2 years ago

Please explain the answer, no links or random words or I will report you.

1 answer:

Ainat [17]2 years ago

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Answer:

y=x-3 is slope-intercept form

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harina [27]

The matrix

$A=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$

has eigenvalues $\lambda$ such that

$\det(A-\lambda I)=\begin{vmatrix}\cos\theta-\lambda&-\sin\theta\\\sin\theta&\cos\theta-\lambda\end{vmatrix}=0$

$(\cos\theta-\lambda)^2+\sin^2\theta=0$

$(\cos\theta-\lambda)^2=-\sin^2\theta$

$\cos\theta-\lambda=\pm\sqrt{-\sin^2\theta}$

$\lambda=\cos\theta\pm\sqrt{-\sin^2\theta}$

$\sin^2\theta\ge0$ for all values of $\theta$ , so we need to have $\sin\theta=0$ in order for $\lambda$ to be real-valued. This happens for

$\sin\theta=0\implies\theta=n\pi$

where $n$ is any integer, and over the given interval we have $\theta=0$ and $\theta=\pi$ .

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Alika [10]

Where’s the picture for the circle C ?

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