145 3 Mathematics Term 1 STPM Chapter 3 Matrices Exercise 3.1 1. If A = 1 1 –2 4 0 0 2 2 and B = 1 0 3 –1 5 –3 1 2 , find (a) 3A (b) –B (c) A + B (d) A – B (e) 2A – 3B. 2. If A = 1 2 2 1 2 1 0 2 –3 4 2, B = 1 3 3 6 3 0 9 3 5 –1 2 and C = 1 4 5 7 4 –1 8 4 0 –1 2, find (a) 3A – 6B + 9C (b) 7A – 2(B – C) 3. Given that A = 1 2 –1 1 3 2, B = 1 1 0 –1 4 3 2 2 , C = 1 –3 0 2 1 4 0 –2 –1 1 2 and D = 1 1 2 –3 –4 0 1 2, determine if each of the following matrix multiplications are possible: A2 , AB, AC, AD, BA, B2 , BC, BD, CA, CB, C2 , CD, DA, DB, DC and D2 . For those multiplications that are possible, find the matrix and state its order. Show that (i) (AB)C = A(BC), (ii) (BC)D = B(CD). 4. Given that A = 1 –2 3 0 4 2, B = 1 2 1 –1 3 2 and C = 1 4 –1 1 3 2 0 2 , find (a) AB, (b) BA, (c) BC, (d) AC. Hence, show that (A + B)C = AC + BC. 5. Given that 31 x z y w2 = 1 x –1 6 2w2 + 1 4 z + w x + y 3 2, find x, y, z and w. 6. Given that A = 1 1 –1 2 –2 0 1 2 1 3 2 and B = 1 –2 1 3 0 2 –1 2 –1 1 2, show that AB ≠ BA. 7. If A = 1 2 3 –1 2 2 and B = 1 a b 1 1 2, find the values of a and b such that AB = BA. 8. If A = (2 –2 4), B = (0 1 2), C = 1 5 0 1 2 and D = 1 –1 –1 2 2, find (a) 5CA (b) (5D)(3B) (c) 5AC – (2A – B)D (d) (A + B)(C – D) (e) (C – D)(A + B) (f) AC + BD 9. Find (a) 1 2 –1 3 –2 3 0 21 4 6 3 2 1 5 2 (b) 1 4 6 3 2 1 5 21 2 –1 3 –2 4 0 2 10. If A = 1 2 –1 0 4 3 9 2 , B = 1 1 0 –1 0 –1 0 –1 1 11 2 and C = 1 –1 2 –3 –1 2 –3 2 , find (a) AB (b) CA (c) AC (d) BC (e) C(AB) (f) B(CA)
146 3 Mathematics Term 1 STPM Chapter 3 Matrices The determinant of a matrix Determinant of a 2 × 2 matrix If A = 1 a c b d 2, the determinant of the matrix A, written det A, is given by det A = * a c b d * = ad – bc. For the matrices B = 1 2 4 5 7 2 and C = 1 3 –2 6 1 2, det B = * 2 4 5 7 * = 2 × 7 – 5 × 4 = –6 and det C = * 3 –2 6 1 * = 3 × 1 – 6(–2) = 15. Note: The determinant of the matrix A may also be denoted by |A| or ∆. Determinant of a 3 × 3 matrix Consider the matrix A = 1 a11 a21 a31 a12 a22 a32 a13 a23 a33 2 and its determinant det A = a11 a21 a31 a12 a22 a32 a13 a23 a33 . Suppose that we choose any element, say a21, and strike out the row and column that pass through a21. We will then obtain a 2 × 2 determinant from the remaining elements, i.e. * a12 a32 a13 a33 *. This determinant is called the minor of the element a21. For example, for the determinant * 1 4 7 2 5 8 3 6 9 * , the minor for the element 4 is * 2 8 3 9 * = 18 – 24 = –6. The minor of any element can be given either a ‘+’ or ‘–’ sign, according to the position of the chosen element in the matrix, such as follows: * + – + – + – + – + * A minor of any chosen element, together with its sign, is called the cofactor of the element. Hence, for the matrix with determinant * 1 4 7 2 5 8 3 6 9 * , the minor for 4 is * 2 8 3 9 * = –6 and the cofactor for 4 is – * 2 8 3 9 * = – (–6) = 6. The cofactor for 3 is + * 4 7 5 8 * = 32 – 35 = –3. In general, for * a11 a21 a31 a12 a22 a32 a13 a23 a33 * , let A11, A12 and A13 represent the cofactors of a11, a12 and a13 respectively.
147 3 Mathematics Term 1 STPM Chapter 3 Matrices Then A11 = * a22 a32 a23 a33 *, A12 = –* a21 a31 a23 a33 * and A13 = * a21 a31 a22 a32 *. With these, det A = a11A11 + a12A12 + a13A13 = a11* a22 a32 a23 a33 * – a12* a21 a31 a23 a33 * + a13* a21 a31 a22 a32 * = a11(a22 a33 – a23a32) – a12(a21a33 – a23a31) + a13 (a21a32 – a22a31). Hence, the determinant of the matrix D = 1 1 4 3 2 5 –2 3 6 –4 2 is det D = * 1 4 3 2 5 –2 3 6 –4 * = 1* 5 –2 6 –4 * – 2* 4 3 6 –4 * + 3* 4 3 5 –2 * = 1(–20 + 12) – 2(–16 – 18) + 3(–8 – 15) = 1(–8) – 2(–34) + 3(–23) = –8 + 68 – 69 = –9 We say that D is a non-singular matrix because det D ≠ 0. Now, consider the matrix E = 1 1 4 7 2 5 8 3 6 9 2 det E = * 1 4 7 2 5 8 3 6 9 * = 1* 5 8 6 9 * – 2* 4 7 6 9 * + 3* 4 7 5 8 * = 1(45 – 48) – 2(36 – 42) + 3(32 – 35) = 1(–3) – 2(–6) + 3(–3) = –3 + 12 – 9 = 0 We say that E is a singular matrix because det E = 0. A is a singular matrix if det A = 0. A is a non-singular matrix if det A ≠ 0. Example 6 If A = 1 2 –3 1 4 2 and B = 1 1 3 4 –2 2, find (a) AB (b) det A (c) det B (d) det AB Determine if det AB = (det A)(det B). Solution: A = 1 2 –3 1 4 2 and B = 1 1 3 4 –2 2 Matrices INFO
148 3 Mathematics Term 1 STPM Chapter 3 Matrices (a) AB = 1 2 –3 1 4 21 1 3 4 –2 2 = 1 2 + 3 –3 + 12 8 – 2 –12 – 82 = 1 5 9 6 –202 (c) det B = * 1 3 4 –2 * (d) det AB = * 5 9 6 –20 * = –2 – 12 = –100 – 54 = –14 = –154 det AB = –154 = 11 × (–14) = (det A)(det B) Hence, det AB = (det A)(det B). Example 7 Evaluate (a) * –1 2 6 3 4 –5 0 1 –3 * (b) * 2 3 5 1 –3 4 –4 0 2 * Solution: (a) * –1 2 6 3 4 –5 0 1 –3 * = –1* 4 –5 1 –3 * – 3* 2 6 1 –3 * + 0* 2 6 4 –5 * = –1(–12 + 5) – 3(–6 – 6) + 0(–10 – 24) = –(–7) – 3(–12) + 0 = 7 + 36 = 43 (b) * 2 3 5 1 –3 4 –4 0 2 * = 2* –3 4 0 2 * – 1* 3 5 0 2 * – 4* 3 5 –3 4 * = 2(–6 – 0) – 1(6 – 0) – 4(12 + 15) = 2 (–6) – 1(6) – 4(27) = –12 – 6 – 108 = –126 Properties of determinants There are some rules or properties of determinants which can be applied to simplify the tedious calculations involved in evaluating determinants. For our purposes, we shall refer to the matrix A and det A of order 3 × 3. Where rows only are mentioned, the property is equally applicable for columns, and vice versa. 1. If the rows and columns of a determinant are interchanged, the value of the determinant remains unchanged. For example, * 3 2 –2 1 5 3 –4 6 7 * = 3(35 – 18) – 1(14 + 12) – 4(6 + 10) = 51 – 26 – 64 = –39. (b) det A = * 2 –3 1 4 * = 8 – (–3) = 8 + 3 = 11
149 3 Mathematics Term 1 STPM Chapter 3 Matrices Interchanging the rows and columns, we have * 3 1 –4 2 5 6 –2 3 7 * = 3(35 – 18) – 2(7 + 12) – 2(6 + 20) = 51 – 38 – 52 = –39. 2. If any two rows of a determinant are interchanged, the value of the determinant changes sign. For example, * 3 2 –2 1 5 3 –4 6 7 * = –39 Interchanging columns 1 and 2, we have * 1 5 3 3 2 –2 –4 6 7 * = 1(14 + 12) – 3(35 – 18) – 4(–10 – 6) = 26 – 51 + 64 = 39. 3. If the elements of a row of a determinant are multiplied by a constant, k, the value of the determinant is multiplied by the factor k. For example, * 3 1 –2 –2 3 4 2 5 3 * = 3(9 – 20) + 2(3 + 10) + 2(4 + 6) = –33 + 26 + 20 = 13, and * 3 1 –2 –4 6 8 2 5 3 * = * 3 1 –2 2(–2) 2(3) 2(2) 4 6 9 * = 2* 3 1 –2 –2 3 4 2 5 3 * = 2 × 13 = 26. 4. If any two rows of a determinant are identical, its value is zero. For example, * 2 1 6 2 1 6 3 4 2 * = 2(2 – 24) – 2(2 – 24) + 3(6 – 6) = –44 + 44 + 0 = 0, and * 3 1 3 –5 6 –5 2 8 2 * = 3(12 + 40) + 5(2 – 24) + 2(–5 – 18) = 156 – 110 – 46 = 0. If any row of a determinant is a multiple of another row, its value is zero. For example, * 2 3 1 4 6 2 5 7 9 * = * 2 3 1 2(2) 2(3) 2(1) 5 7 9 * = 2* 2 3 1 2 3 1 5 7 9 * = 2 × 0 = 0. 5. If any row of a determinant is added to or subtracted from any other row, the value of the determinant remains unchanged. For example, * 3 1 –2 –2 3 4 2 5 3 * = 13, from above. If row 2 is added to row 3, we have * 3 1 –2 + 1 –2 3 4 + 3 2 5 3 + 5 * = * 3 1 –1 –2 3 7 2 5 8 * = 3(24 – 35) + 2(8 + 5) + 2(7 + 3) = –33 + 26 + 20 = 13.
