Example 27 - Chapter 4 Class 12 Determinants (Term 1)
Last updated at Jan. 16, 2020 by Teachoo
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
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Example 27 You are here
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Chapter 4 Class 12 Determinants (Term 1)
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About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.