Question

can u please help me solve this equation. i have the answer in my book but can u help me solve it
i have a 3x3 matrix
1=p(1,1) + p(2,1)*e^ix + p(3,1)*e^-ix
cos x = p(2,1) + p(2,2)*e^ix +p(3,2)*e^-ix
sin x= p(1,3) +p(2,3)*e^ix + p(3,3)*e^-ix
where p is a 3 by 3 matrix

Answer 1

Answer on question #34945 – Math – Linear Algebra

can u please help me solve this equation. i have the answer in my book but can u help me solve it

i have a 3x3 matrix

$$1 = p(1,1) + p(2,1) * e^{\wedge}ix + p(3,1) * e^{\wedge} - ix$$ $$\cos x = p(2,1) + p(2,2) * e^{\wedge}ix + p(3,2) * e^{\wedge} - ix$$ $$\sin x = p(1,3) + p(2,3) * e^{\wedge}ix + p(3,3) * e^{\wedge} - ix$$

where p is a 3 by 3 matrix

Answer:

As we know

$$e^{ix} = \cos x + i \sin x$$ $$e^{-ix} = \cos x - i \sin x$$

So we get

$$\begin{array}{l} 1 = p(1,1) + p(2,1) * e^{ix} + p(3,1) * e^{-ix} = \\ = p(1,1) + (p(2,1) + p(3,1)) \cos x + i(p(2,1) - p(3,1)) \sin x \\ \end{array}$$

It holds when $p(1,1) = 1, p(2,1) = p(3,1) = 0$.

$$\begin{array}{l} \cos x = p(2,1) + p(2,2) * e^{ix} + p(3,2) * e^{-ix} = \\ = p(2,1) + (p(2,2) + p(3,2)) \cos x + i(p(2,2) - p(3,2)) \sin x \\ \end{array}$$

It holds when $p(2,1) = 0, p(2,2) = p(3,2) = 0.5$.

$$\begin{array}{l} \cos x = p(1,3) + p(2,3) * e^{ix} + p(3,3) * e^{-ix} = \\ = p(1,3) + (p(2,3) + p(3,3)) \cos x + i(p(2,3) - p(3,3)) \sin x \\ \end{array}$$

It holds when $p(2,1) = 0, p(2,2) = \frac{1}{2i}, p(3,2) = -1/2i$.

Therefore we get

$$p = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0.5 & 0.5 \\ 0 & \frac{1}{2i} & -\frac{1}{2i} \end{pmatrix}.$$