150 3 Mathematics Term 1 STPM Chapter 3 Matrices 6. If the elements of a row of a determinant are expressed as the sum of two (or more) terms, then the determinant can be expressed as the sum of two (or more) determinants. Thus, * a1 + d1 b1 c1 a2 + d2 b2 c2 a3 + d3 b3 c3 * = * a1 b1 c1 a2 b2 c2 a3 b3 c3 * + * d1 b1 c1 d2 b2 c2 d3 b3 c3 * . For example, * 5 2 3 8 –2 4 9 3 1 * = * 4 + 1 2 3 3 + 5 –2 4 2 + 7 3 1 * = * 4 2 3 3 –2 4 2 3 1 * + * 1 2 3 5 –2 4 7 3 1 * . 7. The determinant of a square null matrix is 0, i.e. * 0 0 0 0 0 0 0 0 0 * = 0. The determinant of a matrix whose elements are all zeros in any row is zero. For example, * 0 1 6 0 3 2 0 4 –5 * = 0. 8. The determinant of a triangular matrix is the product of the elements on the major diagonal. For example, * 2 4 3 0 –1 6 0 0 5 * = 2 × (–1) × 5 = –10, * 2 0 0 3 4 0 –2 6 –5 * = 2 × 4 × (–5) – 3 × 0 – 2 × 0 = –40. The determinant of a diagonal matrix is equal to the product of the diagonal elements. For example, * 3 0 0 0 5 0 0 0 4 * = 3 × 5 × 4 = 60. Exercise 3.2 1. Evaluate the determinant of each of the following matrices: (a) 1 1 2 –4 1 2 (b) 1 6 5 8 7 2 (c) 1 x + 1 1 + y 1 – y x – 1 2 2. Evaluate the determinants by adding or subtracting of rows or columns: (a) * 2 6 0 3 5 0 1 2 0 * (b) * –2 4 –2 6 7 6 3 1 3 * (c) * 4 2 1 –5 –3 –2 7 3 8 *
151 3 Mathematics Term 1 STPM Chapter 3 Matrices 3. Factorise (a) * x2 x 1 1 * (b) * 2a 8a2 2 4 * (c) * xy xy2 a2 2a * 4. Find the value of x in each of the following: (a) * 1 5 3 x * = –18 (b) * x –15 3 25 * = 60 (c) * 1 2 3 2 x 4 3 3 3 * = 0 5. Evaluate (a) det 1 3 0 5 0 4 0 6 0 2 2 (b) det 1 3 0 5 0 –8 0 6 0 2 2 (c) det 1 5 2 6 5 –4 3 5 1 3 2 6. Without expanding, find the value of x in each of the following: (a) * 2 4 1 3 x 3 1 –2 3 * = 0 (b) * 1 3 1 –1 4 x 2 5 2 * = 3 (c) * x 1 –2 2x –1 3 3x 2 5 * = 13 7. Use the properties of determinants to find the values of x in each of the following: (a) * x + 2 2 2 3 x + 3 3 4 4 x + 4 * = 0 (b) * 1 1 x + 1 –2 x – 2 –2 x + 3 3 3 * = 0 (c) * x + 1 3 x + 1 x + 2 x + 2 2 3 x + 1 x + 3 * = 0 8. Use the properties of the determinants to show the following: (a) * y x 1 x + 1 y + 1 x + y 1 1 1 * = 0 (b) * 1 1 1 x y + 2 2 y + 2 x x + y * = 0 (c) * 2 2 a 2 a 2 a 2 2 * = –(a + 4)(2 – a) 2 The inverse of a matrix A matrix B is called the inverse of a matrix A if AB = BA = I. The inverse of A is written as A–1. Hence, if we replace B with A–1, we get AA–1 = A–1A = I. Inverse of a matrix using elementary row operations The inverse of a non-singular matrix can be found by using some elementary row operations, like interchanging rows, adding and subtracting rows, multiplying by constants, etc. in the matrix. Let A = 1 3 –1 5 1 2 0 0 2 –1 2 be a 3 × 3 matrix and I = 1 1 0 0 0 1 0 0 0 1 2 be the corresponding identity matrix.
152 3 Mathematics Term 1 STPM Chapter 3 Matrices By writing the columns of the matrices A and I side by side, we have what is called an augmented matrix, represented by (A | I) where (A | I) = 1 3 –1 5 1 2 0 0 2 –1 * 1 0 0 0 1 0 0 0 1 2. An augmented matrix is usually used for the purpose of performing the same elementary row operations on each of the given matrices. We now use the elementary row operations on the augmented matrix in a systematic manner, the whole idea being to transform the left half of the augmented matrix into an identity matrix, and the right half then becomes the inverse of matrix A, i.e. to transform (A | I) to the form (I | A–1). We can do this by making use of any of the following elementary row operations: 1. Any two rows may be interchanged; 2. The elements of any row may be multiplied (or divided) by the same non-zero number; 3. Any row may be changed by adding to its elements a multiple of the corresponding elements of another row. Let aij be the element in the i th row and j th column of the augmented matrix (A | I) = 1 3 –1 5 1 2 0 0 2 –1 * 1 0 0 0 1 0 0 0 1 2 . We first create a 1 for the element a11 by adding row 1 with 2 times row 2: ⇒ (A | I) = 1 1 –1 5 5 2 0 4 2 –1 * 1 0 0 2 1 0 0 0 1 2 Next, we create 0’s below it by adding row 2 with row 1, and row 3 with 5 times row 2: ⇒ (A | I) = 1 1 0 0 5 7 10 4 6 9 * 1 1 0 2 3 5 0 0 1 2 Now, column 1 is done, we create a 1 for the element a22 by dividing row 2 by 7: ⇒ (A | I) = 1 1 0 0 5 1 10 4 6 7 9 * 1 1 7 0 2 3 7 5 0 0 1 2 Next, we create 0’s above and below this 1 by subtracting multiples of row 2 from row 1 and row 3 (5 and 10 times respectively): ⇒ (A | I) = 1 1 0 – 2 7 2 7 – 1 7 0 2 0 1 6 7 1 7 3 7 0 0 0 3 7 – 10 7 5 7 1
153 3 Mathematics Term 1 STPM Chapter 3 Matrices Now that column 2 is done, we create a 1 for the element a33 by multiplying row 3 by 7 3 : ⇒ (A | I) = 1 1 0 – 2 7 2 7 – 1 7 0 2 0 1 6 7 1 7 3 7 0 0 0 1 – 10 3 5 3 7 3 Next, we create 0’s above this 1 by adding 2 7 times row 3 to row 1, and subtracting 6 7 times row 3 from row 2 respectively: ⇒ (A | I) = 1 1 0 0 – 2 3 1 3 2 3 2 0 1 0 3 –1 –2 0 0 1 – 10 3 5 3 7 3 The left half is now a 3 × 3 identity matrix, and the right half is the inverse of matrix A, i.e. A–1. Hence, A–1 = 1 – 2 3 1 3 2 3 2 3 –1 –2 – 10 3 5 3 7 3 . We can confirm this result by showing that AA–1 = A–1A = I. Now, AA–1 = 1 3 –1 5 1 2 0 0 2 –1 21 – 2 3 1 3 2 3 2 3 –1 –2 – 10 3 5 3 7 3 = 1 –2 + 3 + 0 1 – 1 + 0 2 – 2 + 0 2 2 3 + 6 – 20 3 – 1 3 – 2 + 10 3 – 2 3 – 4 + 14 3 – 10 3 + 0 + 10 3 5 3 + 0 – 5 3 10 3 + 0 – 7 3 = 1 1 0 0 0 1 0 0 0 1 2
154 3 Mathematics Term 1 STPM Chapter 3 Matrices A–1A = 1 – 2 3 1 3 2 3 2 3 –1 –2 – 10 3 5 3 7 3 1 3 –1 5 1 2 0 0 2 –1 2 = 1 –2 – 1 3 + 10 3 – 2 3 + 2 3 + 0 0 + 2 3 – 2 3 2 9 + 1 – 10 3 – 2 + 0 0 – 2 + 2 –10 – 5 3 + 35 3 – 10 3 + 10 3 + 0 0 + 10 3 – 7 3 = 1 1 0 0 0 1 0 0 0 1 2 Hence, we have AA–1 = A–1A = I. (For ease of work, we let R1 , R2 , R3 stand for row 1, row 2 and row 3 respectively of the augmented matrix. In the subsequent steps that follow, for example in R2 = R2 – R3 , the LHS stands for the R2 in the current step while the RHS stands for the R2 and R3 in the previous step.) Example 8 Using elementary row operations, find the inverse of the following matrices. (a) A = 1 2 3 1 –2 2, (b) B = 1 –1 0 –5 3 –6 –3 –3 5 1 2. Solution: (a) The augmented matrix is (A | I) = 1 2 3 1 –2 1 0 0 1 2 = 1 1 3 1 2 –2 1 2 0 0 1 2 R1 = R1 × 1 2 (R2 unchanged) = 1 1 0 1 2 – 7 2 * 1 2 – 3 2 0 1 2 (R1 unchanged) R2 = R2 – 3R1 = 1 1 0 1 2 1 * 1 2 3 7 0 – 2 7 2 (R1 unchanged) R2 = R2 × (– 2 7 ) = 1 1 0 0 1 * 2 7 3 7 1 7 – 2 7 2 R1 = R1 – 1 2 R2 (R2 unchanged) Hence, A–1 = 1 2 7 3 7 1 7 – 2 7 2 = 1 7 1 2 3 1 –2 2 .
155 3 Mathematics Term 1 STPM Chapter 3 Matrices (b) The augmented matrix is (B | I) = 1 –1 0 –5 3 –6 –3 –3 5 1 * 1 0 0 0 1 0 0 0 1 2 = 1 1 0 –5 –3 –6 –3 3 5 1 * –1 0 0 0 1 0 0 0 1 2 R1 = R1 × (–1) (R2 unchanged) (R3 unchanged) = 1 1 0 0 –3 –6 –18 3 5 16 * –1 0 –5 0 1 0 0 0 1 2 (R1 unchanged) (R2 unchanged) R3 = R3 + 5R1 = 1 1 0 0 –3 1 –18 3 – 5 6 16 –1 0 –5 0 – 1 6 0 0 0 1 2 (R1 unchanged) R2 = R2 ÷ (–6) (R3 unchanged) = 1 1 0 1 2 –1 – 1 2 0 2 0 1 – 5 6 0 – 1 6 0 0 0 1 –5 –3 1 R1 = R1 + 3R2 (R2 unchanged) R3 = R3 + 18R2 = 1 1 0 0 3 2 1 – 1 2 2 0 1 0 – 25 6 – 8 3 5 6 0 0 1 –5 –3 1 R1 = R1 – 1 2 R2 R2 = R2 + 5 6 R3 (R3 unchanged) Hence, B–1 = 1 3 2 1 – 1 2 2 – 25 6 – 8 3 5 6 –5 –3 1 = 1 6 1 9 –25 –30 6 –16 –18 –3 5 6 2 . Example 9 Given that A = 1 1 –1 2 0 2 and AB = 1 0 2 1 –3 2, find B. Solution: AB = 1 0 2 1 –3 2 A–1 AB = A–11 0 2 1 –3 2 (Multiply both sides by A–1) i.e. B = A–11 0 2 1 –3 2
156 3 Mathematics Term 1 STPM Chapter 3 Matrices For A = 1 1 –1 2 0 2, the augmented matrix is (A | I) = 1 1 –1 2 0 * 1 0 0 1 2 = 1 1 0 2 2 * 1 1 0 1 2 (R2 unchanged) R2 = R2 + R1 = 1 1 0 0 2 * 0 1 –1 1 2 R1 = R1 – R2 (R2 unchanged) = 1 1 0 0 1 * 0 1 2 –1 1 2 2 (R1 unchanged) R2 = R2 × 1 2 Hence, A–1 = 1 0 1 2 –1 1 2 2 = 1 2 1 0 1 –2 1 2 and B = 1 2 1 0 1 –2 1 21 0 2 1 –3 2 = 1 2 1 0 – 4 0 + 2 0 + 6 1 – 3 2 = 1 2 1 –4 2 6 –2 2 = 1 –2 1 3 –1 2 . Example 10 Show that the matrix C = 1 2 –2 1 –1 4 –3 3 1 0 2 is non-singular. Use elementary row operations to find the inverse matrix C–1. Verify your result. Solution: C = 1 2 –2 1 –1 4 –3 3 1 0 2 det C = * 2 –2 1 –1 4 –3 3 1 0 * = 2 * 4 –3 1 0 * – (–1)* –2 1 1 0 * + 3 * –2 1 4 –3 * = 2(0 + 3) + (0 – 1) + 3(6 – 4) = 6 – 1 + 6 = 11 ≠ 0. Hence, the matrix C is non-singular.
