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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 row 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 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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