BBA102_BCM102_Business Mathematics and Statistics

Annuities 45 A = P1R P1 = A/R Putting V = 1/R, we get P1 = AV Similarly, if P2 is the present value of the second payment of ` A that is made at the end of two years, then A = P2R2 P2 = A/R2 = AV2 Proceeding in this manner, we get P3 = A/R3 = AV3 ......................... ......................... Pn = A/Rn = AVn The present value of the annuity is P = P1 + P2 + …+Pn = A/R + A/R2 + ...... + Pn = A/R + A/R2 + ................. + A/Rn = AV + AV2 + ....... + AVn = AV(1 + V + V2 + ...... + Vn–1) §1 Vn · … (3) ? P = AV ©¨ 1  V ¹¸ A §1 Rn · Also P = R ©¨ 1  R1 ¸¹ = A §1 Rn · ¨© R 1 ¸¹ A §Rn 1· … (4) ? P = Rn ©¨ R  1 ¹¸ CU IDOL SELF LEARNING MATERIAL (SLM)

46 Business Mathematics and Statistics A ª (1  i)n 1º Also P = (1 i)n «¬ 1 i 1 »¼ A ª(1 i)n  1º = « » i ¬ (1  i)n ¼ = A ¬ª«1  (1 1 º i  i)n ¼» ? P= A ¬ª1  (1  i) n ¼º … (5) i Example 3: Find the present value of an annuity of ` 300 per annum for 5 years at 4%. Answer to the nearest paisa. Given that log 1.04 = .0170333 and log 821.923 = 2.9148335. If the above annuity be a perpetual one, what will be the present value? [I.C.W.A.] Solution: Using formula (5) P= A [1 – (1 + i)–n] i 300 = [1 – (1.04)–5] .04 Now (1.04)–5 =Antilog {–5 log (1.04)} = Antilog {–5 × .0170333} = Antilog {–.0851665} = Antilog {1.9148335} = .821923 300 P = .04 (1 – (.821923) 300 = .04 × .178077 53.423100 = .04 5342.31 =4 CU IDOL SELF LEARNING MATERIAL (SLM)

Annuities 47 = 1335.5775 = 1335.5775 ? Present value is ` 1335.58 If the above annuity be a perpetual one, then putting n Ÿf in formula (5) we get A P= i 300 P= .04 30000 =4 = ` 7,500 3.4 Perpetuity The present value of a perpetuity is obtained by making n infinite. A … (6) ? P = i In particular if xA is the present value of annuity A, then the annuity is said to be worth “x years’ purchase”. A Here xA = i 1 100 ? x = 1 ¨©§10r0 ¸·¹ r … (6) i= 100 … (7) The No. of years purchase = Rate per cent Therefore, the number of years’ purchase of a perpetual annuity is obtained when we divide 100 by the rate percent. A freehold estate yields a perpetual annuity. This is called the rent. In order to know the rate per cent at which interest is reckoned, we divide 100 by the number of years’ purchase. CU IDOL SELF LEARNING MATERIAL (SLM)

48 Business Mathematics and Statistics Example 4: A freehold estate is purchased for ` 27,500. Find the rent for which it should be let so that the owner may receive 4% on the purchase money. Solution: The present value of perpetual annuity of amount A is ` 27,500, interest being 4%. r 41 i= 100 100 25 Present value of a perpetuity is given by A P= i A ? 27500 = ©¨§ 1 ¹·¸ = 25A 25 27500 A = 25 = 1100 Therefore rent should be ` 1,100. Example 5: A freehold estate worth ` 1,200 a year is sold for ` 40,000. What is the rate of interest? Solution: Using the formula — A P= i 1200 40,000 = i 1200 r i = 40000 100 120000 r = 40000 = 3 The rate of interest is 3%. CU IDOL SELF LEARNING MATERIAL (SLM)

Annuities 49 3.5 Annuity Due To find the amount of an annuity due of ` A at the end of n years. As the first payment is made immediately, it earns interest for n years. ©§¨1 r · n 100 ¹¸ ? M1 = A  = A(1 + i)n i.e., M1 = ARn. This is the amount of the first payment at the end of n years. The second payment which is made at the end of the first year, earns interest for (n-1) years. M2 = ARn–l. This is the amount of the second payment. Similarly, M3 = ARn–2 is the amount of the third payment. Proceeding in this manner we get the amount of the last payment as M3 = AR ? the total amount of the annuity due is M1 + M2 + M3 + …… + Mn ? M = M1 + M2 + M3 + … + Mn = ARn + ARn–1 + ARn–2 + … + AR = AR (Rn–1 + Rn–2 + …… + 1) = AR (1 + R + R2 + ……+Rn–1) § Rn 1· … (8) ? M = AR ¨© R  1 ¹¸ ­(1 i)n  1½ = A(1 + i) ® ¾ ¯ (1  i) 1 ¿ ?M= A(1 1) [1 1)n  1] … (9) i CU IDOL SELF LEARNING MATERIAL (SLM)

50 Business Mathematics and Statistics 3.6 Present Value of an Annuity Due To find the present value of an annuity due of Rupees A (say), for n years with an effective rate of interest r per cent per annum. Suppose P1 is the present value of the first payment of Rupees A. Since the first payment is made at the beginning of the first year, there is no interest on it. ? P1 = A If P2 is the present value of the second payment, then interest is calculated for one year. A = P2R i.e., P2 = A = AF–1 = AV R A Similarly P3 = R2 = AR–2 = AV2 Thus the present value of the last payment is A Pn = Rn1 = AR–n+1 = AVn–1 The present value of the annuity is P = P1 + P2 + P3 + …… + Pn = A + AV + AV2 + …… + AVn–1 = A(1 + V + V2 + …… Vn–1) P= A A §1 Vn · … (10) ¨© 1 V ¹¸ Also P = A §1 Rn · since V = R–1 ¨© 1  R 1 ¸¹ = A § 1 1 · ¨ 1 Rn ¸ ¨ 1 ¸ ¨© ¸¹ R = A A § Rn  1 · ¨©§ R ¸·¹ ©¨ R ¹¸ R n 1 CU IDOL SELF LEARNING MATERIAL (SLM)

Annuities 51 A §Rn 1· … (11) i.e., P = Rn1 ¨© R  1 ¸¹ A ­(1  i)n  1½ Also, P = ® ¾ (1  i)n1 ¯ 1i 1 ¿ A ­(1 i)n 1½ P= ® ¾ … (12) i ¯ (1  i)n1 ¿ Formula (12) can also be written as A … (13) P = [(1 + i) – (1 + i)–n-1)] i Example 6: A man purchases a machine on instalment basis. As per the agreement he is required to pay ` 1,000 every year for 16 years and the first instalment is to be paid immediately. If the rate of compound interest works out to be 5% per annum, what is the present worth of the agreement? Solution: Using formula (13) A P = {(1 + i) – (1 + i) –(n–1)} i = 1000 {1.05 – (1.05)–16–1)} .05 1000 = {1.05 – (1.05)–15} .05 Now (1.05)–15 = Antilog {–15 log (1.05)} = Antilog {–15 × .0212} = Antilog {–.3180} = Antilog (1.6820) = 4808 1000 P = ¨§©1050 ¹¸· (1.05 – .4808) CU IDOL SELF LEARNING MATERIAL (SLM)