157 3 Mathematics Term 1 STPM Chapter 3 Matrices The augmented matrix is (C | I) = 1 2 –2 1 –1 4 –3 3 1 0 * 1 0 0 0 1 0 0 0 1 2 = 1 1 –2 1 2 4 –3 3 1 0 * 1 0 0 0 1 0 –1 0 1 2 R1 = R1 – R3 (R2 unchanged) (R3 unchanged) = 1 1 0 0 2 8 –5 3 7 –3 * 1 2 –1 0 1 0 –1 –2 2 2 (R1 unchanged) R2 = R2 + 2R1 R3 = R3 – R1 = 1 1 0 0 2 1 –5 3 7 8 –3 * 1 1 4 –1 0 1 8 0 –1 – 1 4 2 2 (R1 unchanged) R2 = R2 × 1 8 (R3 unchanged) = 1 1 0 5 4 1 2 – 1 4 – 1 2 2 0 1 7 8 1 4 1 8 – 1 4 0 0 11 8 1 4 5 8 3 4 R1 = R1 – 2R2 (R2 unchanged) R3 = R3 + 5R2 = 1 1 0 5 4 1 2 – 1 4 – 1 2 2 0 1 7 8 1 4 1 8 – 1 4 0 0 1 2 11 5 11 6 11 (R1 unchanged) (R2 unchanged) R3 = R3 × 8 11 = 1 1 0 0 3 11 – 9 11 – 13 11 2 0 1 0 1 11 – 3 11 – 8 11 0 0 1 2 11 5 11 6 11 R1 = R1 – 5 4 R3 R2 = R2 – 7 8 R3 (R3 unchanged) Hence, C–1 = 1 3 11 – 9 11 – 13 11 2 1 11 – 3 11 – 8 11 2 11 5 11 6 11 = 1 11 1 3 1 2 –9 –3 5 –13 –8 6 2
158 3 Mathematics Term 1 STPM Chapter 3 Matrices To verify, we have CC–1 = 1 2 –2 1 –1 4 –3 3 1 0 2 . 1 11 1 3 1 2 –9 –3 5 –13 –8 6 2 = 1 11 1 6 – 1 + 6 –6 + 4 + 2 3 – 3 + 0 –18 + 3 + 15 18 – 12 + 5 –9 + 9 + 0 –26 + 8 + 18 26 – 32 + 6 –13 + 24 + 0 2 = 1 11 1 11 0 0 0 11 0 0 0 11 2 = 1 1 0 0 0 1 0 0 0 1 2 C–1C = 1 11 1 3 1 2 –9 –3 5 –13 –8 6 21 2 –2 1 –1 4 –3 3 1 0 2 = 1 11 1 6 + 18 – 13 2 + 6 – 8 4 – 10 + 6 –3 – 36 + 39 –1 – 12 + 24 –2 + 20 – 18 9 – 9 + 0 3 – 3 + 0 6 + 5 + 0 2 = 1 11 1 11 0 0 0 11 0 0 0 11 2 = 1 1 0 0 0 1 0 0 0 1 2 Exercise 3.3 1. Determine if an inverse exists for each of the following matrices. If an inverse exists, find the matrix. (a) 1 1 0 1 1 2 (b) 1 3 2 5 3 2 (c) 1 3 4 5 7 2 (d) 1 5 7 –2 –3 2 (e) 1 3 –2 –6 4 2 (f) 1 –2 3 –1 2 2 (g) 1 4 –2 –2 1 2 (h) 1 3 0 0 –2 2 For the inverse, A–1, that exists for the matrix A, show that AA–1 = A–1A = I. 2. If P = 1 5 4 1 2 2 and Q = 1 1 2 –1 4 2, find the matrix X in each of the following cases. (a) PX = Q (b) P = XQ (c) P–1XP = Q 3. Given that A, B and X are 2 × 2 matrices, and that A–1 exists for the matrix A, express X in terms of A, B and A–1 in each of the following cases. (a) AX = B (b) XA = B (c) AX = AB (d) AX = BA
159 3 Mathematics Term 1 STPM Chapter 3 Matrices 4. Show that 31 2 1 3 –1 21 0 3 1 1 24 –1 = 1 0 3 1 1 2 –1 1 2 1 3 –1 2 –1 . 5. Given that T = 1 5 4 4 3 2, find T–1. If M = 1 –14 –12 20 17 2 and D = 1 2 0 0 1 2, prove that M = TDT–1. Hence, show that M3 = TD3 T–1. 6. Given that A = 1 2 3 1 2 2 and B = 1 1 0 0 2 2, find matrix X if AXA–1 = B. 7. Find A2 if it is given that A = 1 2 –5 –1 3 2 . Given that I is the 2 × 2 identity matrix, and 0 is the 2 × 2 null matrix, find the values of m and n such that A2 + mA + nI = 0. By using the earlier result above, find A–1 and A3 . 8. If A = 1 1 0 3 7 5 0 3 2 –1 2 and B = 1 5 –6 15 –7 10 –21 1 2 –5 2, find AB. Hence, find A–1. 9. If A, P and D are all 2 × 2 non-singular matrices, and that AP = PD, prove that A2 = PD2 P–1 and A3 = PD3 P–1. 10. For each of the following matrices, find its inverse, if it exists, using elementary row operations. (a) A = 1 1 2 6 –1 0 –2 3 4 22 2 (b) B = 1 –1 0 1 0 5 4 4 –2 –1 2 (c) C = 1 2 3 5 –1 2 –2 4 –1 9 2 (d) D = 1 1 –1 2 2 1 5 0 0 1 2 (e) E = 1 2 5 –1 3 3 2 1 4 5 2 11. If A = 1 1 2 0 2 3 –1 1 –1 3 2 and B = 1 4 –3 –1 2 0 1 1 5 6 2, find (a) AB, (b) BA, (c) A–1, (d) B–1, (e) A–1B–1, (f) B–1A–1. Deduce that B–1A–1 = (AB) –1 and A–1B–1 = (BA) –1.
160 3 Mathematics Term 1 STPM Chapter 3 Matrices 3.2 Systems of Linear Equations Matrix representation of a linear system Matrices are useful in rewriting a system of linear equations in a very simple form. The algebraic properties of matrices may then be used to solve the equations. Consider the system of linear equations x + 3y + 5z = 3 2x + 4y + 6z = 7 3x + 2y + 4z = 1 Let A = 1 1 2 3 3 4 2 5 6 4 2, X = 1 x y z 2 and B = 1 3 7 1 2. A is known as the matrix of the coefficients, and B the matrix of the constants. X is the column matrix of the variables x, y and z. Using matrix multiplications, the system of linear equations above can be rewritten as the matrix equation AX = B. Given the matrices A and B, where A = 1 1 2 3 3 4 2 5 6 4 2 and B = 1 3 7 1 2, we have defined the augmented matrix (A | B) as (A|B) = 1 1 2 3 3 4 2 5 6 4 * 3 7 1 2 . The row-echelon form of a matrix A matrix is said to be in row-echelon form when it satisfies the following conditions: 1. The first non-zero element in each row, called the leading entry, is in a column to the right of the leading entry in the previous row. 2. Rows with all zero elements, if any, are below rows having a non-zero element. A typical augmented matrix in the row-echelon form is as follows: 1 a11 0 0 a12 a22 0 a13 a23 a33 * b1 b2 b3 2 . Solving a system of linear equations using Gaussian elimination The algorithm of solving a system of linear equations is as follows: 1. Eliminate x from all equations below R1 . 2. Eliminate y from all equations below R2 . This will put the system into triangular form. 3. Using back-substitution, each unknown can be solved for. This algorithm is known as the Gaussian elimination.
161 3 Mathematics Term 1 STPM Chapter 3 Matrices Example 11 Solve the following system of equations using augmented matrices: x + y – z = –2 2x – y + z = 5 –x + 2y + 2z = 1 The coefficients of x, y, z and the constants on both sides of the equation can be put in matrix form A and B, where A = 1 1 2 –1 1 –1 2 –1 1 2 2 and B = 1 –2 5 1 2 The augmented matrix of the system of linear equations above is the matrix (A|B) obtained by appending the columns of the two matrices A and B, as shown below: Equation System x + y – z = –2 2x – y + z = 5 –x + 2y + 2z = 1 Augmented matrix (A|B) = 1 1 2 –1 1 –1 2 –1 1 2 * –2 5 1 2 Each row of the augmented matrix represents one equation of the system. The vertical line represents where the equal sign was in the original system of equations (sometimes it is not included, depending on the writer). To solve a system of equations using the augmented matrix, we use the elementary row operations to transform the augmented matrix into row-echelon form. (Note: Let R1 , R2 , R3 stand for row 1, row 2 and row 3 respectively of the augmented matrix. In the steps we do, for example R2 = R2 – R3 , the LHS stands for the R2 in the current step while the RHS stands for the R2 and R3 in the previous step.) Solution: The augmented matrix for this system is (A|B) = 1 1 2 –1 1 –1 2 –1 1 2 * –2 5 1 2 Let R1 , R2 , R3 stand for row 1, row 2 and row 3 respectively of the augmented matrix (A|B). Then (A|B) = 1 1 0 –1 1 –3 2 –1 3 2 * –2 9 1 2 (R1 unchanged) R2 = R2 – 2R1 (R3 unchanged) = 1 1 0 0 1 –3 3 –1 3 1 * –2 9 –1 2 (R1 unchanged) (R2 unchanged) R3 = R1 + R3 = 1 1 0 0 1 –3 0 –1 3 4 * –2 9 8 2 (R1 unchanged) (R2 unchanged) R3 = R2 + R3 The matrix is now of the row-echelon form. Converting back to a system of equations, we have now x + y – z = –2 3y + 3z = 9 4z = 8 Hence, we have z = 2.
162 3 Mathematics Term 1 STPM Chapter 3 Matrices Substituting into the first two equations, we have y = z – 3 = 2 – 3 = –1, and x = –y + z – 2 = –(–1) + 2 – 2 = 1. Hence, the solution to the system of linear equations is the ordered triple (1, –1, 2). We see that the system of linear equations above has a unique solution. Consider the following example: Example 12 Solve the following system of equations using Gaussian elimination: 3x – 4y + 4z = 7 x – y – 2z = 2 2x – 3y + 6z = 5 Solution: The augmented matrix for this system is 1 3 1 2 –4 –1 –3 4 –2 6 * 7 2 5 2 = 1 1 3 2 –1 –4 –3 –2 4 6 * 2 7 5 2 Interchanging R1 and R2 = 1 1 0 2 –1 –1 –3 –2 10 6 * 2 1 5 2 (R1 unchanged) R2 = R2 – 3R1 (R3 unchanged) = 1 1 0 0 –1 –1 –1 –2 10 10 * 2 1 1 2 (R1 unchanged) (R2 unchanged) R3 = R3 – 2R1 = 1 1 0 0 –1 –1 0 –2 10 0 * 2 1 0 2 (R1 unchanged) (R2 unchanged) R3 = R2 – R3 Row 3 represents the equation 0 = 0, which is always true. This indicates that the system is dependent and has an infinite number of solutions. Writing back into equation form, we have x – y – 2z = 2 ................................a –y + 10z = 1 or y = 10z – 1.....................b Substituting b into a : x – (10z – 1) – 2z = 2 x – 10z + 1 – 2z = 2 x = 12z + 1 Hence, z can be any real number t, where t ∈ R. The solution of the system is the ordered triple of the form (12t + 1, 10t – 1, t), where t ∈ R. We say that the system of linear equations above has infinitely many solutions.
163 3 Mathematics Term 1 STPM Chapter 3 Matrices Example 13 Solve the following system of equations using Gaussian elimination: 2x – 2y – z = 2 4x – 4y – 3z = 2 x – y = 5 Solution: The augmented matrix for this system is 1 2 4 1 –2 –4 –1 –1 –3 0 * 2 2 5 2 = 1 2 0 1 –2 0 –1 –1 –1 0 * 2 –2 5 2 (R1 unchanged) R2 = R2 – 2R1 (R3 unchanged) = 1 2 0 0 –2 0 0 –1 –1 –1 * 2 –2 –8 2 (R1 unchanged) (R2 unchanged) R3 = R1 – 2R3 = 1 2 0 0 –2 0 0 –1 –1 0 * 2 –2 –6 2 (R1 unchanged) (R2 unchanged) R3 = R3 – R2 The third row says that 0x + 0y + 0z = –6, i.e. 0 = –6 which is not true. This indicates that the system of equations has no solutions. Hence, given any system of linear equations, there are three possibilities. 1. There is a unique solution. 2. There are infinitely many solutions. The equations in the system are said to be dependent. 3. There is no solution. The system is said to be inconsistent. Finding the unique solution using the inverse of a matrix A system of linear equations can be written in matrix form AX = B, where A is the matrix of the coefficients, B the matrix of the constants, and X the matrix of the variables x, y and z. We can use the inverse of matrix A to solve the equations if it exists. AX = B Multiplying both sides of the equation with A–1, we have A–1 AX = A–1B IX = A–1 B X = A–1 B By comparing the corresponding elements in both X and A–1B, the unique solution of the system of equations is obtained.
