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Exercise 1.8 · Q15

Q.If A=(cos⁡θsin⁡θ−sin⁡θcos⁡θ)A=\begin{pmatrix}\cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{pmatrix} and A(adj⁡A)=(k00k)A(\operatorname{adj}A)=\begin{pmatrix}k & 0\\ 0 & k\end{pmatrix}, then k=k=

(1) 0
(2) sin⁡θ\sin\theta
(3) cos⁡θ\cos\theta
(4) 1
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By the fundamental adjoint identity A(adj⁡A)=∣A∣IA(\operatorname{adj}A)=|A|I, the scalar kk multiplying I2I_2 on the right side must equal ∣A∣|A|; we just need to compute the determinant.

Step 1. Write down AA. A=(cos⁡θsin⁡θ−sin⁡θcos⁡θ)A=\begin{pmatrix}\cos\theta&\sin\theta\\ -\sin\theta&\cos\theta\end{pmatrix}.

Step 2. Compute ∣A∣|A|. ∣A∣=cos⁡θ(cos⁡θ)−sin⁡θ(−sin⁡θ)=cos⁡2θ+sin⁡2θ=1|A|=\cos\theta(\cos\theta)-\sin\theta(-\sin\theta)=\cos^2\theta+\sin^2\theta=1. …

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