52 Business Mathematics and Statistics = 100000 u .5692 5 = 100000 u . 5692 5 10000 = ` 11,384 The present worth of the agreement is ` 11,384. 3.7 Deferred Annuity Amount of a deferred annuity: To find the amount of an annuity of Rupees A for n years, which is deferred for m years, the effective rate of interest being r per cent per annum. As the amount is deferred for m years, the first payment falls due at the end of (m + 1) years, and there is no interest for the period of m years. The amount at the end of (m + n) years would be the same as the amount of immediate annuity of n years. A M = [(1 + i)n – 1] i 3.8 Present Value of a Deferred Annuity To find the present value of an annuity of Rupees A for n years, deferred for m years. The effective rate of interest is r per cent per annum. The first payment of ` A falls due at the end of (m + 1) years. Therefore the present value of annuity of ` A would be: A = Rm1 because A = P1Rm+1, where P1 is the present value. Thus P1 = A A AVm1 R m1 (1 i)m1 CU IDOL SELF LEARNING MATERIAL (SLM)

Annuities 53 If P2 is the present value of the second payment of ` A which falls due at the end of (m + 2) years, then A= P2 Rm+2 AVm2 ? P2 = AA R m2 (1  i)m2 Proceeding in this manner, A A AV m 2 Pn = Rmn (1  i)m2 where Pn is the present value of the last payment of ` A which falls due at the end of (m + n) years. Thus, the present value of the annuity is: P = P1 + P2 + P3+ ……Pn = AVm+1+ AVm+2 + ……+AVm+n = AVm+1 (1 + V + ……Vn–1) = AVm+1 §1 Vn · ©¨ 1  V ¹¸ A §¨ 1  1 · R m1 ¨ Rn ¸ ¨ 1 ¸ = ¨ 1 ¸ 1 ¸ © R¹ = A § Rn 1· ©¨§ R 1¹·¸ R m1 ¨© Rn ¸¹ R A §Rn 1· … (14) ? P = R  1 ¨© R mn ¸¹ … (15) A ­ (1  i)n1 ½ Also, P = ® ¾ (1  i  1) ¯ (1  i)m n ¿ A ­(1 i)n 1½ i.e., P = ® ¾ i ¯ (1  i)mn ¿ CU IDOL SELF LEARNING MATERIAL (SLM)

54 Business Mathematics and Statistics Example 7: What is the present value of an annuity of ` 100 payable for 12 years but deferred for 3 years, the effective rate of interest being 5% p.a.? Solution: Using formula (15), A ­(1 i)n 1½ P= ® ¾ i ¯ (1  i)m n ¿ 100 ­(1.05)12 1½ = ® ¾ .05 ¯ (1.05)15 ¿ 100 = ¨©§1050 ·¸¹ [(1.05)12 – 1] × (1.05)–15 10000 = 5 (1.805 – 1) × 0.4808 10000 = (1.805 – 1) × 0.4808 5 = 10000 u 805 u 4808 5 1000 10000 = ` 774.088 ? The present value is ` 774.09 approx. 3.9 Endowment Fund This fund is created for the payment of an annuity for a definite period. Therefore the formula for this fund is the same as the present value of an annuity. This fund is invested for a certain period at compound interest. Further, this fund being the same as the present value of a deferred perpetuity, we have, A1 … (16) P = i (1 i)n where n = number of years for which it is deferred. CU IDOL SELF LEARNING MATERIAL (SLM)

Annuities 55 Example 8: A person wishes to institute a scholarship of the value of ` 1200 to be awarded every year to a deserving student. Assuming that money is worth 5% p.a. compound interest, find the amount he should provide. Solution: Here the formula for the present value of a perpetuity is applicable. Therefore A P= i 1200 1200 = .05 §¨© 1 ¹·¸ = 24,000 20 The person concerned has to provide ` 24,000. Example 9: Find the present value of a perpetual annuity of ` 800, the first payment of which is to be made in six years time, interest being reckoned at 4% per annum. Answer to the nearest paisa. Solution: Using the formula (16) A1 P = i (1 i)n and substituting A = 800 i = .04 n= 6–1=5 we get 800 u (1.04)5 P= .04 = Antilog [log 800-5 log (l.04)–log (.04)] = Antilog [2.9031 – 5(0.170) – 2.6021] = Antilog (4.2160) = 16440 The present value is ` 16,440. CU IDOL SELF LEARNING MATERIAL (SLM)

56 Business Mathematics and Statistics 3.10 Annuity in the case of which payments are made other than annually If an annuity A is payable t times every year and interest is compounded t times a year, then the value of each payment is A/t. The ‘interest periods’ in n years is nt. The interest on unit sum for an interest period (1/t) is i §¨©1010 ·¸¹ were i is the rate per cent per annum. t Example 10: What will an annuity of ` 400 (annual value) amount to in 4 years at 4% compound interest, payable quarterly? Solution: Using the formula: M = A {(1 + i)n – 1} i 400 ­°¯°®§¨©1  .04 ·¹¸ 4u 4  1½°¾ 4 °¿ = ¨©§ .04 ·¸¹ 4 ^ `400 1.01)16 1 = ©§¨1010 ¹·¸ = 100 × 400 {(1.01)16 – 1} Now (1.01)16 = Antilog {16 log 1.01} = Antilog {16 (.0043)} = Antilog {0.0688} = 1.171 M = 100 × 400 (1. 171 – 1) 171 = 100 × 400 × 1000 40000 = 1000 × 171 The amount is ` 6840. CU IDOL SELF LEARNING MATERIAL (SLM)

Annuities 57 3.11 Sinking Fund This fund is created for the purpose of discharging a known liability or for replacing a wasting asset. Further, it is very often created for redemption of debentures. If M is the amount of liability to be discharged after n years and A is the sum invested at the end of every year, then A M = {(1 + i)n – 1} i Example 11: A man wishes to create a fund of ` 50,000 for a house at the time of retirement which is due after 10 years. If the rate of compound interest is 5% per annum, find the amount that he should deposit each year to receive the fund at the time of his retirement. A Solution: Here M = i {(l + i)n – 1} is the formula to be used. ? 50,000 = A {(1.05)10 – 1} .05 A = ¨§© 1 ¸¹· (1.6289 – 1) 20 = 20A × (0.6289) 50000 ? A = 20 u 0.06289 2500 = 0.6289 = Antilog (log 2500 – log 0.6289) = Antilog (3.3979 – 1.7986) = Antilog (3.5993) = 3975 The man has to deposit ` 3,975 every year. 3.12 Repayment of Loan by Instalments In problems relating to repayment of loan by instalments, we should use the formula for the present value of an annuity. In problems connected with investment or saving by instalment or with CU IDOL SELF LEARNING MATERIAL (SLM)

58 Business Mathematics and Statistics the creation of ‘Depreciation Fund’ by instalment, we should use the formula for the amount of annuity. Example 12: A man wishes to pay back his debt of ` 2,522 due after 3 years by three equal yearly instalments. Find the amount of each instalment, money being worth 5% per annum compound interest. Solution: Here we use the formula for the amount of annuity since the debt is due after 3 years. A M = {(1 + i)n – 1} i A = .05 {(1.05)3 – 1} A ? 2522 = (1.05 – 1) {(1.05)2 + 1.05 + 1} .05 = A (3.1525) 2522 ? A = 3.1525 = ` 800 ? The amount of each instalment is ` 800. Additional Worked Examples Example 13: A man purchases a house and takes a mortgage on it for ` 60,000 to be paid off in 12 years by equal annual payments. If the interest rate is 5% compounded annually, what amount will he be required to pay each year? Solution: Using formula (5), i.e., P = A {1 – (1 + i)–n} we get i A 60,000 = .05 {1 – (1.05)–12} A = ©§¨ 1 ¸·¹ {1 – (1.05)–12} 20 CU IDOL SELF LEARNING MATERIAL (SLM)