164 3 Mathematics Term 1 STPM Chapter 3 Matrices Example 14 Using the inverse of a matrix, solve the simultaneous equations 3x + 7y = 24 and 2x + 5y = 17. Solution: 3x + 7y = 24 2x + 5y = 17 Writing in matrix form, we have AX = B …………… 1 where A = 1 3 2 7 5 2, B = 1 24 17 2 and X = 1 x y 2 . The augmented matrix is (A | I) = 1 3 2 7 5 * 1 0 0 1 2 = 1 1 2 2 5 * 1 0 –1 1 2 R1 = R1 – R2 (R2 unchanged) = 1 1 0 2 1 * 1 –2 –1 3 2 (R1 unchanged) R2 = R2 – 2R1 = 1 1 0 0 1 * 5 –2 –7 3 2 R1 = R1 – 2R2 (R2 unchanged) Hence, A–1 = 1 5 –2 –7 3 2 . Multiplying both sides of the matrix equation 1 above with A–1, we have A–1AX = A–1B IX = A–1B X = A–1B 1 x y 2 = 1 5 –2 –7 3 21 24 17 2 = 1 120 – 119 –48 + 51 2 = 1 1 3 2 Hence, x = 1 and y = 3. Example 15 If A = 1 2 1 3 1 –2 2 1 –3 4 2, verify that A–1 = 1 9 1 2 13 –8 2 –5 1 1 –7 5 2 . Hence, solve the simultaneous equations 2x + y + z = 1 x – 2y – 3z = 1 3x + 2y + 4z = 5
165 3 Mathematics Term 1 STPM Chapter 3 Matrices Solution: 1 2 1 3 1 –2 2 1 –3 4 2. 1 9 1 2 13 –8 2 –5 1 1 –7 5 2 = 1 9 1 4 + 13 – 8 2 – 26 + 24 6 + 26 – 32 4 – 5 + 1 2 + 10 – 3 6 – 10 + 4 2 – 7 + 5 1 + 14 – 15 3 – 14 + 20 2 = 1 9 1 9 0 0 0 9 0 0 0 9 2 = 1 1 0 0 0 1 0 0 0 1 2 Also, 1 9 1 2 13 –8 2 –5 1 1 –7 5 21 2 1 3 1 –2 2 1 –3 4 2 = 1 9 1 4 + 2 + 3 26 – 5 – 21 –16 + 1 + 15 2 – 4 + 2 13 + 10 – 14 –8 – 2 + 10 2 – 6 + 4 13 + 15 – 28 –8 – 3 + 20 2 = 1 9 1 9 0 0 0 9 0 0 0 9 2 = 1 1 0 0 0 1 0 0 0 1 2 Hence, A–1 = 1 9 1 2 13 –8 2 –5 1 1 –7 5 2 . Rewriting the simultaneous equations in matrix form, 1 2 1 3 1 –2 2 1 –3 4 21 x y z 2 = 1 1 1 5 2 i.e. A1 x y z 2 = 1 1 1 5 2 1 x y z 2 = A–1 1 1 1 5 2 = 1 9 1 2 13 –8 2 –5 1 1 –7 5 21 1 1 5 2 = 1 9 1 9 –27 18 2 = 1 1 –3 2 2 Hence, x = 1, y = –3 and z = 2.
166 3 Mathematics Term 1 STPM Chapter 3 Matrices Example 16 If A = 1 1 3 –1 –2 1 4 1 2 1 2, find A–1. Hence, solve the simultaneous equations x – 2y + z = 1 3x + y + 2z = 4 –x + 4y + z = 2. Solution: The augmented matrix is (A | I) = 1 1 3 –1 –2 1 4 1 2 1 * 1 0 0 0 1 0 0 0 1 2 = 1 1 0 0 –2 7 2 1 –1 2 * 1 –3 1 0 1 0 0 0 1 2 (R1 unchanged) R2 = R2 – 3R1 R3 = R3 + R1 = 1 1 0 0 –2 1 2 1 – 1 7 2 * 1 – 3 7 1 0 1 7 0 0 0 1 2 (R1 unchanged) R2 = R2 × 1 7 (R3 unchanged) = 1 1 0 5 7 1 7 2 7 0 2 0 1 – 1 7 – 3 7 1 7 0 0 0 16 7 13 7 – 2 7 1 R1 = R1 + 2R2 (R2 unchanged) R3 = R3 – 2R2 = 1 1 0 5 7 1 7 2 7 0 2 0 1 – 1 7 – 3 7 1 7 0 0 0 1 13 16 – 2 16 7 16 (R1 unchanged) (R2 unchanged) R3 = R3 × 7 16 = 1 1 0 0 – 7 16 6 16 – 5 16 2 0 1 0 – 5 16 2 16 1 16 0 0 1 13 16 – 2 16 7 16 R1 = R1 – 5 7 R3 R2 = R2 + 1 7 R3 (R3 unchanged) Hence, A–1 = 1 – 7 16 6 16 – 5 16 2 – 5 16 2 16 1 16 13 16 – 2 16 7 16 = 1 16 1 –7 –5 13 6 2 –2 –5 1 7 2
167 3 Mathematics Term 1 STPM Chapter 3 Matrices Rewriting the simultaneous equations in matrix form, 1 1 3 –1 –2 1 4 1 2 1 21 x y z 2 = 1 1 4 2 2 A1 x y z 2 = 1 1 4 2 2 1 x y z 2 = A–1 1 1 4 2 2 = 1 16 1 –7 –5 13 6 2 –2 –5 1 7 21 1 4 2 2 = 1 161 –7 + 24 – 10 –5 + 8 + 2 13 – 8 + 14 2 = 1 16 1 7 5 19 2 = 1 7 16 5 16 19 16 2 Hence, x = 7 16 , y = 5 16 and z = 19 16 . Exercise 3.4 1. Using the augmented matrices and elementary row operations, solve the following simultaneous equations. (a) 2x – y = 4 (b) 5x + 7y = –3 3x – y = 5 2x + 3y = –1 (c) 2x – y = 1 (d) 3p – 5q = 7 5x – 2y = 2 2p – 4q = 6 (e) r + s = 11 (f) 5u + 2v = 4 4r – s = 9 2u – v = 7 2. Transform the augmented matrix into row-echelon form, and find the point of intersection, if it exists, for each of the pairs of straight lines represented by the following equations. (a) 3x + y + 1 = 0 (b) 4x – 10 y = 7 4x + 3y – 2 = 0 5y – 2x = 3 (c) y = 2x + 3 (d) 3x + 4y = 5 x – 2y + 3 = 0 x – 3y = 6
168 3 Mathematics Term 1 STPM Chapter 3 Matrices 3. Write the augmented matrix for each of the following sets of simultaneous equations. Hence, solve for x and y in terms of k, stating the condition for the value of k in each case. (a) (2k + 1)x – y = 1 (b) x – ky = 1 (k – 1)x + y = 2 kx – 4y = 2 4. By applying the Gaussian elimination, solve the following systems of equations. (a) x – 4y – z = 6 (b) 2x + 4y – 2z = 2 (c) x + 3y – 2z = 5 2x – y + 3z = 0 4x + 9y – 3z = 8 3x + 5y + 6z = 7 –3x + 2y – z = –4 –2x – 3y + 7z = 10 2x + 4y + 3z = 8 5. By transforming the augmented matrix into row-echelon form, determine whether each of the following systems of linear equations has a unique solution. (a) 3x + 2y – z = 1 (b) 2x + y – 3z = –4 (c) x + 2y – z = 6 2x – 2y + 4z = –2 4x + 9y – z = 3 2x + 3y + z = 11 –2x + y – 2z = 0 3x + 5y – 2z = 5 4x + 5y + 5z = 21 6. Verify that the inverse of matrix A = 1 1 2 0 1 0 2 0 –1 –3 2 is A–1 = 1 8 1 2 6 4 3 –3 –2 –1 1 –2 2. Hence, solve the simultaneous equations x + y = 2 2x – z = 1 2y – 3z = 1. 7. If A = 1 1 2 1 1 –1 –2 1 2 –1 2 and B = 1 5 4 –3 –1 –2 3 3 0 –3 2, find AB. Hence, solve the simultaneous equations. (a) x + y + z = 4 (b) 5x – y + 3z = 3 2x – y + 2z = 5 4x – 2y = 0 x – 2y – z = –3 –3x + 3y – 3z = –9 8. If A = 1 2 1 2 1 –2 1 –3 1 –1 2, find A–1. Hence, solve the equation A1 x y z 2 = 1 4 1 0 2. 9. If B = 1 3 1 4 –1 1 –1 –1 1 1 2, find B–1. Hence, solve the equation B1 x y z 2 = 1 2 4 7 2. 10. Show that each of the following systems of equations has a unique solution. Using the inverse of a matrix, solve each of the following systems of equations. (a) x – y + 2z = –1 (b) x – y – 4z = 1 4x + y + z = 13 2x + 5y – z = 2 5x – y + 8z = 5 3x + 2y – 3z = –1
169 3 Mathematics Term 1 STPM Chapter 3 Matrices (c) 4x + 2y – z = 24 (d) 4x – 7y + 6z = –18 2x + 3y + 2z = 17 5x + y – 4z = –9 6x – 5y + 7z = –21 3x – 2y + 3z = 12 Summary 1. An identity matrix is a square matrix where each element is 0, except for the elements on the major diagonal, which are all equal to 1. 2. A diagonal matrix is a square matrix where each element, apart from the elements on the major diagonal, is 0. 3. Two matrices are equal if the order of the matrices are the same and each corresponding element in the two matrices are equal. 4. For any matrices A, B and C of the same order, A + B = B + A (Commutative rule of addition) (A + B) + C = A + (B + C) (Associative rule of addition) 5. If A is the matrix 1 a c b d 2 and k is a constant, then kA = 1 ka kc kb kd 2. 6. If A is a matrix of order m × p and B is a matrix of order p × n, the product of the matrices, AB, exists and is a matrix of order m × n. 7. The multiplication of three matrices A, B and C obeys the associative rule ABC = (AB)C = A(BC). 8. A matrix B is called the inverse of a matrix A if AB = BA = I. 9. For a system of linear equations a1 x + b1 y + c1 z = d1 a2 x + b2 y + c2 z = d2 a3 x + b3 y + c3 z = d3 the augmented matrix is 1 a1 a2 a3 b1 b2 b3 c1 c2 c3 * d1 d2 d3 2. 10. An augmented matrix is said to be in row-echelon form if it is of the form 1 a11 0 0 a12 a22 0 a13 a23 a33 * b1 b2 b3 2 . 11. A system of linear equations has a unique solution, infinitely many solutions, or no solution. The system has dependent equations if it has infinitely many solutions. The system is said to be inconsistent if it has no solution.
170 3 Mathematics Term 1 STPM Chapter 3 Matrices STPM PRACTICE 3 1. If A = 1 1 1 0 1 2, B = 1 2 1 0 1 2 and C = 1 –1 3 2 2 2, show that (a) (AB)C = A(BC), (b) A(B + C) = AB + AC 2. Express each of the following as a single matrix. (a) 1 2 –1 0 3 4 1 2 + 6 1 1 3 1 3 2 0 2 3 1 2 (b) 1 0 1 1 0 0 1 21 2 –2 1 –1 1 3 21 1 –1 2 3. Find the inverse of each of the following matrices. (a) 1 3 5 7 10 2 (b) 1 –4 8 3 –6 2 (c) 1 6 3 –5 –2 2 (d) 1 2 3 7 5 2 4. Matrices M and N are given by M = 1 –3 20 –8 4 –15 0 –12 8 5 2 , N = 1 3 –5 2 –1 6 0 3 –2 1 2 Find the real numbers x and y for which M = xN + yI, where I is the 3 × 3 identity matrix. 5. Matrix Q is given by: Q = 1 –3 –1 1 –2 0 –1 1 –2 1 2 (a) Find Q2 + 2Q – 8I. (b) Show that Q(Q2 + 2Q – 8I) = 9I, where I is the 3 × 3 identity matrix and deduce the inverse of Q. 6. The matrices A and B are given by A = 1 p2 q2 r 2 p q r 1 1 1 2 , B = 1 4p2 r 2 q2 4p r q 4 1 1 2 . (a) Show that det A = (p – q)(q – r)(p – r). (b) Deduce det B. 7. If 1 a c b d 21 x y 2 = k1 x y 2, prove that k satisfies the equation k2 – (a + d)k + (ad – bc) = 0. If the roots of this quadratic equation are a and b, find the value of a + b and ab in terms of a, b, c and d. Hence, or otherwise, prove that 1 a c b d 21 b a – a b b – a 2 = 1 b a – a b b – a 21 a 0 0 b2. 8. (a) If A and B are square matrices such that AB = A and BA = B, where and are non-zero scalars, prove that A2 = A and B2 = B. (b) If A is a 2 × 2 matrix such that A2 = A, where is a non zero scalar, prove that A is either nonsingular, or A = I, where I is an identity matrix.
