Find the determinant of the matrix [x, p, 0; 1-x, -2, 3; 1, 1+x, 4] in terms of x.

Since there are two zeros in the first row, this simplifies the calculation.

$$|M| = x \begin{vmatrix} 2 & 3 \\ 1+x & 4 \end{vmatrix} - 0 \begin{vmatrix} 1-x & 3 \\ 1 & 4 \end{vmatrix} + 0 \begin{vmatrix} 1-x & 2 \\ 1 & 1+x \end{vmatrix}$$ $$|M| = x((2)(4) - (3)(1+x)) - 0 + 0$$

$$|M| = x(8 - (3 + 3x))$$ $$|M| = x(8 - 3 - 3x)$$ $$|M| = x(5 - 3x)$$ $$|M| = 5x - 3x^2$$

The determinant of the matrix is $\boxed{5x - 3x^2}$.

3) Find the non-zero positive value of $x$ which satisfies the equation:
Given the equation $\begin{vmatrix} x & 1 & 0 \\ 1 & x & 1 \\ 0 & 1 & x \end{vmatrix} = 0$.

$$= x((x)(x) - (1)(1)) - 1((1)(x) - (1)(0)) + 0$$ $$= x(x^2 - 1) - 1(x - 0)$$ $$= x^3 - x - x$$ $$= x^3 - 2x$$

$$x^3 - 2x = 0$$ $$x(x^2 - 2) = 0$$

This gives three possible solutions:

$$x = 0 \quad \text{or} \quad x^2 - 2 = 0$$ $$x^2 = 2$$ $$x = \pm \sqrt{2}$$

The non-zero positive value is $\sqrt{2}$.
The non-zero positive value of $x$ is $\boxed{\sqrt{2}}$.

4) Find the value of $K$ if:
Given the equation $\begin{vmatrix} -2 & 1 & 1 \\ 2 & 1 & K \\ 1 & 3 & -1 \end{vmatrix} = 23$.

$$= -2((1)(-1) - (K)(3)) - 1((2)(-1) - (K)(1)) + 1((2)(3) - (1)(1))$$ $$= -2(-1 - 3K) - 1(-2 - K) + 1(6 - 1)$$ $$= 2 + 6K + 2 + K + 5$$ $$= 7K + 9$$

$$7K + 9 = 23$$ $$7K = 23 - 9$$ $$7K = 14$$ $$K = \frac{14}{7}$$ $$K = 2$$

The value of $K$ is $\boxed{2}$.

5) Find the transpose of $A$:
Given the matrix $A = \begin{pmatrix} 4 & 2 & 1 \\ 3 & -1 & 2 \\ 1 & 0 & -1 \end{pmatrix}$.

The first row of $A$ becomes the first column of $A^T$, the second row of $A$ becomes the second column of $A^T$, and so on.

$$A^T = \begin{pmatrix} 4 & 3 & 1 \\ 2 & -1 & 0 \\ 1 & 2 & -1 \end{pmatrix}$$

The transpose of $A$ is $\boxed{\begin{pmatrix} 4 & 3 & 1 \\ 2 & -1 & 0 \\ 1 & 2 & -1 \end{pmatrix}}$.