If A=begin{bmatrix}-2 & 1&-4 -2 & 4&-1 1 &-1 &-4 end{bmatrix} text{ and } B=begin{bmatrix}-2 & 4&2 -4 & -1&1 4 &1 &1 end{bmatrix} then AB=? BA=? True or false : AB=BA for any two square matrices A and B of the same size.

Let B be a 4x4 matrix to which we apply the following operations:
1. double column 1,
2. halve row 3,
3. add row 3 to row 1,
4. interchange columns 1 and 4,
5. subtract row 2 from each of the other rows,
6. replace column 4 by column 3,
7. delete column 1 (column dimension is reduced by 1).
(a) Write the result as a product of eight matrices.
(b) Write it again as a product of ABC (same B) of three matrices.

Find a basis for the space of $2\times 2$ diagonal matrices.
$\text{Basis }=\{\left[\begin{array}{cc}& \\ & \end{array}\right],\left[\begin{array}{cc}& \\ & \end{array}\right]\}$

Simplification of ${\displaystyle {\cos }^4\left(x\right)+{\sin }^4\left(x\right)}$
${\displaystyle {\left(\sin x\right)}^4+{\left(\cos x\right)}^4=\frac{{(1-\cos 2x)}^2}{4}+\frac{{(1+\cos 2x)}^2}{4}}$
${\displaystyle =\frac{1-2\cos 2x+{\left(\cos 2x\right)}^2+1+2\cos 2x+{\left(\cos 2x\right)}^2}{4}}$
${\displaystyle =\frac{1+{\left(\cos 2x\right)}^2}{2}}$
its correct?

Let ${\displaystyle x_1\text{ and }{x}_2}$ be the roots of the equation : ${\displaystyle x^2-2\sqrt{2}x+1=0}$
Calculate ${\displaystyle \arctan \left(x_1\right)\cdot \arctan \left(x_2\right)}$

Prove:
${\displaystyle \cos \left(3\theta \right)=4{\cos }^3\left(\theta \right)-3\cos \left(\theta \right)}$

Let u have the cosine series representation
$u=\sum _{k_1=0}^{\mathrm{\infty }}\sum _{k_2=0}^{\mathrm{\infty }}{a}_{\underset{\_}{k}}\cos \left(\frac{2\pi k_{1}x}{L_1}\right)\cos \left(\frac{2\pi k_{2}y}{L_2}\right)$
What are the coefficients $A_{\underset{\_}{k}}$ in
$u^2=\sum _{k_1=0}^{\mathrm{\infty }}\sum _{k_2=0}^{\mathrm{\infty }}{A}_{\underset{\_}{k}}\cos \left(\frac{2\pi k_{1}x}{L_1}\right)\cos \left(\frac{2\pi k_{2}y}{L_2}\right)$
in terms of the $a_{\underset{\_}{k}}$?

Explain how to add complex numbers.
${\displaystyle 5+2i}$
${\displaystyle 8+7i}$