The given problem is:

Matrices $A$ and $B$ are:

$A = \begin{bmatrix} x & 1 \\ y & 2 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & x \\ -y & -2 \end{bmatrix}$

We are tasked to find:

1. The order of the null matrix (i.e., $AB$ must be the zero matrix).
2. The possible values of $x$ and $y$.

Step 1: Matrix Multiplication of $A$ and $B$

To calculate $AB$, we perform the multiplication of these two matrices:

$AB = \begin{bmatrix} x & 1 \\ y & 2 \end{bmatrix} \begin{bmatrix} 1 & x \\ -y & -2 \end{bmatrix}$

Multiplying the matrices:

$AB = \begin{bmatrix} (x \cdot 1 + 1 \cdot (-y)) & (x \cdot x + 1 \cdot (-2)) \\ (y \cdot 1 + 2 \cdot (-y)) & (y \cdot x + 2 \cdot (-2)) \end{bmatrix}$

Simplifying the elements of $AB$:

$AB = \begin{bmatrix} (x - y) & (x^2 - 2) \\ (y - 2y) & (yx - 4) \end{bmatrix}$

$AB = \begin{bmatrix} x - y & x^2 - 2 \\ -y & yx - 4 \end{bmatrix}$

Step 2: Null Matrix Condition

For $AB$ to be a null matrix, all elements of $AB$ must be zero:

$\begin{bmatrix} x - y & x^2 - 2 \\ -y & yx - 4 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

This gives us the following system of equations:

1. $x - y = 0$
2. $x^2 - 2 = 0$
3. $-y = 0$
4. $yx - 4 = 0$

Step 3: Solve the System of Equations

From equation (3), we have: $y = 0$

Substitute $y = 0$ into equation (1): $x - 0 = 0 \implies x = 0$

Now, check with the remaining equations:

• Equation (2): $x^2 - 2 = 0$ gives $x^2 = 2$, so $x = \pm \sqrt{2}$.
• Equation (4): If $y = 0$, this becomes $0 \cdot x - 4 = 0$, which is inconsistent.

Thus, there seems to be a contradiction unless $y \neq 0$.

Conclusion

We found that $x = 0$ and $y = 0$ leads to contradictions. Therefore, valid answers require ( x \approx \pm sqrt