The state vector $|\psi \rangle$ is defined in the matrix approach $$|\psi \rangle=\begin{pmatrix} \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}} \end{pmatrix}$$ Find the average value of the observable $F$ in this state, if it corresponds to an operator with the matrix $\begin{pmatrix} 5 &1-2i \\ 1+2i&-4 \end{pmatrix}$
My attempt: The average value of the observable represented by the operator $F$ in the state $|\psi \rangle$ can be found using the formula $$\langle F \rangle = \langle\psi|F|\psi\rangle$$
Let's write everything down in component form.
$$|\psi \rangle=\begin{pmatrix} \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}} \end{pmatrix}, \quad F=\begin{pmatrix} 5 &1-2i \\ 1+2i&-4 \end{pmatrix}$$
Now find the average value of the operator $F$ in the state $|psi \rangle$
$$F|\psi\rangle = \begin{pmatrix} 5 &1-2i \\ 1+2i&-4 \end{pmatrix} \cdot \begin{pmatrix} \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}} \end{pmatrix} = \begin{pmatrix} 5\frac{1}{\sqrt{2}}+(1-2i)\left(-\frac{1}{\sqrt{2}}\right)\\ (1+2i)\frac{1}{\sqrt{2}}-4\frac{1}{\sqrt{2}} \end{pmatrix} = \begin{pmatrix} (5-1+2i)\frac{1}{\sqrt{2}}\\ (1+2i-4)\frac{1}{\sqrt{2}} \end{pmatrix} = \begin{pmatrix} (4+2i)\frac{1}{\sqrt{2}}\\ (-3+2i)\frac{1}{\sqrt{2}} \end{pmatrix}$$
$$\langle\psi|F|\psi\rangle = \begin{pmatrix} \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \end{pmatrix} \cdot \begin{pmatrix} (4+2i)\frac{1}{\sqrt{2}}\\ (-3+2i)\frac{1}{\sqrt{2}} \end{pmatrix} = (4+2i)\frac{1}{2} - (3-2i)\frac{1}{2} = \frac{1+4i}{2} = \frac{1}{2}+2i$$
The average value of the observed $F$ in this state $|psi \rangle$ I got $\frac{1}{2}+2i$
Was my reasoning and task correct? I would be grateful if you could check or correct the errors
It should be $$F|\psi\rangle = \begin{pmatrix} 5 &1-2i \\ 1+2i&-4 \end{pmatrix} \cdot \begin{pmatrix} \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}} \end{pmatrix} = \begin{pmatrix} 5\frac{1}{\sqrt{2}}+(1-2i)\left(-\frac{1}{\sqrt{2}}\right)\\ (1+2i)\frac{1}{\sqrt{2}}+4\frac{1}{\sqrt{2}} \end{pmatrix} = \begin{pmatrix} (5-1+2i)\frac{1}{\sqrt{2}}\\ (1+2i+4)\frac{1}{\sqrt{2}} \end{pmatrix} = \begin{pmatrix} (4+2i)\frac{1}{\sqrt{2}}\\ (5+2i)\frac{1}{\sqrt{2}} \end{pmatrix}$$
$$\langle\psi|F|\psi\rangle = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix} \cdot \begin{pmatrix} (4+2i)\frac{1}{\sqrt{2}}\\ (5+2i)\frac{1}{\sqrt{2}} \end{pmatrix} = (4+2i)\frac{1}{2} - (5+2i)\frac{1}{2} = -\frac{1}{2}$$