Annuities 59 Now (1.05)–12 =Antilog {–12 log (1.05)} l = Antilog {–12 × .02119} i = Antilog {–0.25428} = Antilog ( 1 .74572) = 0.55683 ? 60,000 = 20A (l – 0.55683) i.e., 60,000 = 20A × 0.44317 = (8.8634)A 60,000 A = 8.8634 60,000* [ 8.8634 = Antilog [log 60,000 – log 8.8634] = Antilog [4.77815 – 0.94760] = Antilog (3.83055) = 6,769.49 = ` 6,769.49 Each year the person concerned has to pay ` 6,769.49. Example 14: A penon purchased a refrigerator and paid ` 5,000 immediately. He cleared off his debt completely by paying ` 5,000 after one year and another sum of ` 5,000 after two years. If he paid compound interest at 3Y2% per annum, find the selling price of the refrigerator to the nearest rupee. Solution: Using formula (5) P = A {1 – (1 + i)–1} we get i = 5000 ®­°1  ¨©§1  3½ · 2 °½ ©§¨130½0 ¹¸· °¯ 100 ¸¹ ¾ °¿ 5000 °­®1  §©¨ 207 · 2 °½ ¯° 100 ¹¸ ¾ = §¨© 7 ¸¹· ¿° 200 CU IDOL SELF LEARNING MATERIAL (SLM)

60 Business Mathematics and Statistics = 10, 00, 000 §©¨1  40, 000 · 7 42,849 ¸¹ = 10, 00, 000 u 2,849 7 42,849 = ` 9,498.47 Since ` 5,000 was paid immediately, the selling price of the refrigerator is ` 5,000 + ` 9,498.47 = ` 14,498.47 Example 15: A man borrows ` 5,000 at 4% compound interest. If the principal and interest are to be repaid by 10 equal annual instalments, what is the amount of each instalment? It is given that Log 1.04 = .0170333 and log 675565 = 5.829667 Solution: Using formula (5) i.e., P = A {1 – (I + i)–n} we get i 5000 = A {1 – (1.04)–10} .04 Now (1.04)–10 = Antilog {–10 log (1.04)} = Antilog {–10 (.0170333)} = Antilog {–.170333} = Antilog ( 1 .829667) = .675565 A 5000 = .04 (l – 0.675565) A = .04 (0.324435) 5000 u .04 A = 0.324435 = ` 616.4561 The amount of each instalment is ` 616.46 approximately. CU IDOL SELF LEARNING MATERIAL (SLM)

Annuities 61 3.13 Summary ‘Annuity’ is a fixed sum of money paid regularly at equal intervals of time. Various types of Annuities are: Immediate Annuity, Annuity Due, and Deferred Annuity. Endowment Fund is created for the payment of Annuity for a definite time period. Sinking Fund is created for the purpose of discharging a known liability or for replacing a wasting asset. Repayment of loans by instalments is based on the basis of calculation of annuity. So, also the examples on ‘Depreciation Fund’. 3.14 Key Words/Abbreviations I.A. = Immediate Annuity, A.D. = Annuity Due, D.A. = Deferred Annuity, P.W. = Present Worth, E.F. Endowment Fund, S.F. = Sinking Fund, D.F. = Depreciatioin Fund 3.15 Learning Activity Study the illustrative examples that explain the calculation of I.A., A.D., D.A., P.W., E.P., S.F., D.F.,M.S. Study the importance of each of the funds. .............................................................................................................................................................. ........................................................................................................................................................... 3.16 Unit End Questions (MCQ and Descriptive) A. Descriptive Type: Short Answer Type Questions 1. A person borrows ` 6000 at 5% compound interest and agrees to pay both the principal and the interest in 5 equal annual instalments at the end of each year. Find the amount of each instalment. [Ans: ` 1,385.85] 2. A person deposits his whole fortune of ` 20,000 in a bank at 5% compound interest per annum and settles to withdraw ` 1,800 per year for his personal expenses. If he begins to spend from the end of the first year and goes on spending at this rate, show that he will be ruined before the end of the 17th year. [I.C.W.A.] CU IDOL SELF LEARNING MATERIAL (SLM)

62 Business Mathematics and Statistics 3. A man procures a machine for ` 1,750. He agrees to pay ` 800 cash immediately and the balance in 9 equal annual instalments. If interest (compound) is calculated at 4½% p.a., find the amount of each instalment. [Ans: ` 130.87] 4. How many year’s purchase should be given for a free hold estate, interest being calculated at 4½%? [Ans: 222/9] 5. A sinking fund is created for the redemption of debentures of ` 1,00,000 at the end of 25 years. How much money should be provided out of profits each year for the sinking fund, if the investment can earn interest @ 4% per annum? [Ans: ` 2,408.19] [Burdwan Uni.] 6. A sinking fund is to be created for the purpose of replacing, after 20 years, some machinery worth ` 1,00,000. How much money should be set aside each year out of the profits for the sinking fund, if the rate of compound interest is 5% p.a.?[Ans: ` 3021.15] [Burdwan Uni.] 7. A person borrows ` 4,000 on the condition that he will repay the money with compound interest at 5% p.a. in 6 equal annual instalments, the first one being payable at the end of the first year; find the value of each instalment. [Ans: ` 787.71] [Burdwan Uni.] 8. A man purchases a house and takes a mortgage on it for ` 60,000 to be paid off in 12 years by equal annual payments. If the interest rate is 5 per cent compounded annually, what amount will he be required to pay each year? [Ans: ` 6,769.41] [I.C.W.A.] 9. A person proposes to make an endowment on 1st July 1999 to the University by depositing a sum of money to its banking account stipulating: (i) the payment of a scholarship of ` 1,000 per annum for 10 years, and (ii) the award of book prizes of the value of ` 500 every year for 20 years. The money deposited together with compound interest @ 5% per annum is to exhaust itself at the end of the said 20 years. Assuming that the payments and awards stipulated are to be made on 1st July every year commencing with the year 2000, determine the amount of endowment required to be made on 1st July 1999 for this purpose. [Ans: ` 13,957] [Calcutta Uni.] 10. Find the present value of a perpetual annuity of ` 10 payable at the end of the first year, ` 20 at the end of the second year, ` 30 at the end of the third year and so on and in this manner increasing ` 10 each year; interest being taken at 5% per annum. [Ans: ` 4,200] CU IDOL SELF LEARNING MATERIAL (SLM)

Annuities 63 11. A man retires at the age of 60 years and his employer gives him a pension of ` 1,200 a year paid in half-yearly instalrnents for the rest of his life. Reckoning his expectation of life to be 13 years and that interest is at 4% p.a. payable half-yearly, what single sum is equivalent to this pension? [Ans: ` 12,075] [Calcutta Uni.] 12. The population of a country increases every year by 2.4% of the population at the beginning of that year. In what time will the population double itself? Answer to the nearest year. [Ans: 29 years (approx)] [I.C.W.A.I] 13. A man wishes to have ` 2,500.00 available in a bank account when his daughter’s first year college expenses begin. How much must he deposit now at 3.5% compounded annually, if the girl is to start in college six years hence? [Ans: ` 2,034] [I.C.W.A.I.] 14. When a boy is born, ` 500.00 is placed to his credit in an account that pays (i) 6% compounded annually, (ii) 6% compounded quarterly, (iii) 6% compounded monthly. If the account is not disturbed, what amount will there be to his credit on his twentieth birthday? [Ans: (i) ` 1,603 (ii) ` 1,626 (iii) ` 1,596] [I.C.W.A.I.] 15. A machine depreciates in value each year at the rate of 10% of its value at the beginning of a year. The machine was purchased for ` 10,000. Obtain, to the nearest rupee, its value at the end of the tenth year. [Ans: ` 3,483] [I.C.W.A.I.] 16. A machine depreciates at the rate of 7% of its value at the beginning of a year. If the machine was purchased for ` 8500, what is the minimum number of complete years at the end of which the worth of the machine will be less than or equal to half of its original cost price? [Ans: 10 years] [I.C.W.A.I.] 17. A truck purchased by a transport company at ` 60,000 depreciates at the rate of 10% p.a. and its maintenance cost for the first year is ` 2,000 which increases by 2% every year. If the scrap value realized when sold is ` 35,429.40, find the minimum average annual return from the truck the company should get so as not to sustain any loss. [Ans: ` 6,995.74] [I.C.W.A.I.] CU IDOL SELF LEARNING MATERIAL (SLM)

