\displaystyle{A}-{B}={\left(\begin{array}{cc} -{7}&{x}-{6}\\{2}{y}+{3}&-{1}\end{array}\right)} Explanation: In a matrix problem like this, all you have to do is subtract each element of \displaystyle{B} ...

For the first half: Not quite, but it's pretty close. You still need to distribute the scalars and recombine everything: \alpha_1[x_1+x_3]_B + \alpha_2[x_2+x_3]_B + \alpha_3[x_3]_B = 0 \\ [\alpha_1x_1+\alpha_1x_3]_B + [\alpha_2x_2+\alpha_2x_3]_B + [\alpha_3x_3]_B = 0 \\ [\alpha_1x_1+\alpha_1x_3 + \alpha_2x_2+\alpha_2x_3 + \alpha_3x_3]_B = 0 \\ [\alpha_1x_1 + \alpha_2x_2 + (\alpha_1 +\alpha_2 + \alpha_3)x_3]_B = 0 \\ \alpha_1[x_1]_B + \alpha_2[x_2]_B + (\alpha_1 +\alpha_2 + \alpha_3)[x_3]_B = 0 \\ ...

5 vectors, none of which are scalar multiples of any of the others, in a 4 dimensional space.

To visualize the solution, it helps to graph it in the x_1 x_2-plane for various c_1, c_2. I will map out most of the details and you can add the rest. Start with the first eigenvalue/vector ...

Let's call your matrix A = \left( \begin{array}{cc} 2 & 3 \\ 1 & 4 \end{array} \right) We want a matrix X_{2\times 2} = \begin{pmatrix} a & b\\ c&d\end{pmatrix} such that AX = XA. AX = \begin{pmatrix} 2a + 3c & 2b+3d\\ a + 4c&b+4d\end{pmatrix} ...

You are looking for the Wigner d-functions . They relate angular momentum eigenstates through rotation. As you can see in the link the definition is d^{(j)}_{m,m'}(\theta) = \langle jm|e^{-i\theta J_y}|j m' \rangle ...