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Example 27 - Chapter 4 Class 12 Determinants (Term 1)
Last updated at Jan. 16, 2020 by Teachoo
Examples
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Example 5 Important
Example 6 Deleted for CBSE Board 2023 Exams
Example 7 Deleted for CBSE Board 2023 Exams
Example 8 Deleted for CBSE Board 2023 Exams
Example 9 Important Deleted for CBSE Board 2023 Exams
Example 10 Important Deleted for CBSE Board 2023 Exams
Example 11 Deleted for CBSE Board 2023 Exams
Example 12 Deleted for CBSE Board 2023 Exams
Example 13 Deleted for CBSE Board 2023 Exams
Example 14 Important Deleted for CBSE Board 2023 Exams
Example 15 Important Deleted for CBSE Board 2023 Exams
Example 16 Important Deleted for CBSE Board 2023 Exams
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Example 18 Important
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Example 26 Important
Example 27 You are here
Example 28 Important
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Example 30
Example 31 Important Deleted for CBSE Board 2023 Exams
Example 32 Important Deleted for CBSE Board 2023 Exams
Example 33 Important
Example 34 Important Deleted for CBSE Board 2023 Exams
Chapter 4 Class 12 Determinants
Serial order wise
Transcript
A = 2532 = 2 × 2 – 5 × 3 = 4 – 15 = – 11 Since 𝑨 ≠ 0, System is consistent and the system of equations has a unique solution AX = B X = A-1 B Step 3 Calculating X = A-1 B Here, A-1 = 1|A| adj (A) adj A = 2532 = 2−5−32 Now, A-1 = 1|A| adj A Putting values = 1−11 2−5−32 & B = 17 Now X = A-1 B 𝑥𝑦 = −111 2−5−32 17 = −111 2 1+ −57−3 1+2(7) = −111 2−35−3+14 𝑥𝑦 = −111 −3311 = −33 × −11111 × −111 𝑥𝑦 = 3−1 Hence x = 3 & y = – 1