64 Business Mathematics and Statistics 18. A sum of money invested at compound interest amounts to ` 21,632 at the end of the second year and to ` 22,497.28 at the end of the third year. Find the rate of interest and the sum invested. [Ans: 4% p.a., ` 20,000] [ I.C.W.A.I.] 19. A man wants to invest ` 5,000 for four years. He may invest the amount at 10% per annum compound interest, interest accruing at the end of each quarter of the year or he may invest it at 10½% per annum compound interest, interest accruing at the end of each year. Which investment will give him slightly better return? [Ans: 2nd investment will give him slightly better return] [I.C.W.A.I.] 20. Mr. Brown was given the choice of two payment plans on a piece of property. He may pay ` 10,000 at the end of 4 years, or ` 12,000 at the end of 9 years. Assuming money can be invested annually at 4% per year converted annually, what plan should Mr. Brown choose? [Ans: 2nd Plan] [ I.C.W.A.I] 21. The population of developing countries increases every year by 2.3% of the population at the beginning of that year. In what time will the population double itself? Answer to the nearest year. [Ans: 31 years] [I.C.W.A.I.] 22. In a certain population the annual birth and death rates per 1000 are 39.4 and 19.4 respectively. Find the number of years in which the population will be doubled assuming that there is no immigration or emigration. [Ans: 35 years] [I.C.W.A.I.] 23. To accumulate a fund for his son’s higher education a person invests a sum of ` 100 on the son’s first birthday and an equal amount on each of the subsequent birthdays. If the interest is compounded half yearly at the rate of 6% per annum, find the amount accumulated just after the investment has been made on the l8th birthday. [Ans: ` 2,341.44] 24. A person borrows a sum of ` 5,000 at 4% compound interest. If the principal and interest are to be repaid in 10 equal annual instalments, find the amount of each instalment, first payment being made after 1 year. [Ans: ` 616.52] [I.C.W.A.] 25. A company sets aside as reserve the sum of ` 20,000 annually to enable it to payoff a debenture issue of ` 2,39,000 at the end of 10 years. Assuming that the reserve accumulates at 4% per annum compound interest, find the surplus after paying off the debenture stock. [Ans: ` 1,122.14] [l.C.W.A.] CU IDOL SELF LEARNING MATERIAL (SLM)

Annuities 65 B. Multiple Choice/Objective Type Questions 1. What is the amount to be set aside every year so as to yield ` 175 at the end of 8 years at 5% C.I. p.a.? (a) ` 18.32 (b) ` 24.50 (c) ` 28 (d) ` 30 (e) None of the above 2. What is the present value of an immediate annuity of ` 150 per year for 10 years at 3% p.a. compound interest? (a) ` 11,050 (b) ` 1,279.53 (c) ` 1,830 (d) ` 1,980 (e) None of the above 3. A person borrows ` 6,000 at 5% C.I., p.a. and agrees to pay both the priincipal and interest in 5 equal annual instalments at the end of each year. (a) ` 1,519.26 (b) ` 1,250.40 (c) ` 1,385.68 (d) ` 1,630.90 (e) None of these. Answers (1) (a); (2) (b); (3) (c) 3.19 References References of this unit have been given at the end of the book. ˆˆˆ CU IDOL SELF LEARNING MATERIAL (SLM)

66 Business Mathematics and Statistics UNIT 4 MATRICES Structure 4.0 Learning Objectives 4.1 Introduction 4.2 Different Types of Matrices 4.3 Addition of Matrices 4.4 Multiplication of Matrices 4.5 Application of Matrices 4.6 Orthogonal Matrix 4.7 Inverse of a Matrices 4.8 Cofactor Matrix 4.9 Adjoint Matrix 4.10 Method of Solving Simultaneous Equations Using Matrix Property 4.11 Summary 4.12 Key Words/Abbreviations 4.13 Learning Activity 4.14 Unit End Questions (MCQ and Descriptive) 4.14 References CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 67 4.0 Learning Objectives After studying this unit, you will be able to: z Explain the meaning and definition of a ‘Matrix’ and the various types of matrices. z Learn ‘Matix Operations of Addition and Multiplication’ along with the illustrative exmples. z Elaborate the Orthogonal Matrix, inverse of a Matrix, Co-factor Matrix, Adjoint Matrix and the method of calculation through illustrative examples. z Illustrate the method of solving simultaneous equations using Matrix Property. 4.1 Introduction A matrix is a rectangular array of m rows and n columns arranged within two brackets as shown below: ª x11 x12 x13 .......... x1n º « » « x21 x22 x23 .......... x2n » « ..... ..... ..... .......... ..... » « » « ..... ..... ..... .......... ..... » « ..... ..... ..... .......... ..... » «» ¬ xm1 xm2 xm3 .......... xmn ¼ Capital letters such as A,B, C .......... can be used to denote matrices. The above matrix becomes a square matrix of order n if m = n Further, ª 2 1 4 9º « 2» If A = ¬« 6 3 1 6 ¼» 3 4 5 Then it is said to be a matrix of order 3 × 4. ª3 4 4 º « 6» Also B = ¬« 2 4  2 »¼ is a square matrix of order 3. 3 8 CU IDOL SELF LEARNING MATERIAL (SLM)

68 Business Mathematics and Statistics C = [ 2 9 1 6 ] is a row matrix of order 1 × 4. ª4º « » « 1 » D = « 9 » is a column matrix of order 4 × 1. « » ¬ 6 ¼ 4.2 Different Types of Matrices (1) Null Matrix: A matrix having all its elements zero is called a null matrix. It may be a rectangular or square matrix. A null matrix is denoted by O. ª0 0º ª0 0 0 0º «¬ 0 0 »¼ « » O= and O= « 0 0 0 0 » 0 0 ¬« 0 0 »¼ are null matrices. Triangular matrix: A square matrix A = (aij)n is called an upper triangular matrix, if aij = 0 for all i > j, i.e., if all elements below the main diagonal are zero. For examples, the matrices A = GHF 2 64IJK X = HFGG –1 3 –255KJJI 0 0 2 0 0 are upper triangular matrices of order 2 and 3 respectively. A square matrix A = (aij)n × n is called an lower triangular matrix, if aij = 0 for all i < j, i.e.,, if all element above the main diagonal are zero. For example, the matrices A = HFGG 3 0 006JJKI B = GFHGG 2 0 0 0000KIJJJ 1 5 –1 3 0 2 –1 7 2 –4 1 7 3 are lower triangular matrices of order 2 and 4 respectively. CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 69 (2) Diagonal Matrix: A matrix having elements, other than the elements in its principal diagonal all zero is called a diagonal matrix. ª2 0 0º ª4 0º « » and B = ¬« 0 5 »¼ are diagonal matrices.. For example, A= « 0 8 0 » 0 «¬ 0 6 ¼» (3) Scalar Matrix: If the elements of a diagonal matrix are all equal, it is known as a scalar matrix. ª4 0º ª8 0 0º A = «¬ 0 4 »¼ « » For example, and B = « 0 8 0 » are scalar matrices. 0 «¬ 0 8 ¼» (4) Unit Matrix: If each of the diagonal elements of a square matrix is equal to 1 (unity) and the other elements are all zero, then such a matrix is called a Unit matrix or Identity matrix. Unit matrix is denoted by the letter I. ª1 0 0º ª1 0º « » ¬« 0 1 ¼» For example, I= « 0 1 0 » and I = 0 ¬« 0 1 »¼ are unit matrices. (5) Equal Matrices: Two matrices of the same order are said to be equal if their corresponding elements are equal. ª a11 a12 º ª b11 b12 º For example, A = ¬« a21 a22 ¼» and B = ¬« b21 b22 »¼ are equal if a11 = b11 , a12 = b12 , a21 = b21 , a22 = b22. CU IDOL SELF LEARNING MATERIAL (SLM)

