how to design LPH and HPF from APF
Since you don't say what response you're actually getting from Matlab, this is all guesswork. 

So let's do the basics: A little formula juggling.

$$\begin{align}A_0(z)&amp;=1\\
A_1(z)&amp;=\frac{-a+z^{-1}}{1-az^{-1}}\\[4em]
H_{LP}(z)&amp;=A_0(z)+A_1(z)\\
   &amp;= 1 + \frac{-a+z^{-1}}{1-az^{-1}}\\
   &amp;= \frac{{1-az^{-1}}-a+z^{-1}}{1-az^{-1}}\\
   &amp;= \frac{(1-a)\quad +\quad (1-a)z^{-1}}{1-az^{-1}}\\[4em]
H_{HP}(z)&amp;=A_0(z)-A_1(z)\\
   &amp;= 1 - \frac{-a+z^{-1}}{1-az^{-1}}\\
   &amp;= \frac{{1-az^{-1}}+a-z^{-1}}{1-az^{-1}}\\
   &amp;= \frac{(1+a)\quad +\quad (-1-a)z^{-1}}{1-az^{-1}}
\end{align}$$

So, taken from these formulas, the filter coefficients for $H_{LP}$ are $b=[1-a,1-a]$, $a=[1,-a]$, and  for $H_{HP}$ it's $b=[1+a,-1-a]$, $a=[1,-a]$ for the canonical form of recursive filters

$$H= \frac{\sum\limits_{i=0}^N b_i z^{-i}}{\sum\limits_{i=0}^M a_i z^{-i}}\text.$$

Plugging that into scipy's scipy.signal.freqz for different $a$ yielded (abscissa: normalized frequency $\in[0,\pi[$, ordinate: magnitude):

$H_{LP,a}$ for $a\in\{0,0.1,\dots,0.9,1\}$:



$H_{HP,a}$ for $ a\in\{0,0.1,\dots,0.9,1\}$:



which both very nicely match the terms low and high pass filters, respectively.