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If $$ A=\begin{bmatrix} \sin\theta & \csc\theta & 1 \\ \sec\theta & \cos\theta & 1 \\ \tan\theta & \cot\theta & 1 \\ \end{bmatrix} $$ then prove that there does not exist a real value of $\theta$ for which characteristics roots of $A$ are $-1,1,3$

i tried to solve as follows, sum of eigen value $$=\sin\theta +\cos\theta + 1=-1+1+3=3$$ $$\sin\theta +\cos\theta = 2$$ but what to do next.

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    You’re almost there. Is there any value of $\theta$ for which both the sine and cosine are equal to one? – 2017-02-18
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    Do you know how to find the amplitude of $\sin\theta+\cos\theta$? – 2017-02-18
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    i don't know how to find amplitude of $sin\theta + cos\theta$ – 2017-02-18
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    Hint: $\cos\frac{\pi}{4} = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$ – 2017-02-18

2 Answers 2

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\begin{eqnarray} a\sin\theta+b\cos\theta&=&\sqrt{a^2+b^2}\left[\frac{a}{\sqrt{a^2+b^2}}\sin\theta+\frac{b}{\sqrt{a^2+b^2}}\cos\theta\right]\\ &=&\sqrt{a^2+b^2}\left[\sin\phi\sin\theta+\cos\phi\cos\theta\right]\\ &=&\sqrt{a^2+b^2}\cos(\theta-\phi) \end{eqnarray}

Where $$\phi=\arcsin\frac{a}{\sqrt{a^2+b^2}}$$

Amplitude of sinusoidal

So the result is a sinusoidal with a phase shift and an amplitude of $\sqrt{a^2+b^2}$

When $a=b=1$ what will be the amplitude?

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    amplitude will be $\sqrt2$. but how can i prove that $\theta$ has no real value. – 2017-02-18
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    That is correct. So the sum $\sin\theta+\cos\theta$ can never be larger than $\sqrt{2}$ for real values of $\theta$, correct? – 2017-02-18
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    Therefore $\cos\theta+\sin\theta=2$ has no solution for real values of $\theta$ since $2>\sqrt{2}$. – 2017-02-18
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$$\cos x+\sin x=2\implies(\cos x+\sin x)^2=4\implies\cos^2+2\cos x\sin x+\sin^2x=4\\\implies\sin2x=3\ (!)$$