Example 25 - Chapter 3 Class 12 Matrices (Important Question)
Last updated at April 16, 2024 by Teachoo
Chapter 3 Class 12 Matrices
Ex 3.1, 9 (MCQ) Important
Example 18 Important
Example 19
Ex 3.2, 7 (i)
Ex 3.2, 12 Important
Ex 3.2, 16 Important
Ex 3.2, 17 Important
Ex 3.2, 20 Important
Example 22 Important
Ex 3.3, 4 Important
Ex 3.3, 10 (i) Important
Ex 3.3, 12 (MCQ)
Question 15 Important Deleted for CBSE Board 2024 Exams
Question 17 Important Deleted for CBSE Board 2024 Exams
Example 25 You are here
Question 3 Important Deleted for CBSE Board 2024 Exams
Misc 6 Important
Misc 8 Important
Misc 9 (MCQ)
Chapter 3 Class 12 Matrices
Last updated at April 16, 2024 by Teachoo
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)]