Combining lenses

[E92M3]

1. The problem statement, all variables and given/known data
2 thin lenses separated by 10cm. First has a focal length of 20cm, the second has a focal length of 30cm.


2. Relevant equations
Lens transfer matrix for each lens:
\begin{bmatrix}
1-\frac{D_2d_{21}}{n_{t1}} &-D_1-D_2+\frac{D_1D_2d_{21}}{n_{t1}} \\
\frac{d_{21}}{n_{t1}} & 1-\frac{D_2d_{21}}{n_{t1}}
\end{bmatrix}


3. The attempt at a solution
Since the thickness of the 2 lens is assumed to be zero we have:
A_1=\begin{bmatrix}
1 & \frac{-1}{f_1}\\
0 & 1
\end{bmatrix}
For the first lens and
A_2=\begin{bmatrix}
1 & \frac{-1}{f_2}\\
0 & 1
\end{bmatrix}
for the second. Therefore we have:
A=A_2TA_1=A_2=\begin{bmatrix}
1 & \frac{-1}{f_2}\\
0 & 1
\end{bmatrix}\begin{bmatrix}
1 & 0\\
d & 1
\end{bmatrix}\begin{bmatrix}
1 & \frac{-1}{f_1}\\
0 & 1
\end{bmatrix}=\begin{bmatrix}
1-\frac{d}{f_2} & \frac{-1}{f_1}-\frac{1}{f_2}+\frac{d}{f_1f_2}\\
d & 1-\frac{d}{f_1}
\end{bmatrix}=\begin{bmatrix}
\frac{2}{3} & \frac{-1}{15}cm^{-1}\\
10cm & \frac{1}{2}
\end{bmatrix}=\begin{bmatrix}
A_{11} & A_{12}\\
A_{21} & A_{22}
\end{bmatrix}

Since the system is assumed to be in air, the focal length is the same on both sides of the system. SO we have:

f=\frac{1}{-A_{12}}=15cm
\overline{H_1O_1}=(1-A_{11})f=5cm
\overline{H_2O_2}=(1-A_{22})f=7.5cm

Where H denotes the principle planes and O denotes the lenses.
So here's the problem, recall that the lenses are separated by only 10cm, this forces principle plane #2 to be in front (left of) principle plane #1. How can this be true? What did i do wrong?