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Example 2
Example 3 Important
Example 4
Example 5 Important
Example 6
Example 7
Example 8
Example 9
Example 10
Example 11 Important
Example 12
Example 13 Important
Example 14
Example 15
Example 16 Important
Example 17
Example 18 Important
Example 19
Example 20 Important
Example 21
Example 22 Important
Example 23 Deleted for CBSE Board 2022 Exams
Example 24 Important Deleted for CBSE Board 2022 Exams
Example 25 Important Deleted for CBSE Board 2022 Exams
Example 26 Important
Example 27 Important
Example 28 You are here
Last updated at Jan. 17, 2020 by Teachoo
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)]