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Example 33 - Use product to solve x-y+2z=1 2y-3z=1 3x-2y+4z=2  - Find solution of equations- Equations given

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  1. Chapter 4 Class 12 Determinants
  2. Serial order wise
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Example 33 Use product 1﷮−1﷮2﷮0﷮2﷮−3﷮3﷮−2﷮4﷯﷯ −2﷮0﷮1﷮9﷮2﷮−3﷮6﷮1﷮−2﷯﷯ to solve the system of equations x – y + 2z = 1 2y – 3z = 1 3x – 2y + 4z = 2 Consider the product 1﷮−1﷮2﷮0﷮2﷮−3﷮3﷮−2﷮4﷯﷯ −2﷮0﷮1﷮9﷮2﷮−3﷮6﷮1﷮−2﷯﷯ = 1 −2﷯+ −1﷯ 9﷯+2(6)﷮1 0﷯+ −1﷯ 2﷯+2(1)﷮1 1﷯+ −1﷯ −3﷯+2(−2)﷮0 −2﷯+2 9﷯+(−3)(6)﷮0 0﷯+2 2﷯+(−3)(1)﷮0 1﷯+0 −3﷯+(−3)(−2)﷮3 −2﷯+ −2﷯ 9﷯+(6)﷮3 0﷯+ −2﷯ 2﷯+4(1)﷮3 1﷯+ −2﷯ −3﷯+4(−2)﷯﷯ = −2−9+12﷮0−2+2﷮1+3−4﷮0+18−18﷮0+4−3﷮0−6+6﷮−6−18+24﷮0−4+4﷮3+6−8﷯﷯ = 1﷮0﷮0﷯ 0﷮1﷮0﷯ 0﷮0﷮1﷯﷯ We know that AA-1 = I So −2﷮0﷮1﷮0﷮2﷮−3﷮6﷮1﷮−2﷯﷯ is inverse of 1﷮−1﷮2﷮0﷮2﷮−3﷮3﷮−2﷮4﷯﷯ i.e. 𝟏﷮−𝟏﷮𝟐﷮𝟎﷮𝟐﷮−𝟑﷮𝟑﷮−𝟐﷮𝟒﷯﷯﷮−𝟏﷯= −𝟐﷮𝟎﷮𝟏﷮𝟗﷮𝟐﷮−𝟑﷮𝟔﷮𝟏﷮−𝟐﷯﷯ Given equations are x – y + 2z = 1 2y – 3z = 1 3x – 2y + 4z = 2 Step 1 Write the equation as AX = B 1﷮−1﷮2﷮0﷮2﷮−3﷮3﷮−2﷮4﷯﷯ 𝑥﷮𝑦﷮𝑧﷯﷯ = 1﷮1﷮2﷯﷯ Here A 1﷮−1﷮2﷮0﷮2﷮−3﷮3﷮−2﷮4﷯﷯, X = 𝑥﷮𝑦﷮𝑧﷯﷯ & B = 1﷮1﷮2﷯﷯ Now, AX = B X = A-1 B Here A = 1﷮−1﷮2﷮0﷮2﷮−3﷮3﷮−2﷮4﷯﷯ So, A-1 = 1﷮−1﷮2﷮0﷮2﷮−3﷮3﷮−2﷮4﷯﷯﷮−1﷯= −2﷮0﷮1﷮9﷮2﷮−3﷮6﷮1﷮−2﷯﷯ X = A-1 B Putting Value 𝑥﷮𝑦﷮𝑧﷯﷯ = −2﷮0﷮1﷮9﷮2﷮−3﷮6﷮1﷮−2﷯﷯ 1﷮1﷮2﷯﷯ = −2﷮0﷮1﷮9﷮2﷮−3﷮6﷮1﷮−2﷯﷯ 1﷮1﷮2﷯﷯ 𝑥﷮𝑦﷮𝑧﷯﷯ = −2 1﷯+0 1﷯+1(2)﷮9 1﷯+2 1﷯−3(2)﷮6 1﷯+1 1﷯−2(2)﷯﷯ = −2+0+2﷮9+2−6﷮6+1−4﷯﷯ 𝑥﷮𝑦﷮𝑧﷯﷯ = 0﷮5﷮3﷯﷯ Hence x = 0 , y = 5 & z = 3

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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