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Question
Find the inverse of A = `[(sec theta, tan theta, 0),(tan theta, sec theta, 0),(0, 0, 1)]`
Solution
|A| = `[(sec theta, tan theta, 0),(tan theta, sec theta, 0),(0, 0, 1)]`
= sec θ(sec θ – 0) – tan θ(tan θ – 0) + 0
= sec2θ – tan2θ
= 1 ≠ 0
∴ A−1 exists.
Consider AA−1 = I
∴ `[(sec theta, tan theta, 0),(tan theta, sec theta, 0),(0, 0, 1)]` A−1 = `[(1, 0, 0),(0, 1, 0),(0, 0, 1)]`
Applying R1 → (sec θ)R1 – (tan θ)R2, we get
`[(sec^2theta - tan^2theta, sectheta tantheta - sectheta tantheta, 0),(tantheta, sectheta, 0),(0, 0, 1)]` A−1 = `[(sectheta, -tantheta, 0),(0, 1, 0),(0, 0, 1)]`
∴ `[(1, 0, 0),(tantheta, sectheta, 0),(0, 0, 1)]` A−1 = `[(sectheta, -tantheta, 0),(0, 1, 0),(0, 0, 1)]`
Applying R2 → R2 – tanθ R1, we get
`[(1, 0, 0),(0, sectheta, 0),(0, 0, 1)]` A−1 = `[(sectheta, -tantheta, 0),(-sectantheta, 1 + tan^2theta, 0),(0, 0, 1)]`
∴ `[(1, 0, 0),(0, sectheta, 0),(0, 0, 1)]` A−1 = `[(sectheta, -tantheta, 0),(-sectantheta, sec^2theta, 0),(0, 0, 1)]`
Applying R2 → `(1/sectheta)` R2, we get
`[(1, 0, 0),(0, 1, 0),(0, 0, 1)]` A−1 = `[(sectheta, -tantheta, 0),(-tantheta, sectheta, 0),(0, 0, 1)]`
∴ A−1 = `[(sectheta, -tantheta, 0),(-tantheta, sectheta, 0),(0, 0, 1)]`
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