171 3 Mathematics Term 1 STPM Chapter 3 Matrices 9. A matrix is given by P = 1 2 1 5 0 –1 2 1 0 4 2. By using the elementary row operations, find the inverse of P. 10. A matrix W is given by W = 1 5 1 –2 –2 –1 0 0 –1 –1 2. By performing elementary row operations on the augmented matrix (W | I), where I is the 3 × 3 identity matrix, find W–1. Hence, solve the equation 1 5 1 –2 –2 –1 0 0 –1 –1 21 x y z 2 = 1 –4 5 6 2 . 11. A system of linear equations is given by: x + 2y + z = 3, 4x + 5y + az = 11, x + 8y + 3z = b, where a and b are real numbers. Show that the augmented matrix for the system may be reduced to 1 1 0 0 2 –1 0 1 a – 4 2a – 6 3 –1 b – 5 2. Hence, determine the values of a and b so that the system of linear equations has (a) no solution, (b) infinitely many solutions, (c) a unique solution. 12. The variables x, y and z satisfy the system of linear equations 3x + 2y + z = 1, 9x + 6y + 4z = a2 , 6x + 4y – 2z = a, where a is a real constant. (a) Write a matrix equation for the system of linear equations. (b) Reduce the augmented matrix to row-echelon form and show that the system of linear equations does not have a unique solution. (c) Determine the values of a for which the system of linear equations has infinitely many solutions, and find the solutions in the case when a is negative. (d) Find the set values of a for which the system of linear equations is inconsistent. 13. A system of linear equations is given by x + 3y + z = n, 2x + 5y + mz = 2n 3x + my + 13z = 4n, where m and n are real constants. (a) Write the augmented matrix for the system and reduce it to row-echelon form. (b) Determine the values of m and n such that the system has (i) no solution, (ii) a unique solution. (c) Find the set values of m and n such that for the system has infinitely many solutions. Using the value of n and larger value of m obtained, find the solutions of the system.
Mathematics Term 1 STPM Chapter 4 Complex Numbers CHAPTER 4 COMPLEX NUMBERS Learning Outcome (a) Identify the real and imaginary parts of a complex number. (b) Use the conditions for the equality of two complex numbers. (c) Find the modulus and argument of a complex number in cartesian form and express the complex number in polar form. (d) Represent a complex number geometrically by means of an Argand diagram. (e) Find the complex roots of a polynomial equation with real coefficients. (f) Perform elementary operations on two complex numbers expressed in cartesian form. (g) Perform multiplication and division of two complex numbers expressed in polar form. (h) Use de Moivre's theorem to find the powers and roots of a complex number. Argand diagram – gambar rajah Argand argument – hujah cartesian form – bentuk kartesan complex number – nombor kompleks complex root – punca kompleks conjugate – konjugat imaginary – khayalan modulus – modulus polar form – bentuk kutub real number – nombor nyata Bilingual Keywords
Mathematics Term 1 STPM Chapter 4 Complex Numbers 4 173 4.1 Complex Numbers Real and imaginary numbers A real number x is defined as a number which can be represented as a point on the real number line, i.e. –∞ , x , ∞ for x ∈ R. For any real number x ∈ R. x2 > 0. If x2 = –1, then x = –1 is not a real number and cannot be represented as a point on the real number line. We introduce an imaginary unit i, where i = –1 or i2 = –1. Hence, we have, for example –4 = 4(–1) = 4i 2 = 2i; –75 = 75(–1) = 75i 2 = 3(25)i 2 = 3(5i)2 = 5 3i; – 4 25 = 4 25 (–1) = 4 25 i 2 = 2 5 i. The numbers such as 2i, 5 3i and 2 5 i above are known as imaginary numbers. A real number and an imaginary number may be added or subtracted to each other, but the combination cannot be simplified. This combined number is called a complex number, for example, 3 + 5i, 2 – 3i, – 3 2 + 7i are complex numbers. In general, a complex number in the cartesian form is the sum of two parts, a real part x, with an imaginary part yi, and is commonly represented by z, i.e. z = x + yi. The set of complex numbers, C, is defined by C = {z : z = x + yi, x, y ∈ R}. A complex number cannot be represented as a point on the real number line. Hence, complex numbers do not obey the rules on order, as real numbers do. If y = 0 in z = x + yi, then z = x is an entirely real number. If x = 0, then z = yi is an entirely imaginary number. The set of real numbers R is a subset of the set of complex numbers C, i.e, R , C. Hence, in set form N , Z , Q , R , C ↑ ↑ ↑ ↑ ↑ set of natural numbers set of integers set of rational numbers set of real numbers set of complex numbers Algebraic operations on complex numbers The operations of addition, subtraction, multiplication and division on real numbers can be extended to the complex numbers. Addition and subtraction If z = x + yi and w = u + vi are two complex numbers, with x, y, u, v ∈ R, then z + w = x + yi + u + vi = (x + u) + (y + v)i
174 Mathematics Term 1 STPM Chapter 4 Complex Numbers 4 and z – w = x + yi – (u + vi) = (x – u) + (y – v)i For example, (3 + 4i) + (5 + 6i) = (3 + 5) + (4 + 6)i = 8 + 10i and (5 + 3i) – (8 + 2i) = (5 – 8) + (3 – 2)i = –3 + i Multiplication If z = 3 + 4i and w = 2 – 3i, then zw = (3 + 4i)(2 –3i) = 3 × 2 – 3 × 3i + 4i × 2 – 4i × 3i = 6 – 9i + 8i – 12i2 = 6 – i + 12 i 2 = –1 = 18 – i If a complex number z = a + bi is multiplied by the complex number w = a – bi, then zw = (a + bi)(a – bi) = a2 – (bi)2 = a2 + b2 i 2 = –1 Hence, zw is a real number and w is known as the complex conjugate of z. If z = a + bi, a, b R, the complex conjugate of z is z* = a – bi with zz* = a2 + b2 The product of a complex number and its conjugate is always a real number. For example, if z = 3 – 4i, its complex conjugate is z * = 3 + 4i. zz* = (3 + 4i)(3 – 4i) = 32 + 42 = 9 + 16 = 25. If v = –1 + 2i, its complex conjugate is v* = –1 – 2i. vv* = (–1 – 2i)(–1 + 2i) = (–1)2 + 22 = 1 + 4 = 5. Division The division of any number by a complex number can be done if the denominator (or divisor) is transformed into a real number. For example, 3 2 + 3i = 3 2 + 3i × 2 – 3i 2 – 3i = 6 – 9i 4 + 9 = 6 13 – 9 13 i 2 + i 3 – 2i = 2 + i 3 – 2i × 3 + 2i 3 + 2i = 6 + 4i + 3i + 2i2 9 + 4 = 6 + 7i – 2 13 = 4 13 + 7 13 i Complex Numbers INFO
Mathematics Term 1 STPM Chapter 4 Complex Numbers 4 175 Equality of complex numbers Let z = x + yi and w = u + vi be two complex numbers such that z = w. Then, x + yi = u + vi This means x – u = (v – y)i Real number Imaginary number Notice that the left hand side is a real number, whereas the right hand side is an imaginary number. This situation is clearly not possible, unless x – u = 0 and v – y = 0 i.e. x = u and y = v Hence, two complex numbers x + yi and u + vi are equal if and only if x = u and y = v. x + yi = u + vi, x, y, u, v R, if and only if x = u and y = v. To solve problems involving the equality of two complex numbers, we need to equate the real and imaginary parts of the equality. Example 1 Given that 3 – 2i = (p + qi)(5 + i), find the values of p and q. Solution: 3 – 2i = (p + qi)(5 + i) = 5p + pi + 5qi + qi 2 = (5p – q) + (p + 5q)i Equating the real parts: 3 = 5p – q ……… Equating the imaginary parts: –2 = p + 5q ……… × 5: 15 = 25p – 5q ……… + : 13 = 26p p = 1 2 Substituting p = 1 2 into : 3 = 5 2 – q q = 5 2 – 3 = – 1 2 Hence, p = 1 2 and q = – 1 2 . Example 2 Find 15 + 8i in the form a + bi, where a, b R. Solution: Let 15 + 8i = a + bi Squaring both sides: 15 + 8i = (a + bi)2 = a2 – b2 + 2abi Equating the real parts: a2 – b2 = 15 ……… Equating the imaginary parts: 2ab = 8 b = 4 a ………
176 Mathematics Term 1 STPM Chapter 4 Complex Numbers 4 Substituting b = 4 a into , a2 – 16 a2 = 15 a4 – 16 = 15a2 a4 – 15a2 – 16 = 0 (a2 – 16)(a2 + 1) = 0 a2 – 16 = 0 or a2 + 1 = 0 a2 = 16 or a2 = –1 a2 = –1 not possible since a2 > for a ∈ R Hence, a2 = 16 ⇒ a = ±4 Substituting a = ±4 into , when a = 4, b = 1 when a = –4, b = –1 Hence, 15 + 8i = 4 + i or –4 – i = ±(4 + i) Example 3 If –1 + 3i 2 + i = h + ki , find the values of h and k. Solution: –1 + 3i 2 + i = h + ki –1 + 3i 2 + i = –1 + 3i 2 + i × 2 – i 2 – i = –2 + 6i + i – 3i2 4 + 1 = –2 + 7i + 3 5 = 1 + 7i 5 Hence, 1 + 7i 5 = h + ki Squaring both sides: (1 + 7i)2 52 = h + ki 1 + 14i + 49i 2 25 = h + ki 1 – 49 + 14i 25 = h + ki – 48 25 + 14 25 i = h + ki Equating the real and imaginary parts: h = – 48 25 , k = 14 25 .
Mathematics Term 1 STPM Chapter 4 Complex Numbers 4 177 Exercise 4.1 1. Express in the form a + bi. (a) 2 + –4 (b) 9 – –25 (c) 2 + –8 (d) –3 – –98 (e) 4 + –80 2. Simplify (a) (3 + 4i) + (5 – 2i) (b) (2 – 8i) – (1 – 3i) (c) (7 + 6i) + (–4 – 3i) (d) (5 + i) + (–10 + 2i) 3. Express in the form a + bi. (a) i(3 + 4i) (b) 2i(1 – 2i) (c) (2 – 3i)2 (d) (5 – i)(2 + i) (e) (1 + 2 i)2 (f) i(1 + i)(2 + i) 4. Express in the form a + bi. (a) 2 1 + i (b) i 1 + 2i (c) 1 + i 1 – i (d) 2 + i 1 – 2i (e) 3 + 4i 2 + 3i (f) 1 + i (2 + i)2 5. If z = 1 – 2i, find (a) z + z*, (b) zz*, (c) z z* , (d) (z*)2 , (e) (z2 )*, (f) 1 zz* . 6. If z1 = 3 – 4i and z2 = 4 – 3i, find (a) z1 + z2 , (b) (z1 + z2 )*, (c) z1 *z2 *, (d) z1 z2 , (e) z1 * z2 * . 7. Find the values of x and y satisfying each of the following equations. (a) x + yi = (3 + i)(2 – 3i) (b) 2 + 5i 1 – i = x + yi (c) (1 + i)(x + yi) = 8 – 2i (d) x + yi = 16 + 22i 6 – i (e) (x + yi)(–2 + 7i) = –11 – 41i 8. Find the square root of each of the following complex numbers. (a) 3 – 4i (b) 7 + 24i (c) 5 + 12i (d) 8 – 6i (e) 24 + 70i 9. Given that z z + 1 = 2 + 3i, find the complex number z, giving your answer in the form x + yi. 10. If (3 + 4i)(1 + 2i) 1 + 3i = a + bi, find the values of a and b. 11. The complex number z and its complex conjugate z* satisfy the equation 2z + z* = 11 + 7i 1 + i . Find z in the form x + yi. 12. Given that z = 2 1 – i + 1 – 2i, express the complex number z in the form x + yi.