70 Business Mathematics and Statistics (6) Transpose of a Matrix: If all the rows and columns of a given matrix are interchanged then the resulting matrix is called the transpose of the given matrix. The transpose of a matrix of order p × n is a matrix of order n × p. The transpose of a matrix A is denoted by ‘A’. ª3 1 2 4º «2 6» If, A= ¬« 1 4 1 6 ¼» then 5 3 ª3 2 1 º « 1 4 5» « » its transpose A' = « 2 1 3 » «¬ 4 6 6 ¼» (7) Coefficient Matrix: Given a set of 3 linear homogenous equations, the matrix formed by the coefficient of these linear equations, is called a coefficient matrix. Given 3x + 2y + 4z = 19 2x – y + z = 3 6x + 2y – z = 17 The coefficient matrix is: ª3 2 4 º « 2 1 1 » ¬« 6 2  1 ¼» (8) Augumented Matrix: With respect to the linear homogenous equations 2x + 3y – z = 9 x + y +z = 9 3x – y – z = – 1 CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 71 The augumented matrix is: ª 3 2 1 9 º «1 1 1 9 » «¬ 3  1  1  1 ¼» (9) Singular and Non-singular Matrix: If the determinant of a matrix A is equal to zero, then it is said to be a ‘singular matrix’. If the determinant of A is not zero then the matrix A is said to be a ‘non-singular matrix’. (10) Symmetric Matrix: A matrix that is equal to its transpose is said to be symmetric. ª2 3 4º ª2 3 4º « » « » For example if A= « 3 7 8 » then A' = « 3 7 8 » ¬« 4 8 5 ¼» «¬ 4 8 5 ¼» and A = A', therefore A is said to be symmetric. If a matrix A = – A', then it is said to be ‘skew-symmetric’. ª 0 7 4º « 3» For example «¬ 7 0 0 ¼» 4 3 4.3 Addition of Matrices Two matrices A and B are said to be conformable for addition if both the matrices have the same number of rows and same number of columns. ª a11 a12 a13 º « » If A = « a21 a22 a 23 » «¬ a31 a32 33 »¼ ª b11 b12 b13 º « » and B = « b21 b22 b23 » «¬ b31 b32 b33¼» CU IDOL SELF LEARNING MATERIAL (SLM)

72 Business Mathematics and Statistics ª a11 + b11 a12 + b12 a13 + b13 º « a22 + b22 » then A + B = « a21 + b21 a32 + b32 a 23 + b23 » ¬« a31 + b31 a33 + b33 ¼» ª a11  b11 a12  b12 a13  b13 º « a22  b22 » and A – B = «¬ a21  b21 a32  b32 a 23  b23 ¼» a31  b31 a33  b33 With reference to addition, the Commutative, Associative and Distributive laws hold good. ª a11 a12 º ª b11 b12 º ª c11 c12 º For example, if A ¬« a21 a22 »¼ and B = ¬« b21 b22 ¼» and C = ¬« c21 c22 »¼ then (i) A + B = B + A (Commutative law) (ii) (A + B) + C = A + (B + C) (Associative law) (iii) K (A + B) = KA + KB (Distributive law) where K is a certain constant. 4.4 Multiplication of Matrices Two matrices A and B are siad to be conformable for multiplication if the number of columns of A is equal to the number of rows of B. If A is an m × p matrix and B is p × n matrix, then AB will be an m × n matrix. Here it is not possible to get the product BA as the number of columns of B is not equal to the number of rows of A, i.e., n m. ª2 1 4 º ª1 2º Example: 12: Given A = «¬ 3 2 1 »¼ « » B = « 3 5 » «¬ 1 4 »¼ ª2u1 +1u 3 4 u1 2 u 2 + 1 u 5  4 u 4º AB = «¬ 3 u 1 + 2 u 3  1 u 1 3 u 2 + 2 u 5  1 u 4 ¼» ª 9 25 º = «¬ 10 20 »¼ CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 73 4.5 Applications of Matrices In this section, applications of matrix addition and subtraction will be illustrated with suitable examples. Example 1: A firm produces two products A and B using three plants P1, P2 and P3. During the month of January, P1 produced 160 unit of A and 80 units of B; P2 produced 90 units of A and 120 units of B; and P3 produced 200 units of A and 170 units of B. During the month of February, P1 produced 140 units of A and 70 units of B; P2 produced 100 units of A and 110 units of B; and P3 produced 180 units of A and 150 units of C. Find the total production of these two products for two months. Solution: Let P be the matrix which represents the production in the month of January. AB §160 80 ·¸P1 ¨ ? P = ©¨¨29000 117200¸¸¹PP32 Let Q be matrix which represents the production in the month of February. A B §140 70 ¸·P1 ¨ ? Q = ¨¨©118000 110 ¹¸¸PP23 150 The total production (say R) is given by R = P + Q §160 80 · §140 70 · A B ¨ ¸¨ ¸ ? R = ¨©¨29000 §300 150 · P1 117200¸¹¸ + ¨¨©110800 111500¸¹¸ ¨ ¸ = ©¨¨139800 322300¸¹¸PP32 ? P1 produced 300 units of A and 150 units of B; P2 produced 190 units of A and 230 units of B; and P3 produced 380 units of A and 320 units of B. Applications of Scalar Multiplication Sometimes, we come across the situations in which a set of data is to be multiplied by a constant. This is illustrated by following example. CU IDOL SELF LEARNING MATERIAL (SLM)

74 Business Mathematics and Statistics Example 2: A firm has four plans P1, P2, P3 and P4 and produces two products X and Y. It is assumed that the daily production is uniform. Number of units produced per pay is as follows: P1 P2 P3 P4 X 100 80 90 110 Y 60 70 40 50 Find the output of each product from each plant during a month of 30 days. Solution: Let A be the matrix which represents the daily production. P1 P2 P3 P4 ? A = §100 80 90 110·X ¨© 60 70 40 50 ¸¹Y ? The total production (say B) in 30 days is given by B = 30A P1 P2 P3 P4 ? B = §3, 000 2, 400 2, 700 3,300·X ©¨1,800 2,100 1, 200 1, 500 ¹¸Y ? This the required production by four plants. Applications of Matrix Multiplication Multiplication of matrices plays a major role in illustrating business applications. In this section we will see several examples of different business applications. Example 3: A furniture manufacturing company manufactures 50 table and 40 chairs in a day. Each table requires 4 hours of labour and 5 units of material. A chair requires 2 hours of labour and 3 units of material. Using matrices, find how many labour hours and material will be required each day? Solution: Let A be the matrix which represents the number of tables and chairs. §50·T ¨¸ ? A = ©40¹C2 u1 Let B be the matrix which represents the input per unit TC ? B = § 4 2·L ¨ ¸ ©5 3¹M2 u2 CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 75 Total input (say C) is given by C = BA T C §280· L L§4 2·§50·T ¨¸ C= 3¹¸©¨40¹¸C = ©370¹M ¨ M©5 ? 280 labour hours and 370 units of material are required. 4.6 Orthogonal Matrix ª a1 b1 c1 º « » If A = « a2 b2 c2 », then «¬ a3 b3 c3 ¼» It is said to be orthogonal if the sum of the squares of the elements of any rwo or of any column is unity and the sum of the products of the corresponding elements of any two rows or of any two columns is zero. 4.7 Inverse of a Matrices If A and B are two square matrices of the same order and if AB = BA = I, where I is the unit matrix then B is said to be the inverse of A and it is denoted by A–1. 4.8 Cofactor Matrix ª a11 a12 a13 º « » If A = « a21 a22 a 23 » ¬« a31 a32 a33 ¼» then the matrix formed by the cofactors. A11 A12 A13 A21 etc of the elements a11 a12 a13 a21 etc; respectively which is represented as CU IDOL SELF LEARNING MATERIAL (SLM)

