# Example 28 - Chapter 3 Class 12 Matrices

Last updated at Jan. 17, 2020 by Teachoo

Last updated at Jan. 17, 2020 by Teachoo

Transcript

Example 28 Let A = [■8(2&−1@3&4)], B=[■8(5&2@7&4)], C = [■8(2&5@3&8)] find a matrix D such that CD – AB = O Order of A = 2 × 2 & Order of B = 2 × 2 Order of AB = 2 × 2 Since we are doing CD – AB Order of CD = Order of AB Order of CD = 2 × 2 Order of C = 2 × 2 So, order of D = × Let D = [■8(a&b@c&d)] Now, given CD – AB = O [■8(2&5@3&8)] [■8(a&b@c&d)] − [■8(2&−1@3&4)][■8(5&2@7&4)] = O [■8(2(a)+5(c)&2(b)+5(d)@3(a)+8(c)&3(b)+8(d))] – [■8(2(5)+(−1)7&2(2)+(−1)(4)@3(5)+4(7)&3(2)+4(4))] = O [■8(2a+5c&2b+5d@3a+8c&3b+8d)] – [■8(10−7&4−4@15+28&6+16)] = O [■8(2a+5c&2b+5d@3a+8c&3b+8d)] – [■8(3&0@43&22)] = O [■8(2a+5c−3&2b+5d−0@3a+8c−43&3b+8d−22)]=[■8(0&0@0&0)] Since matrices are equal, Corresponding elements are equal Hence, 2a + 5c – 3 = 0 3a + 8c – 43 = 0 2b + 5d = 0 3b + 8d – 22 = 0 Solving (1) 2a + 5c – 3 = 0 2a + 5c = 3 2a = 3 – 5c a = (3 − 5𝑐)/2 Putting value of a in (2) 3a + 8c – 43 = 0 3((3−5𝑐)/2) + 8c – 43 = 0 (3(3 − 5𝑐) + 2(8𝑐) − 2(43))/2 = 0 9 – 15c + 16c – 86 = 0 − 15c + 16c – 86 + 9 = 0 c – 77 = 0 c = 77 From (1) 2a + 5c – 3 = 0 Putting value of c = 77 2a + 5 × 77 – 3 = 0 2a + 385 – 3 = 0 2a + 382 = 0 2a = –382 a = (−382)/2 a = −191 From (3) 2b + 5d = 0 2b = − 5d b = ((− 5)/2)d From (4) 3b + 8d – 22 = 0 Putting value of b = ((− 5)/2)d 3((− 5)/2)d + 8d − 22 = 0 (−15𝑑)/2 + 8d – 22 = 0 (−15𝑑 + 16𝑑 − 44)/2 = 0 d – 44 = 0 × 2 d – 44 = 0 d = 44 From (3) 2b + 5d = 0 Putting value of d = 44 2b + 5 × 44 = 0 2b + 220 = 0 2b = –220 b = (−220)/2 b = −110 Hence, a = −191, b = −110 , c = 77 , d = 44 Thus, Matrix D is = [■8(𝑎&𝑏@𝑐&𝑑)] = [■8(−𝟏𝟗𝟏&−𝟏𝟏𝟎@𝟕𝟕&𝟒𝟒)]

Examples

Example 1

Example 2

Example 3

Example 4

Example 5 Important

Example 6

Example 7

Example 8

Example 9

Example 10

Example 11

Example 12

Example 13

Example 14

Example 15

Example 16

Example 17

Example 18 Important

Example 19

Example 20 Important

Example 21

Example 22 Important

Example 23

Example 24 Important

Example 25 Important

Example 26 Important

Example 27 Important

Example 28 You are here

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.