Question

प्राथमिक पंक्ती परिवर्तन (elementary row transformation) वापरून दिलेल्या सारणीचा `[("cos" theta, -"sin" theta, 0),("sin" theta, "cos" theta, 0),(0,0,1)]` व्यस्त (inverse) काढा.

Answer 1

|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 

AA^–1 = I असे मानू.

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

cos θ × R₁ द्वारे, आम्हाला मिळते.

`[("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)]`

R₁ + sin θ × R₂ द्वारे, आम्हाला मिळते.

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

R₂ – sin θ × R₁ द्वारे, आम्हाला मिळते.

`[(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)]`

`(1/("cos"theta)) xx "R"_2` द्वारे, आम्हाला मिळते.

`[(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)]`