76 Business Mathematics and Statistics ª A11 A12 A13 º « » « A 21 A 22 A 23 » ¬« A31 A32 A33 ¼» is called the Cofactor matrix of A. 4.9 Adjoint Matrix The transpose of the Co-factor matrix is called the Adjoint Matrix. Calculation of the Inverse of a matrix A The inverse of A, denoted by A–1 is expressed as the Adjoint of matrix A divided by the determinant of A. Matrix A is non-singular. A–1 is also called ‘reciprocal of matrix A’. Adj A Thus A–1 = A 4.10 Method of Solving Simultaneous Equations Using Matrix Property Suppose a1x + b1y + c1z = P1 a2x + b2y + c2z = P2 a3x + b3y + c3z = P3 Using the matrix notation we can write the above equations as ª a1 b1 c1 º ª x º ª P1 º « » « » « » « a2 b2 c2 » « y » = « P2 » ¬« a3 b3 c3 »¼ ¬« z »¼ «¬ P3 »¼ i.e., AX = P ? A–1 AX = A–1P i.e., IX = A–1P i.e., X = A–1P CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 77 ª 3 2 6 º ª 8 2 5 º « » « 1» Example 13: If A= ¬« 5 6 9 ¼» B= «¬ 3 6 3 »¼ 2 7 3 7 4 Find (i) 2A – 3B (ii) AB + BA Solution: ª  6  24 4  6 12  15 º «  19  9 18  3 » 2A – 3B = «¬ 4  21 12  18 6  9 »¼ 14 12 ª 30 2  3 º « 1 15 » = «¬ 25 6 3 »¼ 2 ª  60 18 1 º ª  9 7 45 º «  121 8» « 19 39 » AB + BA = ¬«  26 82 26 »¼ + ¬« 7 49 3 ¼» 50 59 ª  64 25 46 º « 140 47 » = ¬« 19 131 29 ¼» 109 ª 3 4 1 º ª2 1 3º ª 2 1 4º « 5» « » « 3» Example 14: Given A= «¬ 2 1 4 »¼ B = «¬ 3 2 4 »¼ C= «¬ 4 2 2 »¼ 1 1 5 1 3 1 6 Find the following (i) 2A – 3B + C (ii) A' and B' (iii) C–1 Solution: ª 10 6 3 º « 1» (i) 2A – 3B + C = ¬« 9 6 15 ¼» 12 11 CU IDOL SELF LEARNING MATERIAL (SLM)

78 Business Mathematics and Statistics ª 3 2 1 º ª2 3 5 º « 1 1 » «1 1 » (ii) A' = ¬« 4 5 4 ¼» and B' = ¬« 3 2 3 ¼» 1 4 Adj C 1 ª 14 22 5 º = « 11 0 20 » (iii) C' = C 121 «¬ 26 11 16 »¼ Example 15: Find the inverses of A , B , C where ª 3 2 4 º ª 3 1 4 º « » « 6» A = ¬« 1 6 1 »¼ B= «¬ 2 1 3 »¼ 2 3 5 5 2 ª 1 2 3 º « 2» C= ¬« 4 5 -4 »¼ 2 6 ª 3 2 4 º « 6 1 » Solution: For matrix A = ¬« 1 3 5 »¼ 2 (i) |A| = ª -6 1º 2 ª 1 1 º ª1 -6 º – 3 «¬ 3 - 5 ¼» – «¬ 2 - 5 ¼» – 4 ¬« 2 3 ¼» = – 3 (30 – 3) – 2 (– 5 – 2) – 4 (3 + 12) = – 3 (27) – 2 (– 7) – 4 (15) = – 81 + 14 – 60 = – 141 + 14 = – 127 (ii) The co-factor of Matrix A is ª A11 A12 A13 º « » «¬ A 21 A22 A23 »¼ A31 A32 A33 ª 6 1 º A11 = «¬ 3  5 ¼» = 30 – 3 = 27 CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 79 ª1 1 º A12 = – ¬« 2  5 »¼ = – (– 5 – 2) = 7 ª 1 6 º A13 = «¬ 2 3 ¼» = 3 + 12 = 15 ª 2 4 º A21 = – «¬ 3  15 »¼ = – (– 10 + 12) = – 2 ª 3 4 º A22 = «¬ 2  5 ¼» = 15 + 8 = 23 ª 3 2 º A23 = – ¬« 2 3 ¼» = – (– 9 – 4) = 13 ª 2 4 º A31 = «¬  6 1 »¼ = 2 – 24 = – 22 ª 3 4 º A32 = – «¬ 1 1 ¼» = – (– 3 + 4) = – 1 ª 3 2 º A33 = «¬ 1  6 »¼ = 18 – 12 = 16 Co-factor matrix ª 27 7 15 º « 2 23 13 » A= ¬«  22  1 16 ¼» (iii) The transpose of the factor matrix of A is the Adjoint matrix A, where ª 27  2  22 º «7 1 » Adj A = «¬ 15 23 16 ¼» 13 CU IDOL SELF LEARNING MATERIAL (SLM)

80 Business Mathematics and Statistics (iv) Inverse of the matrix A is ª  27 2 22 º « » Adj A « 127 127 127 » « 7  23 1 » A–1 = | A| = « 127 127 127 » «  15  13  16 » «¬ 127 127 127 »¼ To verify the above solution we should prove that ª1 0 0º «0 0» A–1 A = A A–1 = I = ¬«0 1 1»¼ 0 ª  27 2 22 º « » « 127 127 127 » ª 3 2  4º « 7  23 1 » « 1 6 1 » ¬« 2 3  5»¼ ? A–1 A = « 127 127 127 » «  15  13  16 » ¬« 127 127 127 ¼» ª 81 + 2 + 44  52  12 + 66 108 +2  110 º «  127 127 127 127 127  » « 127 127  14  138 + 3 127 127 » « 21 23 +2 127 127 127 28  23 5 » = « 127 127 127 127 127 127 » « 45  13  32  30  78  48 60  13  80 » «¬127 127 127 127 127 127 127 127 127 »¼ ª127 0 0 º «««1207 » «127 127 127 » ª1 0 0º «0 127 0 » «0 1 0» «¬0 0 1»¼ = 127 127 » = 0 127 » ¬«127 127 127 ¼» Thus it is verified. CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 81 Example 16: For matrix B ª3  1 4 º = «2  1 6 » «¬5 2  3»¼ ª 1 6 º ª2 6 º ª2  1º (i) |B| = 3 «¬ 2  3»¼ + 1 «¬5  3¼» + 4 ¬«5  2¼» = 3 (3 – 12) + 1 ( – 6 – 30) + 4 (4 + 5) = 3 (– 9) + 1 (– 36) + 4 (9) = – 27 – 36 + 36 = – 27 0 (ii) The co-factor Matrix of B is ª B11 B12 B13 º « » ¬« B21 B22 B23 »¼ B31 B32 B33 ª 1 6 º B11 = ¬« 2  3»¼ = 3 – 12 = – 9 ª2 6 º B12 = – «¬5  3¼» = – (– 6 – 30) = 36 ª2  1º B13 = ¬«5 2 ¼» = 4 + 5 = 9 ª 1 4 º B21 = – ¬« 2  3¼» = – (3 – 8) = 5 ª3 4 º B22 = «¬5  3»¼ = – 9 – 20 = – 29 ª3  1º B23 = – ¬«5 2 ¼» = – (6 + 5) = – 11 ª 1 4º B31 = ¬« 1 6¼» = – 6 + 4 = – 2 CU IDOL SELF LEARNING MATERIAL (SLM)

