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Question
A = `[(cos theta, - sin theta),(-sin theta, -cos theta)]` then find A−1
Solution
|A| = `|(cos theta, - sin theta),(-sin theta, - cos theta)|`
= – cos2θ – sin2θ
= –1 ≠ 0
∴ A–1 exists.
A11 = (–1)1+1 M11 = M11 = – cos θ
A12 = (–1)1+2 M12 = – M12 = sin θ
A21 = (–1)2+1 M21 = – M21 = sin θ
A22 = (–1)2+2 M22 = M22 = cos θ
∴ adj (A) = `[(- cos theta, sin theta),(sin theta, cos theta)]^"T"`
= `[(- cos theta, sin theta),(sin theta, cos theta)]`
A−1 = `1/|"A"|` adj (A)
= `[(cos theta, -sin theta),(-sin theta, -cos theta)]`
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