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


Transcript

Example 5 Find the values of a, b, c, and d from the following equation: [■8(2a+b&a−2b@5c−d&4c+3d)] = [■8(4&−3@11&24)] Since matrices are equal. Their corresponding elements are equal 2a + b = 4 5c – d = 11 a – 2b = – 3 4c + 3d = 24 Solving these equations Solving (1) 2a + b = 4 b = 4 – 2a Putting value of b in (3) a – 2b = – 3 Putting b = 4 – 2a a – 2(4 – 2a) = -3 a – 8 + 4a = -3 a + 4a = -3 + 8 5a = 5 a = 5/5 a = 1 From (1) 2a + b = 4 Putting a = 1 2(1) + b = 4 2 + b = 4 b = 4 – 2 b = 2 From (2) 5c – d = 11 5c = 11 + d 5c – 11 = d d = 5c – 11 Solving (4) 4c + 3d = 24 Putting d = 5c – 11 4c + 3(5c – 11) = 24 4c + 15c – 33 = 24 19c – 33 = 24 19c = 24 + 33 19c = 57 c = 57/19 c = 3 Now, d = 5c – 11 Putting c = 3 d = 5(3) – 11 d = 15 – 11 d = 4 Hence, a = 1, b = 2, c = 3 and d = 4

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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.