82 Business Mathematics and Statistics ª3 4º B32 = – ¬«2 6»¼ = – (18 – 8) = – 10 ª3  1º B33 = «¬2  1¼» = – 3 + 2 = – 1 ? Co-factor matrix ª 9 36 9 º «5  11» B= ¬« 2  29  1 ¼»  10 (iii) The transpose of the cofactor matrix of B / Adj. B ª 9 5  2 º « 36  29  10» ¬« 9  11  1 ¼» ? The required inverse of the matix B is ª 9 5 2º « » Adj B « 27  27  27 » B= « 36  29  10 » B–1 = «  27  27  27 » «9  11 1 » ¬«  27  27  27 ¼» To verify the above solution we should prove that B–1 B = B B–1 = I ª 9 5 2º « » « 27 27 27 » ª3  1 4 º «  36 29 10 » «2  1 6 » «¬5 2  3¼» B–1 B = « 27 27 27 » «9 11 1» «¬ 27 27 27 »¼ CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 83 ª 27  10 + 10 9  5 + 4  36  30  6 º « 27 27 27 27 27  27 » «  108  58 27 36  29 + 20 27 174 27 » « + 50 144  30 » = « 27 27 27 27 27 27 27 27 27 » «  27  2+5 9  11 + 2  36  66  3 » ¬« 27 27 27 27 27 27 27 27 27 ¼» ª1 0 0º «0 0» = «¬0 1 1»¼ 0 Thus it is verified. Example 17: For matrix C we have ª 1 2  3º «4 2» C= ¬« 2 5 4 ¼» 6 (i) |C| = 1 (– 20 – 12) – 2 (16 + 4) – 3 (24 – 10) = – 32 – 40 – 42 = – 114 0 (ii) By calculating different cofactors we obtain the following co-factor of matric C. ªC11 C12 C13 º «¬«CC3211 » C= C22 C23 ¼» C32 C33 ª 32  20 14 º = «« 20  2  10»» «¬ 11  14  13¼» CU IDOL SELF LEARNING MATERIAL (SLM)

84 Business Mathematics and Statistics (iii) Adj. of C. ª 32  26  11º = «« 20  2  14»» «¬ 14  10  13»¼ (iv) Inverse of Matrix C ª 32 26 11 º « » « 114 114 114 » « 20 2 14 » Adj. C « » C–1 = C = « 114 114 114 » 10 «  14 13 » «¬ 114 114 114 »¼ To verify C–1 C = C C–1 = I ª 32 26 11 º « » « 114 114 114 » « 20 2 14 » ª 1 2 3º « » « » C–1 C = « 114 114 114 » « 4 5 2 » «  14 10 13 » ¬«2 6 4 ¼» ¬« 114 114 114 ¼» ª 32 + 104  22 64  130  66 96  52  44 º « +  114 114 114   » « 114 114 114 40  10  84 114 114 114 » « 20 8 28 60 4 56 » = «114 114 114 114 114 114 114 114 114» « 14 40 26 28  50  78 42 20 52 » «¬ 114 + 114  114 114 114 114 114  114  114 »¼ ª1 0 0º = ««0 1 0»» «¬0 0 1»¼ CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 85 Thus it is verified. ª8 2 4º Example 18: If A = ««3 5 5»» find A–1 ¬«5 1 3¼» Solution: (i) |A| = 8 (15 – 5) – 2 (9 – 25) + 4 (3 – 25) = 8 (10) – 2 (– 16) + 4 (– 22) = 80 + 32 – 88 = 112 – 88 = 24 0 (ii) Cofactor matrix of A is: ª 10 16 22º « 2 2» A= ¬«10 4 34 »¼ 28 ª10 2 10 º «16 28» (iii) Adj. A = ¬«2 4 34 »¼ 2 ª 10 2 10 º « » Adj.A « 24 24 24 » « 16 4 28 » (iv) A–1 = A = « 24 24 24 » « 22 2 34 » «¬ 24 24 24 »¼ (v) To verify we should prove that A–1 A = A A–1 = I CU IDOL SELF LEARNING MATERIAL (SLM)

86 Business Mathematics and Statistics ª 10 2 10 º « » A–1 A = « 24 24 24 » ª8 2 4º « 16 4 28 » «3 5 5» « ¬«5 1 3¼» ? 24 24 24 » « 22 2 34 » ¬« 24 24 24 »¼ ª 80  6  50 20  10  10 40  10  30 º « +  24 24 24 24 24  » « 24 24 24 32 + 20  28 64 + 20 24 » « 128 12 140 24 24 24 84 » = « 24 24 24 24 24 24 » « 176 + 6 + 170 44 + 10 + 34 88 + 10 + 102 » ¬« 24 24 24 24 24 24 24 24 24 »¼ ª24 0 0 º « » « 24 24 24 » 24 «0 24 0» ª1 0 0º « 0 » = ««0 1 0»» = I = « 24 24 » 0 1»¼ «¬0 «0 24 » ¬«24 24 24 ¼» ? It is proved. ª1 4 2º «2 1» Example 19: Given A= «¬7 5 5¼» find A–1 3 Solution: (i) |A| = 1 (25 – 3) – 4 (10 – 6) + 2 (6 – 30) = 22 – 4 (4) + 2 (–24) = 22 – 16 – 48 = 22 – 64 = – 42 0 CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 87 ª 22 4 24º «14 21 » (ii) Cofactor matrix of A = «¬ 6 7 3 »¼ 3 ª 22 14 6º «4 3» (iii) Adj. A = «¬24 7 3»¼ 21 ª 22 14 6 º « » Adj. A « 42 42 42 » « 4 7 3 » (iv) A–1 = A = « 42 42 42» « 24 21 3 » «¬ 42 42 42¼» (v) To verify we should prove that A–1 A = A A–1 = I ª 22 14 6 º « 423»»» « 42 42 ª1 4 2º « 4 7 «2 5 1» ¬«6 3 5»¼ A–1 A = « 42 42 42» « 24 21 3 » «¬ 42 42 42»¼ ª 22 + 28 + 36 88 + 70 + 18 44 + 14 + 30 º « 42 42  42 42 42 42 42 » « 4 + 14 42 16 + 35  9 8+ 7 42 » « 18 15 » = « 42 42 42 42 42 42 42 42 42 » « 24 42 + 18 96  105 + 9 48 21 15 » «¬ 42  42 42 42 42 42 42  42 + 42 ¼» ª1 0 0º «0 0» = ¬«0 1 1»¼ . Hence verified. 0 ª1 3 2 º «2 5» Example 20: Given A= ¬«1 4 3 »¼ find A–1 6 CU IDOL SELF LEARNING MATERIAL (SLM)

