and det(a) = 82. 8 5 2 We write the A i by replacing the i th column by the constant vector. So, And And A 1 = A 2 = A 3 = 2 2 3 By Cramer s rule 16 2 5 4 5 2 4 2 3 2 16 5 8 4 2 4 2 2 2 2 16 8 5 4 and det(a 1 ) = 410. and det(a 2 ) = 656. and det(a 3 ) = 164. x 1 = det(a 1) det(a) = 410 82 = 5, x 2 = det(a 2) det(a) = 656 82 = 8,
3.5. APPLICATION OF DETERMINANTS 111 x 3 = det(a 3) det(a) = 164 82 = 2. Exercise 3.5.7 (Ex. 30, p. 169) Use Cramer s Rule to solve the system of linear equations: 14x 1 21x 2 7x 3 = 21 4x 1 +2x 2 2x 3 = 2 56x 1 21x 2 +7x 3 = 7 Solution: Here the coefficient matrix is 14 21 7 A = 4 2 2 and det(a) = 1568. 56 21 7 We write the A i by replacing the i th column by the constant vector. So, A 1 = 21 21 7 2 2 2 7 21 7 and det(a 1 ) = 1568. And And A 2 = A 3 = 14 21 7 4 2 2 56 7 7 14 21 21 4 2 2 56 21 7 and det(a 2 ) = 3136. and det(a 3 ) = 1568. By Cramer s rule
112 CHAPTER 3. DETERMINANTS x 1 = det(a 1) det(a) = 1568 1568 = 1, x 2 = det(a 2) det(a) = 3136 1568 = 2, x 3 = det(a 3) det(a) = 1568 1568 = 1. 3.5.3 Area, Volume, and Equations of lines and planes Theorem 3.5.8 We have thoe following three part theorem(s) regarding triangles: 1. The area of the triangle whose vertices are (x 1,y 1 ), (x 2,y 2 ), (x 3,y 3 ) is given by Area = ± 1 x 1 y 1 1 x 2 2 y 2 1 x 3 y 3 1 where the sign ± is chosen to yield a positive area. 2. Three points (x 1,y 1 ), (x 2,y 2 ), (x 3,y 3 ) are collinear if and only if x 1 y 1 1 x 2 y 2 1 = 0. x 3 y 3 1 3. The equation of the line passing through two distince points (x 1,y 1 ), (x 2,y 2 ) is given by x y 1 x 1 y 1 1 = 0. x 2 y 2 1
3.5. APPLICATION OF DETERMINANTS 113 Proof. You can skip the proof, if you want. I will also skip the proof of (1), you can look at the textbook. Now, since three points are collinear if and only if the area of the corresponding triangle is zero, (1) (2). Similarly, a point (x,y) is nn the line through (x 1,y 1 ), (x 2,y 2 ) if and only if (x,y), (x 1,y 1 ), (x 2,y 2 ) are collinear. So, The proof is complete. (2) (3). We have similar theorems for points in 3-space. Theorem 3.5.9 We have thoe following three part theorem(s) regarding tetrahedrons: 1. The volume of the tetrahedron whose vertices are (x 1,y 1,z 1 ), (x 2,y 2,z 2 ), (x 3,y 3,z 3 ), (x 4,y 4,z 4 ) is given by x 1 y 1 z 1 1 V olume = ± 1 x 2 y 2 z 2 1 6 x 3 y 3 z 3 1 x 4 y 4 z 4 1 where the sign ± is chosen to yield a positive area. 2. Four points (x 1,y 1,z 1 ), (x 2,y 2,z 2 ), (x 3,y 3,z 3 ), (x 4,y 4,z 4 ) in 3-space are coplanar if and only if x 1 y 1 z 1 1 x 2 y 2 z 2 1 x 3 y 3 z 3 1 = 0. x 4 y 4 z 4 1
114 CHAPTER 3. DETERMINANTS 3. The equation of the plane passing through three distince points (x 1,y 1,z 1 ), (x 2,y 2,z 2 ), (x 3,y 3,z 3 ) in 3-space is given by x y z 1 x 1 y 1 z 1 1 x 2 y 2 z 2 1 = 0. x 3 y 3 z 3 1 Proof. Skip. Reading assignment: Read Examples 5-8, page 165-, from the textbook. Exercise 3.5.10 (Ex. 48. p. 169) Find the area of the triange passing through the points (1, 1), ( 1, 1), (0, 2). Solution: By theorem 3.5.8 (1), Area = ± 1 x 1 y 1 1 x 2 2 y 2 1 = ± 1 2 x 3 y 3 1 1 1 1 1 1 1 0 2 1 = 6/2 = 3. Exercise 3.5.11 (Ex. 52. p. 169) Determine if the points ( 1, 3), ( 4, 7), (2, 13) are colinear. Solution: Because of theorem 3.5.8, we compute Area = ± 1 x 1 y 1 1 x 2 2 y 2 1 = ± 1 1 3 1 4 7 1 2 x 3 y 3 1 2 13 1 = 0. Since the area is zero, these three points are colinear.