Finding unknown - Matrice

Chapter 3 Class 12 Matrices
Concept wise

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

### Transcript

Example 25 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)]

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.