88 Business Mathematics and Statistics Solution: (i) |A| = (12 – 30) + 3 (6 + 5) + 2 (– 12 – 4) = – 18 + 3 (11) + 2 (–16) = – 18 + 33 – 32 = – 17 0 (ii) Cofactor matrix of A ª18 11 16º « 3 3» A= «¬ 7 1 10 »¼ 3 ª18 3 7 º « 11 9» (iii) Adj. A = ¬«16 1 10»¼ 3 ª18 3 7 º «««1171 » Adj. A 17 17 » 1 9 » (iv) A–1 = A = «17 17 17 » «16 3 10 » ¬«16 17 17 »¼ (v) To verify we should prove that A–1 A = A A–1 = I ª18 3 7 º «««1171 » 17 17 » ª1 3 2 º 1 9 » «2 4 5» ¬«1 6 3 »¼ A–1 A = «17 17 17 » «16 3 10 » ¬«17 17 17 ¼» ª 18 + 6 7 54 + 12 + 42 36  15  21º ««« 1117  17 17 17 17 +  » 17 33  4 + 54 17 17 17 » 2 9 22 5 27 » = «17 17 17 17 17 17 17 17 17 » «16 6 10 48  12 + 60 32 + 15 30 » ¬«17  17  17 17 17 17 17 17  17 ¼» CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 89 ª1 0 0º «0 0» = «¬0 1 1»¼ =I 0 Hence verified. 4.10 Summary In the introduction, we study the meaning and definition of the Matrix and the various types of Matrices such as Null Matrix, Diagonal Matrix, Scalar Matrix, Unit Matrix, Equal Matrices, Transpose of a Matrix, Coefficient Matrix, Augmented Matrix, Singular and Non-singular Matrix, Symmetric Matrix. We learn the operation of Matrix Additiion and Multiplication of Matrices. We understand the meanings of Orthogonal Matrix, Co-factor Matrix, Adjoint Matrix and the method of calculation of the inverse of a matrix by standing the different stages through the illustrative examples. 4.12 Key Words/Abbreviations Matrics called: Null, Diagonal, Scalar, Unit, Transpose, Coefficient, Augmented, Singular, Non-singular, Symmetric, Orthogonal, Inverse, Co-factor, Adjoint. 4.13 Learning Activity Learn the operations of Matrix Addition and Multiplication and the calculation of the Inverse of a Matrix by working out the examples given in the ‘Exercises on Matries’ ............................................................................................................................................................. ............................................................................................................................................................ 4.14 Unit End Questions (MCQ and Descriptive) [Ans: x = 1, y = –2] A. Descriptive Type: Short Answer Type Questions 1. Solve, using determinants 7x – 4y = 15 3x + 5y + 7 = 0 CU IDOL SELF LEARNING MATERIAL (SLM)

90 Business Mathematics and Statistics 2. Evaluate the determinant of the matrix. [Ans: 0] ª a a + 3 a + 6º [Ans: 360] «a + 1 a + 4 a + 7» «¬a + 2 a + 5 a + 8»¼ 3. Evaluate the determinant of the matrix ª35 38 24º «12 15 8 » «¬51 57 40»¼ ª 7 1 1º «10 1» 4. Find the inverse of the matrix ¬« 6 2 2¼» , and hence solve the following sytem equations: 3 7x – y – z = 0, 10x – 2y + z = 8, 6x + 3y – 2z = 7. ª2 4 1º «3 2» 5. Obtain the inverse of the matrix «¬1 1 -3 ¼» and hence solve the following system of 3 equations. 2x + 4y – z = 9, 3x + y + 2z = 7, x + 3y – 3z = 4 6. If the matrix A is given by  ¡ 2 1°¯ , prove that A satisfies the relation A2 – 4A + 3I = ¢¡1 2 ±° 0 where I stands for the unit matrix of order 2. ª4º « » « 0 » 7. Let A and B be two matrices defined by A = [1 –½ –1 3] and B = « 2 » Obtain the ¬«1¼» product AB and BA. Find, if possible, the sum of the two matrices AB and BA. 8. Let two matrices A and B be given by ª1 1 0º ª 2 2 4º A = ««2 3 4»» and B = ««4 2 4»» ¬«0 1 2»¼ ¬« 2 1 5 »¼ Verify that BA = 6I, where I is the unit matrix of order 3. CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 91 9. A company is considering which of the three methods of production it should use in producing three goods A, B and C. The amount of each good produced by each method is shown in the matrix. (C.A.) AB C Method 1 4 8 2 Method 2 5 7 1 Method 3 5 3 9 The vector (or row matrix) (10 4 6) represents the profit per unit for the goods. A, B and C in that order. Using matrix multiplication, find which method maximises total profit. 10. An engineering product requires five kinds of materials, the quantities of which are given in the form of the vector (i.e., a row matrix) q = [17 4 36 18 6]. If p = [5 200 4 50 3] represents a vector of the corresponding prices per unit in Rs. find the total expenses for the manufacture of the product. ª1 2 3º 11. If A = ««3 2 1»» , show that A3 –23A – 40I = 0, where I is the unit matrix of order «¬4 2 1¼» 3 and 0 is the null matrix of order 3. 12. There are two families A and B. Threre are 2 men, 3 women and 1 child in family A and 1 man, 1 woman and 2 children in family B. The recommended daily allowance for calories is: Men :2400, Women: 1900, Children: 1800 and for proteins is: Men: 55 gms, Woman: 45 gms, Children: 33 gms. Represents the above information by matrices. Using multiplication, calculate the total requirements of calories and proteins for each of the two families. ª1 2 3º 13. Find the inverse of the matrix. ««2 4 5»» «¬3 5 6¼» CU IDOL SELF LEARNING MATERIAL (SLM)

92 Business Mathematics and Statistics ª4 1 2º 14. what is the Adjoint matrix of A, Where A = ««2 10 4»»? «¬1 7 1 »¼ 15. Solve the system of linear equations. x+y+z=6 x + 2y + 3z = 14 –x+y–z=–2 using matrix inverse method. ª1 1 1º 16. Give A = ««1 2 3»» find its inverse and hence solve the system of equations: «¬1 4 9»¼ x + y + z = 6, x + 2y + 3z = 14 and x + 4y + 9z = 36. ª1 2 1º 17. If A = ««3 2 1»» verify whether A3 –23A – 40I = 0 where I is the unit matrix of ¬«4 2 1¼» order 3. 18. Using matrices A , B and C verify the rule (A.B) C = A (BC) where ª1 1 0 º ª1 0º « 1»» C = ««0 1»» A= 0 11 B = « 0 1 1 »¼ «¬1 1»¼ 1 21 1 ¬«1 ª 2 4 0º 19. Find the inverse of the matrix ««3 4 4»» ¬« 2 2 5¼» 5 14 5 71 20. Given A = 2 7 3 , B = 8 1 2 3 14 2 34 Find (i) AB + BA (ii) A'B + B'A CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 93 2 1 2   21. Given A=  3 4 5  find the inverse of A' 1 1 3 2 1 1  2 show that AA–1 = I 22. If A =  1 3 2 2 1 23. Solve x+p q r =0 [Ans: x = 0 , x = – (p + q + r)] r x+q p p q x+r 24. Prove that 1 a a2  bc =0 1 b b2  ca 1 c c2 - ab  2 1 1  25. If A = 1 2 1 show that A3 – 6A2 + 9A – 4I = 0  1 1 2  4.14 References References of this unit have been given at the end of the book.  CU IDOL SELF LEARNING MATERIAL (SLM)

94 Business Mathematics and Statistics UNIT 5 DETERMINANTS Structure 5.0 Learning Objectives 5.1 Introduction 5.2 Simultaneous Equations 5.3 The Linear Equations in two Unknowns 5.4 Theorems of Determinants 5.5 Summary 5.6 Key Words/Abbreviations 5.7 Learning Activity 5.8 Unit End Questions (MCQ and Descriptive) 5.9 References 5.0 Learning Objectives After studying this unit, you will be able to: z Explain the meaning and definition of a determinants and note the simple examples to begin with. z Illustrate the method of solving linear equatiions in two unknowns. z Study the solved examples for finding the ‘consistence’. z Elaborate the method of solving a determinant and the 6 different theorems, along the illustrative examples. CU IDOL SELF LEARNING MATERIAL (SLM)