178 Mathematics Term 1 STPM Chapter 4 Complex Numbers 4 Complex roots of a polynomial equation A quadratic equation az2 + bz + c = 0, where a, b, c ∈ R, has either two real roots or two complex roots. If the discriminant b2 – 4ac , 0, the roots are complex numbers and are conjugates of each other. For example, the equation z2 – 4z + 13 = 0 has b2 – 4ac = (–4)2 – 4 × 13 = –36 = 36i2 . Hence, z = 4 ± 36i 2 2 = 4 ± 6i 2 = 2 ± 3i Therefore, we see that the roots to equation z2 – 4z + 13 = 0 are 2 + 3i and 2 – 3i, which are conjugate complex numbers. For a quadratic equation az2 + bz + c = 0 with b2 – 4ac , 0, if p + qi is a complex root, then p – qi is also a root. For a cubic equation f(z) = 0 with real coefficients, it has at least one real root. Let z = a be the real root. Then (z – a) is a factor of f(z), and the other factor must be quadratic, say (pz2 + qz + r), where p, q, r ∈ R. Hence, f(z) = 0 ⇒ (z – a)(pz2 + qz + r) = 0. The roots of pz2 + qz + r = 0 are either both real, or are a pair of conjugate complex numbers. Thus, a cubic equation with real coefficients can have either three real roots, or one real root and two conjugate complex roots. A cubic equation az3 + bz2 + cz + d = 0 where a, b, c, d ∈ R, has either three real roots, or one real root and a pair of conjugate complex roots. Basically, whether a polynomial equation has real or complex roots depends on whether the polynomial function cuts the x-axis, or is entirely above (or below) the x-axis, as shown in the study of polynomial functions in Chapter 1. A polynomial equation with real coefficients of degree n . 3 behaves in the following manner: 1. Its complex roots always occur in conjugate pairs; 2. If n is odd, the polynomial equation has at least one real root; 3. If n is even, it may not have any real roots. Example 4 Find the roots of the cubic equation z3 – 1 = 0. Solution: For the equation z 3 – 1 = 0, z = 1 is a root. Hence, z – 1 is a factor of z 3 – 1. ⇒ (z – 1)(z2 + z + 1) = 0 z – 1 = 0 or z 2 + z + 1 = 0 z = 1 or z = –1 ± 1 – 4 2 z = 1 or z = –1 ± 3 i 2 ⇒ z = 1 or z = – 1 2 + 3 2 i or z = – 1 2 – 3 2 i Therefore, the roots of the given equation z3 = 1 are the real root z = 1, and the complex conjugate pair z = – 1 2 + 3 2 i and z = – 1 2 – 3 2 i The above roots are known as the three cube roots of unity.
Mathematics Term 1 STPM Chapter 4 Complex Numbers 4 179 Note: 1. 1– 1 2 + 3 2 i2 2 = 1 4 – 3 2 i – 3 4 = – 1 2 – 3 2 i; and 2. 1– 1 2 – 3 2 i2 2 = 1 4 + 3 2 i – 3 4 = – 1 2 + 3 2 i; The square of any of the complex root results in its conjugate root. The three roots may be denoted by 1, w and w2 . 3. The sum of the three roots is 1 + 1– 1 2 – 3 2 i2 + 1– 1 2 + 3 2 i2 = 0, i.e. 1 + w + w2 = 0. Example 5 Find the roots of the following polynomial equations. (a) z4 + 14z2 + 45 = 0 (b) z5 – 3z4 – z3 + 3z2 – 6z + 18 = 0 Solution: (a) z4 + 14z2 + 45 = 0 Treating this as a quadratic equation in z2 and factoring the LHS, we have (z2 + 5)(z2 + 9) = 0 Therefore, either z2 + 5 = 0 or z2 + 9 = 0 z2 = –5 or z2 = –9 z = ±i 5 or z = ±3i. Hence, the roots of the equation z4 + 14z2 + 45 = 0 are z = 3i, –3i, i 5 , –i 5 . (Note the 2 pairs of conjugate complex roots.) (b) z5 – 3z4 – z3 + 3z2 – 6z + 18 = 0 Let P(z) = z5 – 3z4 – z3 + 3z2 – 6z + 18 Since the degree of the polynomial is odd, there is at least one real root z = k, where k must be a factor of 18 and P(k) = 0. By choosing k = 3, P(3) = 35 – 3 × 34 – 33 + 3 × 32 – 6 × 3 + 18 = 35 – 35 – 33 + 33 – 18 + 18 = 0 Hence, z = 3 is a real root of the equation P(z) = 0. P(z) = (z – 3)(z4 – z2 – 6) = 0 (z – 3)(z2 – 3)(z2 + 2) = 0 z – 3 = 0, z2 – 3, or z2 + 2 = 0 z = 3, z2 = 3, or z2 = –2 z = 3, ± 3 , or ±i 2 . Hence, the roots of the equation z5 – 3z4 – z3 + 3z2 – 6z + 18 = 0 are z = 3, 3 , – 3 , i 2 , –i 2 . (3 real roots and a pair of conjugate complex roots). Exercise 4.2 1. Solve each of the following quadratic equations. (a) z 2 + 2z + 3 = 0 (b) z 2 – 3z + 4 = 0 (c) 2z 2 + 7z + 10 = 0 (d) z 2 – 5z + 7 = 0 (e) 3z 2 – 4z + 2 = 0
180 Mathematics Term 1 STPM Chapter 4 Complex Numbers 4 2. Form the quadratic equation in z whose roots are: (a) 2 + i, 2 – i (b) 2 – 3i, 2 + 3i (c) 3 2 + i 2 , 3 2 – i 2 (d) 3 5 + 2 5 i, 3 5 – 2 5 i (e) – 4 7 + 5 7 i, – 4 7 – 5 7 i 3. In each of the following cases, find one real root of the cubic equation and show that the equation has no more real roots. Hence, find the other two complex roots of the cubic equation. (a) z3 – 4z2 + 7z – 6 = 0 (b) 2z3 + 9z2 + 18z + 7 = 0 (c) 2z3 – 12z2 + 23z – 15 = 0 (d) 5z3 – 23z2 + 25z + 14 = 0 (e) 3z3 – 13z2 + 19z – 5 = 0 4. Given that (2x + 1) is factor of 2z 3 + kz2 + 16z + 6, show that k = 9. Find the real quadratic factor of 2z 3 + 9z 2 + 16z + 6. Hence, find the roots of the equation 2z 3 + 9z 2 + 16z + 6 = 0. 5. If z = 1 + 2i is a root of the equation z 4 – z 3 + 4z 2 + 3z + 5 = 0, express z 4 – z 3 + 4z 2 + 3z + 5 as the product of two quadratic factors. Hence, find the complex roots of the equation z 4 – z 3 + 4z 2 + 3z + 5 = 0. 6. Find all the roots for each of the following equations: (a) z4 + 6z2 – 16 = 0 (b) 2z4 – z3 – z2 + 3z + 1 = 0 (c) z5 + 3z4 + 8z3 + 24z2 + 16z + 48 = 0 (d) z6 + 3z4 – 16z2 – 48 = 0 (e) z8 – 20z4 + 64 = 0 (f) z9 + 8z6 – z3 – 8 = 0 The Argand diagram A real number can be represented by a point on the real number line. Similarly, a complex number a + bi can be represented as a point on the coordinate plane, with coordinates (a, b). This plane is referred to as the Argand diagram. On the Argand diagram, real numbers are represented on the x-axis, while imaginary numbers on the y-axis. Because of this, the x-axis is also known as the real axis and the y-axis the imaginary axis. O P (a, b) x y Imaginary axis Real axis Figure 4.1 Modulus and argument The modulus of a complex number z = x + yi, written as |z|, is the distance of the point P(x, y) from the origin O. Thus, |z| = OP = x 2 + y 2 If z* is the complex conjugate of z, then zz* = x2 + y2 . Hence, zz* = |z| 2 . O P (x, y) x y r x y θ Figure 4.2 The argument of a complex number z = x + yi, written as arg z, is the angle of inclination, q, of OP with the x-axis (or real axis). This angle q, expressed in radians, is such that –π , q π.
Mathematics Term 1 STPM Chapter 4 Complex Numbers 4 181 From Figure 4.2, tan q = y x arg z = q = tan–1 1 y x 2 If z = x + yi, x, y R, then modulus of z = |z| = x 2 + y 2 and argument of z = arg z = tan–1 1 y x 2 To obtain the correct argument, it is important to note the position of the point representing the complex number on the Argand diagram. Example 6 Find the modulus and argument of each of the complex numbers z1 = 1 + i 3 , z2 = –1 + i 3 , z3 = –1 – i 3 and z4 = 1 – i 3 . Show the points P1 , P2 , P3 , P4 representing the complex numbers z1 , z2 , z3 , z4 respectively on an Argand diagram. Solution: The modulus of each of the complex numbers is |z1 | = |z2 | = |z3 | = |z4 | = 1 + 3 = 2 arg z1 = tan–1 3 1 = tan–1 3 = π 3 arg z2 = tan–1 3 –1 = tan–1 (– 3 ) = 2π 3 O π – 3 x y P1(1, 3) fiff O 2 – π 3 x y P2(–1, 3) fiff arg z3 = tan–1 – 3 –1 = tan–1 3 = – 2π 3 (point is in 3rd quadrant) arg z4 = tan–1 – 3 1 = tan–1 (– 3) = – π 3 (point is in 4th quadrant) O 2 – – π 3 x y P3(–1, – 3) fiff O π – – 3 x y P4(1, – 3) fiff Polar form of a complex number From the Argand diagram in Figure 4.2, the point P(x, y) can also be represented by polar coordinates (r, q) with r > 0 and –π , q < π. Since x = r cos q and y = r sin q, where r is the modulus and q the argument, we can also write the complex number z = x + yi as z = r(cos q + i sin q). This is the modulus-argument form of a complex number, also known as the polar form of a complex number.
182 Mathematics Term 1 STPM Chapter 4 Complex Numbers 4 Example 7 Find the modulus and argument of each of the following complex numbers, and express them in polar form. (a) z = 3 – 4i (b) w = 5 + 12i (c) v = –2 + i 5 Solution: (a) |z| = 32 + (–4)2 (b) |w| = 52 + 122 = 9 + 16 = 25 + 144 = 25 = 169 = 5 = 13 arg z = tan–1 1– 4 3 2 arg w = tan–1 12 5 = –0.927 rad = 1.176 rad z = 5[cos(–0.927) + i sin(–0.927)] w = 13 [cos(1.176) + i sin(1.176)] (c) v = –2 + i 5 |v| = (–2)2 + ( 5)2 = 4 + 5 = 3 arg v = tan–1 1 5 –2 2 = tan–1 (–1.118) = 2.30 rad v = 3 [cos(2.30) + i sin(2.30)] Exercise 4.3 1. Plot each of the following complex numbers on separate Argand diagrams, and obtain the modulus and argument in each case. (a) z = 3i (b) z = –2 + i (c) z = 2 – 4i (d) z = 5 + 7i (e) z = –4 – 4i (f) z = (–4 – 3i)* 2. Express each of the following complex numbers in the form r(cos q + i sin q), where –π , q < π. (a) 1 – i (b) 3 + i (c) 3 + 4i (d) 5 – 12i (e) 2 – i 3 (f) 3 2 + 5i 3. Given the complex numbers z1 = 2 + i and z2 = 3 + 2i and that z3 = 1 + z1 2 + 5z2 z1 , (a) express z3 in the form x + yi, (b) find the modulus and argument of z3 and express z3 in polar form. 4. The complex number z is given by z = (1 + 2i)2 + 3 + i 1 + i . (a) Express z in the form x + yi. (b) Find the modulus and argument of z, and hence, express z in polar form. 5. The two complex numbers z1 , z2 are represented on an Argand diagram. Show that |z1 + z2 | < |z1 | + |z2 |. If |z1 | = 5 and z2 = 5 + 12i, show that the greatest value of |z1 + z2 | is 18 and find its least value.
