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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

Find the inverse of A = cossinsincos[cosθ-sinθ0sinθcosθ0001] by elementary row transformations. - Mathematics and Statistics

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प्रश्न

Find the inverse of A = `[("cos" theta, -"sin" theta, 0),("sin" theta, "cos" theta, 0),(0,0,1)]` by elementary row transformations.

बेरीज
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उत्तर

|A| = `|("cos" theta, -"sin" theta, 0),("sin" theta, "cos" theta, 0),(0,0,1)|`

= cos θ(cos θ – 0) + sin θ(sin θ – 0) + 0

= cos2θ + sin2θ

= 1 ≠ 0

∴ A-1 exists.

Consider AA–1 = I

∴ `[("cos" theta, -"sin" theta, 0),("sin" theta, "cos" theta, 0),(0,0,1)] "A"^-1 = [(1,0,0),(0,1,0),(0,0,1)]`

By cos θ × R1, we get,

`[("cos"^2theta, -"sin"theta "cos"theta,0),("sin"theta, "cos"theta, 0),(0,0,1)] "A"^-1 = [("cos"theta,0,0),(0,1,0),(0,0,1)]`

By R1 + sin θ × R2, we get,

`[(1,0,0),("sin"theta,"cos"theta,0),(0,0,1)] "A"^-1 = [("cos"theta,"sin"theta,0),(0,1,0),(0,0,1)]`

By R2 – sin θ × R1, we get,

`[(1,0,0),(0,"cos"theta,0),(0,0,1)] "A"^-1 = [("cos"theta,"sin"theta, 0),(-"sin"theta"cos"theta,"cos"^2theta,0),(0,0,1)]`

By `(1/("cos"theta)) xx "R"_2`, we get,

`[(1,0,0),(0,1,0),(0,0,1)] "A"^-1 = [("cos"theta,"sin"theta,0),(-"sin"theta,"cos"theta,0),(0,0,1)]`

`∴ "A"^-1 = [("cos"theta,"sin"theta,0),(-"sin"theta,"cos"theta,0),(0,0,1)]`

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Elementry Transformations
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