Example 5 - Chapter 3 Class 12 Matrices (Term 1)
Last updated at Jan. 17, 2020 by Teachoo
Examples
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Example 5 Important You are here
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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
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Chapter 3 Class 12 Matrices (Term 1)
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 Solving (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