Mathematics Term 1 STPM Chapter 4 Complex Numbers 4 183 6. Given that z1 = 2 + i and that z2 = –2 + 4i, find, in the form a + bi, the complex number z if 1 z = 1 z1 + 1 z2 . Find the modulus and argument of z, giving your answer for arg z in the interveral –π , arg z < π, correct to three decimal places. Hence, express z in polar form. 7. In the Argand diagram, P, Q, R represent the complex numbers 1 + 0i, z and z2 respectively, where z = 2 + i 2. (a) Write down the modulus and argument of z. (b) Find the area of the triangle OPQ, where O is the origin, and find the area of triangle OQR. 8. Two complex numbers z1 and z2 are given by z1 = 10 – 2i and z2 = 2 – 3i. Show that z1 z is of the form k(1 + i), where k is real. If z = x + yi is any non-zero complex number such that z1 z = a(1 + i), where a is real, show that 3x + 2y = 0. Hence, find the possible values of arg z, where –π , arg z < π, giving your answers in radians, correct to three significant figures. Multiplication and division of complex numbers in polar form Let z1 = r1 (cos q1 + i sin q1 ) and z2 = r2 (cos q2 + i sin q2 ) be two complex numbers in polar form. Then, the product of z1 and z2 is z1 z2 = r1 r2 (cos q1 + i sin q1 )(cos q2 + i sin q2 ) = r1 r2 [cos q1 cos q2 + i sin q1 cos q2 + i cos q1 sin q2 + i 2 sin q1 sin q2 ] = r1 r2 [cos q1 cos q2 – sin q1 sin q2 + i(sin q1 cos q2 + cos q1 sin q2 )] = r1 r2 [cos(q1 + q2 ) + i sin(q1 + q2 )] i.e. z1 z2 is a complex number with modulus r1 r2 and argument q1 + q2 . The quotient of z1 and z2 (with z2 ≠ 0) is z1 z2 = r1 (cos q1 + i sin q1 ) r2 (cos q2 + i sin q2 ) = r1 (cos q1 + i sin q1 ) r2 (cos q2 + i sin q2 ) × cos q2 – i sin q2 cos q2 – i sin q2 = r1 (cos q1 cos q2 + i sin q1 cos q2 – i cos q1 sin q2 – i2 sin q1 sin q2 ) r2 (cos2 q2 – i2 sin2 q2 ) = r1 r2 · (cos q1 cos q2 + sin q1 sin q2 ) + i(sin q1 cos q2 – cos q1 sin q2 ) (cos2 q2 + sin2 q2 ) = r1 r2 [cos(q1 – q2 ) + i sin(q1 – q2 )] i.e. z1 z2 is a complex number with modulus r1 r2 and argument q1 – q2 . If z1 and z2 are two complex numbers, |z1 z2 | = |z1 ||z2 |, arg (z1 z2 ) = arg z1 + arg z2 ; u z1 z2 u = |z1 | |z2 | , arg z1 z2 = arg z1 – arg z2 , (with z2 ≠ 0)
184 Mathematics Term 1 STPM Chapter 4 Complex Numbers 4 Example 8 Find the modulus and argument of the complex numbers z1 = 1 + i and z2 = 3 – i, and express z1 and z2 in polar form. Hence, find z1 z2 and z1 z2 , expressing them in polar form. Solution: We have |z1 | = 12 + 12 = 2 , and arg z1 = tan–1 1 = π 4 . Hence, z1 = 2 1cos π 4 + i sin π 4 2. |z2 | = ( 3)2 + (–1)2 = 3 + 1 = 2, and arg z2 = tan–1 1– 1 3 2 = – π 6 . Hence, z2 = 2 3cos 1– π 6 2 + i sin1– π 6 24 . z1 z2 = 2 2 1cos π 4 + i sin π 4 23cos1– π 6 2 + i sin1– π 6 24 = 2 2 3cos1 π 4 – π 6 2 + i sin1 π 4 – π 6 24 = 2 2 1cos π 12 + i sin π 12 2 . 2 1cos π 4 + i sin π 4 2 z1 z2 = 23cos1– π 6 2 + i sin1– π 6 24 z1 z2 = 2 2 3cos1 π 4 + π 6 2 + i sin1 π 4 + π 6 24 = 1 2 1cos 5 12 π + i sin 5 12 π2 . Example 9 If z1 = –2 2 + 2 2i and z2 = 3 + 3 3i, find the modulus and argument of z1 and z2 . Hence, find the modulus and argument of z1 z2 and z1 z2 , and express z1 z2 and z1 z2 in polar form. Solution: z1 = –2 2 + 2 2i |z1 | = (–2 2)2 + (2 2)2 = (4 × 2) + (4 × 2) = 16 = 4 arg z1 = tan–1 2 2 –2 2 = tan–1 (–1) = 3 4 π z2 = 3 + 3 3i |z2 | = 32 + (3 3)2 = 9 + 27 = 36 = 6 arg z2 = tan–1 3 3 3 = tan–1 3 = 1 3 π
Mathematics Term 1 STPM Chapter 4 Complex Numbers 4 185 |z1 z2 | = |z1 ||z2 | = 4 × 6 = 24 arg z1 z2 = arg z1 + arg z2 = 3 4 π + 1 3 π = 13 12 π u z1 z2 u = |z1 | |z2 | = 4 6 = 2 3 arg z1 z2 = arg z1 – arg z2 = 3 4 π – 1 3 π = 5 12 π Hence, z1 z2 = 24 1cos 13 12 π + i sin 13 12 π2, and z1 z2 = 2 3 1cos 5 12 π + i sin 5 12 π2. De Moivre's theorem We have seen that the product of two complex numbers z1 = r1 (cos q1 + i sin q1 ) and z2 = r2 (cos q2 + i sin q2 ) is given by z1 z2 = r1 r2 [cos (q1 + q2 ) + i sin (q1 + q2 )] Suppose z1 = z2 = z and q1 = q2 = q. Then z2 = r 2 (cos 2q + i sin 2q) By repeated use of the product formula to the above result, we have z2 × z = r 2 × r[cos (2q + q) + i sin (2q + q)] i.e. z3 = r 3 (cos 3q + i sin 3q). This can be repeated n times, where n is a positive integer, i.e. zn = [r(cos q + i sin q)]n = r n [cos nq + i sin nq] This is called the De Moivre's theorem. De Moivre's theorem: If z = r(cos q + i sin q), then for all real values of n (i.e. n ∈ R) zn = [r(cos q + i sin q)]n = r n [cos nq + i sin nq] Powers and roots of a complex number Using De Moivre's theorem, we can find the nth power of a complex number, by taking the nth power of the modulus and multiplying the argument by n. Example 10 Using De Moivre's theorem, find 1 3 2 + 1 2 i2 10 . Solution: Let z = 3 2 + 1 2 i. Then, |z| = 3 4 + 1 4 = 1, arg z = tan–1 1 3 = π 6 . Hence, z = 1cos π 6 + i sin π 6 2 z10 = 1cos π 6 + i sin π 6 2 10
186 Mathematics Term 1 STPM Chapter 4 Complex Numbers 4 = cos 10 6 π + i sin 10 6 π = cos 5 3 π + i sin 5 3 π = cos π 3 – i sin π 3 = 1 2 – 3 2 i To find the nth roots of a complex number, let w be the nth root of a complex number z, i.e. w = n z or wn = z. Let w = s(cos f + i sin f) and z = r(cos q + i sin q) Using De Moivre's theorem, we have s n (cos nf + i sin nf) = r(cos q + i sin q) This means that s n = r or s = r —1 n i.e. cos nf = cos q and sin nf = sin q. Since the sine and cosine functions are periodic and have period 2π, we can write nf = q + 2kπ or f = q + 2kπ n . Thus, w = r —1 n 3cos 1 q + 2kπ n 2 + i sin 1 q + 2kπ n 24 for k = 0, 1, 2, …, n – 1. Roots of a complex number: Let z = r(cos q + i sin q) and let n be a positive integer (i.e. n ∈ Z+ ). Then z has the n distinct nth roots wk = r —1 n 3cos 1 q + 2kπ n 2 + i sin 1 q + 2kπ n 24 where k = 0, 1, 2, …, n – 1. Note: 1. For each nth root of z = r(cos q + i sin q), its modulus is r —1 n . 2. The argument of each nth root exceeds the argument of the previous root by 2π n . 3. The nth roots of z lie equally spaced on the circle with radius r —1 n . Example 11 Find the six sixth roots of z = –27, and show the points representing the roots on the Argand diagram. Solution: Putting the equation in the form z = 27(cos π + i sin π), we have wk = 27—1 6 1cos π + 2kπ 6 + i sin π + 2kπ 6 2, with k = 0, 1, 2, 3, 4, 5. Hence, the 6 sixth roots of –27 are:
Mathematics Term 1 STPM Chapter 4 Complex Numbers 4 187 O 3 x y w4 w1 w2 w0 w3 w5 – 3fiff fiff w0 = 27—1 6 1cos π 6 + i sin π 6 2 = 3 1 3 2 + 1 2 i2 = 3 2 + 3 2 i w1 = 27—1 6 1cos π 2 + i sin π 2 2 = 3 (0 + i) = 3i w2 = 27—1 6 1cos 5π 6 + i sin 5π 6 2 = 3 1– 3 2 + 1 2 i2 = – 3 2 + 3 2 i w3 = 27—1 6 1cos 7π 6 + i sin 7π 6 2 = 3 1– 3 2 – 1 2 i2 = – 3 2 – 3 2 i w4 = 27—1 6 1cos 3π 2 + i sin 3π 2 2 = 3 10 – 1 2 i2 = – 3 2 i w5 = 27—1 6 1cos 11π 6 + i sin 11π 6 2 = 3 1 3 2 – 1 2 i2 = 3 2 – 3 2 i All the points representing the roots lie on the circle with radius 3 in the Argand diagram. Example 12 Use De Moivre's theorem to express sin 5q and cos 5q in terms of sin q and cos q, and hence, show that tan 5q = 5t – 10t 3 + t 5 1 – 10t 2 + 5t 4 , where t = tan q. Solution: Using De Moivre's theorem, cos 5q + i sin 5q = (cos q + i sin q) 5 = cos5 q + 5 cos4 q(i sin q) + 10 cos3 q(i sin q) 2 + 10 cos2 q(i sin q) 3 + 5 cos q(i sin q) 4 + (i sin q) 5 = cos5 q + 5i cos4 q sin q – 10 cos3 q sin2 q – 10i cos2 q sin3 q + 5 cos q sin4 q + i sin5 q = cos5 q – 10 cos3 q sin2 q + 5 cos q sin4 q + i(5 cos4 q sin q – 10 cos2 q sin3 q + sin5 q) Comparing the real and imaginary parts, cos 5q = cos5 q – 10 cos3 q sin2 q + 5 cos q sin4 q sin 5q = 5 cos4 q sin q – 10 cos2 q sin3 q + sin5 q tan 5q = sin 5q cos 5q = 5 cos4 q sin q – 10 cos2 q sin3 q + sin5 q cos5 q – 10 cos3 q sin2 q + 5 cos q sin4 q Dividing both numerator and denominator by cos5 q: tan 5q = 5 tan q – 10 tan3 q + tan5 q 1 – 10 tan2 q + 5 tan4 q = 5t – 10t 3 + t 5 1 – 10t 2 + 5t 4 , where t = tan q
188 Mathematics Term 1 STPM Chapter 4 Complex Numbers 4 Exercise 4.4 1. Use De Moivre's theorem to express each of the following complex numbers in the form cos nq + i sin nq. (a) 1cos π 3 + i sin π 3 2 8 (b) 1cos 2π 3 + i sin 2π 3 2 –5 (c) 1cos π 6 – i sin π 6 2 5 (d) 1cos 3π 8 + i sin 3π 8 2 –4 2. Express each of the following complex numbers in the form (cos q + i sin q) n . (a) cos 7q + i sin 7q (b) cos 3q – i sin 3q (c) cos 5q 3 – i sin 5q 3 (d) cos 13q 6 + i sin 13q 6 3. If z represents each of the following complex numbers, express z in polar form. Hence, find z5 in each case, giving your answer in polar form. (a) 1 + i (b) 3 + 3i (c) 3 2 – i 2 (d) –5 – 12i (e) 2 – 2 3i (f) 3 2 + 5i 4. Use De Moivre's theorem to simplify each of the following expressions. (a) 1cos 2q 3 + i sin 2q 3 21cos 5q 4 + i sin 5q 4 2 (b) cos 5q + i sin 5q cos 3q – i sin 3q (c) 1cos q 2 + i sin q 2 2 4 (cos q + i sin q) 3 1cos π 4 + i sin π 4 2 3 1cos 3π 8 + i sin 3π 8 2 6 (d) (e) 1cos π 3 – i sin π 3 2 5 1cos 2π 3 + i sin 2π 3 2 2 5. Express 1 2 + 1 2 i in polar form. Hence, find 1 1 2 + 1 2 i2 10 . 6. Use De Moivre's theorem to find the square roots of the following complex numbers. (a) 1 + i (b) 3 – 4i (c) 5 – 12i (d) 3 – i (e) –2 + 2i (f) 2 + i 7. Find the cube roots of each of the complex numbers given in Question 6. 8. Write down the modulus and argument of z = 1 + i 3. Given than z = 1 + i 3, express zn in polar form. On an Argand diagram, mark and label the points corresponding to n = –2, –1, 1 and 3. 9. If z denotes the complex number 1 2 (cos 1 3 π + i sin 1 3 π), write down z2 and z3 in polar form. Show, on an Argand diagram, the points A, B, C and D representing the numbers 1, 1 + z, 1 + z + z2 and 1 + z + z2 + z3 respectively. 10. Express the complex numbers 3 + i and 2 – 2i in the form r(cos q + i sin q), where r . 0 and –π , q < π. Solve the equation (2 – 2i)z3 = 3 + i, giving each answer in polar form. 11. Use De Moivre's theorem to show that cos 4q = 8 cos4 q – 8 cos2 q + 1. Hence, find all acute angles q for which cos 4q + 2 cos2 q = 0. 12. Show that sin 3q = 3 sin q – 4 sin3 q. Hence, show that sin 3q – cos 2q = (1 – s)(4s 2 + 2s – 1), where s = sin q.
