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Question
Find the inverse of A = `[("cos" theta, -"sin" theta, 0),("sin" theta, "cos" theta, 0),(0,0,1)]` by elementary column transformations.
Solution
|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 A-1A = I
∴ `"A"^-1 [("cos"theta, -"sin"theta,0),("sin"theta,"cos"theta,0),(0,0,1)] = [(1,0,0),(0,1,0),(0,0,1)]`
By (cos θ) × C1, we get,
`"A"^-1 = [("cos"^2theta,-"sin"theta,0),("sin"theta"cos"theta,"cos"theta,0),(0,0,1)] = [("cos"theta,0,0),(0,1,0),(0,0,1)]`
By C1 - sin θ × C2, we get,
`"A"^-1 [(1,-"sin"theta,0),(0,"cos"theta,0),(0,0,1)] = [("cos"theta,0,0),(-"sin"theta,1,0),(0,0,1)]`
By C2 + sin θ × C1, we get,
`"A"^-1 [(1,0,0),(0,"cos"theta,0),(0,0,1)] = [("cos"theta,"sin"theta"cos"theta,0),(-"sin"theta,"cos"^2theta,0),(0,0,1)]`
By `(1/("cos"theta))"C"_2`, we get,
`"A"^-1[(1,0,0),(0,1,0),(0,0,1)] = [("cos"theta,"sin"theta,0),(-"sin"theta,"cos"theta,0),(0,0,1)]`
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