Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Example 5 - Chapter 3 Class 12 Matrices
Last updated at June 8, 2023 by Teachoo
Examples
Example 1
Example 2
Example 3 Important
Example 4
Example 5 Important You are here
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 Important
Example 24 Important
Example 25
Question 1 Deleted for CBSE Board 2024 Exams
Question 2 Important Deleted for CBSE Board 2024 Exams
Question 3 Important Deleted for CBSE Board 2024 Exams
Chapter 3 Class 12 Matrices
Serial order wise
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