Mathematics Term 1 STPM Chapter 4 Complex Numbers 4 189 Summary 1. A complex number in the cartesian form is a number in the form of the sum of two parts, a real part x, with an imaginary part yi, and is commonly represented by z, i.e. z = x + yi. 2. The set of complex numbers, C, is defined by C = {z : z = x + yi, x, y ∈ R}. 3. If z = a + bi, a, b ∈ R, the complex conjugate of z is z* = a – bi with zz* = a2 + b2 . 4. Two complex numbers x + yi and u + vi are equal if and only if x = u and y = v. 5. If p + qi is a complex root of the quadratic equation az2 + bz + c = 0, with b2 – 4ac , 0, then p – qi is also a root. 6. A cubic equation az3 + bx2 + cz + d = 0 where a, b, c, d ∈ R, has either three real roots, or one real root and a pair of conjugate complex roots. 7. A complex number a + bi can be represented by a point (a, b) on a complex plane, referred to as the Argand diagram. 8. The modulus of a complex number z = x + yi, written as |z|, is the distance of the point P(x, y) from the origin O. Thus |z| = OP = x2 + y2 . 9. The argument of a complex number z = x + yi, written as arg z, is the angle of inclination, q, of OP with the positive x-axis. Thus arg z = q = tan–1 1 y x 2 10. The polar form of a complex number z = x + yi is z = r(cos q + i sin q). 11. The product of two complex numbers z1 = r1(cos q1 + i sin q1) and z2 = r2(cos q2 + i sin q2) is z1 z2 = r1 r2 [cos (q1 + q2 ) + i sin (q1 + q2 )]. 12. The quotient of two complex numbers z1 = r1 (cos q1 + i sin q1 ) and z2 = r2 (cos q2 + i sin q2 ) (with z2 ≠ 0) is z1 z2 = r1 r2 [cos (q1 – q2 ) + i sin (q1 – q2 )] 13. De Moivre's theorem: If z = r(cos q + i sin q), then for all real values of n (i.e. n ∈ R), zn = [r(cos q + i sin q)]n = r n (cos nq + i sin nq). 14. Let z = r(cos q + i sin q) and n be a positive integer (i.e. n ∈ Z+ ) Then z has the n distinct nth roots wk = r —1 n 3cos 1 q + 2kπ n 2 + i sin 1 q + 2kπ n 24 where k = 0, 1, 2, 3, …, n – 1.
190 Mathematics Term 1 STPM Chapter 4 Complex Numbers 4 STPM PRACTICE 4 1. Given that (1 + 5i)p – 2q = 3 + 7i, find the values of p and q when (a) p and q are real, (b) p and q are respectively a complex number and its conjugate. 2. Given that z = x + yi and w = z + 8i z – 6 , z ≠ 6. If w is totally imaginary, show that x2 + y2 + 2x – 48 = 0. 3. Find 2 pairs of values of x, y R such that (x + yi)2 = –3 + 4i. Indicate, on an Argand diagram, the points that represent the complex numbers that you obtained, and also the point that represents –3 + 4i. Show that the triangle formed from these points is right-angled. 4. Find the modulus and argument of the complex numbers z1, z2 and z3 if z1 = 1 – i, z2 = z1 3 , and z3 = 3 – i 3 + i . Show, on an Argand diagram, the points that represent z1 , z2 and z3 . 5. If z = x + yi and z2 = a + bi, x, y, a, b R, prove that 2x2 = (a2 + b2 ) + a By solving the equation z4 + 6z2 + 25 = 0 for z2 , express each root of the equation in the form x + yi. 6. (a) Express the complex numbers 1 + i and 1 – i in the form r(cos q + i sin q), where r is the modulus and q the argument of the complex number. Find the modulus and argument of (1 + i)5 (1 – i)7 . (b) Given that z1 = 2 + i and z2 = –2 + 4i, find, in the form a + bi, the complex number z such that 1 z = 1 z1 + 1 z2 . Find the modulus and argument of z, giving your answer for arg z in the range –π , arg z π, correct to 3 decimal places. 7. The complex number z satisfies the equation |z| = |z + 2|. Show that the real part of z is –1. The complex number also satisfies the equation |z| = 3. The two possible values of z are represented by the points P and Q on an Argand diagram. Draw a sketch showing the positions of P and Q, and calculate the two possible values of arg z, giving your answers in radians, correct to 3 significant figures. 8. Use de Moivre's theorem to show that cos 4q = 8 sin4 q – 8 sin2 q + 1. Hence show that one of the roots of the equation 8z4 – 8z2 + 1 = 0 is sin π 8 and express the other roots in polar form. Deduce that sin π 8 = 1 2 2 – 2 and find an exact expression for sin 11 8 π. 9. (a) The complex numbers z and w are such that w = 1 + ia, z = –b – i. where a and b are real and positive. Given that wz = 3 – 4i, find the exact values of a and b. (b) The complex numbers z and w are such that |z| = 2, arg z = – 2 3 π, |w| = 5, arg w = 3 4 π. Find the exact values of (i) the real part of z and the imaginary part of z, (ii) the modulus and argument of w z2 .
Mathematics Term 1 STPM Chapter 4 Complex Numbers 4 191 10. Given that z = cos q + i sin q, show that z + 1 z = 2 cos q. Find a similar expression for z2 + 1 z2 . Show that z2 – z + 2 – 1 z + 1 z2 = 4 cos2 q – 2 cos q. Hence, solve the quartic equation z4 – z3 + 2z2 – z + 1 = 0, giving the roots in the form a + bi. 11. Express the complex numbers 3 + i and 2 – 2i in the form r(cos q + i sin q), where r . 0 and –π , q < π. Solve the equation (2 – 2i)z3 = 3 + i, giving each answer in the polar form. 12. Use de Moivre's theorem to show that cos 5q = cos q (16 cos4 q –20 cos2 q + 5). By considering the equation cos 5q = 0, show that cos2 1 π 10 2 = 5 + 5 8 . 13. Use de Moivre's theorem to show that sin 5q = a cos4 q sin q + b cos2 q sin3 q + c sin5 q where a, b and c are integers to be determined. Hence, show that sin 5q sin q = 16 cos4 q – 12 cos2 q + 1 (q ≠ kπ, where k ∈ Z+ ). By means of the substitution z = 2 cos q, find, in trigonometric form, the roots of the equation z4 – 3z2 + 1 = 0. Hence, show that cos2 1 1 5 π2 + cos2 1 2 5 π2 = 3 4 . 14. The complex number z is given by z = 12 – 2i (a) Find |z| and arg z. (b) Using de Moivre’s theorem, show that z4 = –128(1 + 3i). (c) Express z* z3 in the form x + yi, where z* is the conjugate of z, and x, y ∈ R. 15. Express the complex number –4 + i 48 in polar form. Hence, solve the equation z3 = –4 + i 48. 16. The complex numbers z and w are given by z = 2 + i and w = 1 + iz z – 1 . (a) Find w in the form x + yi, where x, y R. State the real and imaginary part of w. (b) Express w in polar form. (c) Using de Moivre’s theorem, determine the cube roots of w. Give your answer in Cartesian form. 17. The complex numbers z and w are given by z = 4 – 4i and w = –3 + i 3. Express z and w in polar form. Hence, express the complex number w8 z6 in the form r(cos q + i sin q), where r . 0 and –π , q < π. 18. (a) Solve the equation z4 = –8 – i8 3, giving roots in Cartesian form. (b) Show the roots on an Argand diagram. 19. A complex number z has modulus 64 and argument – π 3 . (a) Find the real and imaginary part of z. (b) Determine z 1 —2 in Cartesian form and show the roots on Argand diagram.
192 Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 CHAPTER 5 ANALYTIC GEOMETRY Learning Outcome (a) Transform a given equation of a conic into the standard form. (b) Find the vertex, focus and directrix of a parabola. (c) Find the vertices, centre and foci of an ellipse. (d) Find the vertices, centre, foci and asymptotes of a hyperbola. (e) Find the equation of parabolas, ellipses and hyperbolas satisfying prescribed conditions (excluding eccentricity). (f) Sketch conics. (g) Find the cartesian equation of a conic defined by parametric equations. (h) Use the parametric equations of conics. asymptote – asimptot centre – pusat circle – bulatan directrix – direktriks ellipse – elips focus – fokus hyperbola – hiperbola parabola – parabola parametric equations – persamaan berparameter radius – jejari vertex – bucu Bilingual Keywords
Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 193 5.1 Analytic Geometry Loci and equations If a point P(x, y) on the coordinate plane moves, either in a straight line or on a curve, such that its movement is restricted by certain conditions, the set of such points is called the locus of P. The relation between x and y, satisfying the conditions, is called the cartesian equation of the locus of P. Example 1 The coordinates of two points are A(4, 5) and B(–2, 1). Find the equation of the locus of a point P(x, y) such that (a) AP = BP, (b) AP = 2BP. Solution: (a) If AP = BP, AP2 = BP2 (x – 4)2 + (y – 5)2 = (x + 2)2 + (y – 1)2 x2 – 8x + 16 + y2 – 10y + 25 = x2 + 4x + 4 + y2 – 2y + 1 –12x – 8y + 36 = 0 3x + 2y = 9 (b) If AP = 2BP, AP2 = 4BP2 (x – 4)2 + (y – 5)2 = 4[(x + 2)2 + (y – 1)2 ] x2 – 8x + 16 + y2 – 10y + 25 = 4(x2 + 4x + 4 + y2 – 2y + 1) 3x2 + 24x + 3y2 + 2y – 21 = 0 If the coordinates of a moving point P(x, y) are such that x and y are functions of another variable t, then t is called a parameter, where t ∈ R. We write x = f(t) and y = g(t), and these are known as the parametric equations of the locus of P. Any point on the locus is said to have coordinates [f(t), g(t)], known as the parametric coordinates. For example, consider the parametric equations x = 2t …………… 1 and y = 3t + 1 ……… 2 Substituting 1 into 2, y = 31 x 2 2 + 1 2y = 3x + 2 Hence, we see that the pair of equations x = 2t, y = 3t + 1 represents a straight line 2y = 3x + 2. For any value of t, we will have the coordinates of a point (x, y) satisfying the equation 2y = 3x + 2. For example, when t = 1, x = 2 and y = 4, which satisfy the equation 2y = 3x + 2. P (x, y) A (4, 5) B (–2, 1) y x O
194 Mathematics Term 1 STPM Chapter 5 Analytic Geometry 5 Example 2 Find the cartesian equation of the locus of P having parametric equations x = t 2 and y = 2t + 1. Solution: x = t 2 …………… 1 y = 2t + 1 ……… 2 From 2, y – 1 = 2t (y – 1)2 = 4t 2 i.e. (y – 1)2 = 4x From 1, x = t 2 Hence, the cartesian equation of the locus of P is (y – 1)2 = 4x. Example 3 Show that the cartesian equation of the locus with parametric equations x = 1 – 3t and y = 2 + 3 t is y = 11 – 2x 1 – x . Solution: x = 1 – 3t ………… 1 y = 2 + 3 t ………… 2 From 1, 3t = 1 – x t = 1 – x 3 1 t = 3 1 – x ………… 3 Substituting 3 into 2, y = 2 + 3( 3 1 – x ) = 2 – 2x + 9 1 – x = 11 – 2x 1 – x Hence, the cartesian equation of the locus is y = 11 – 2x 1 – x . Exercise 5.1 1. Find the locus of a point whose distance from the origin is twice its distance from the point (0, 2). 2. Find the locus of a point whose distance from the point (3a, 0) is equal to its distance from the y-axis. 3. Find the locus of a point such that its distance from the point (0, 2) is twice its distance from the point (0, –1). 4. Find the locus of a point which is equidistant from the points (4, 4) and (–4, –1). Show that it represents the equation of the perpendicular bisector of the line joining the points.