Non-singularity of a certain block matrix

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We are given a matrix of the form

$$\left[\begin{array}{cccccccccccccc} 0&0&0&0&0&0&0&0&0&q_{1}&0&1&0&0\\ 0&2p_{2}q_{2}&0&0&0&0&0&0&0&q_{2}&0&0&1&0\\ 0&0&2p_{3}q_{3}&0&0&0&0&0&0&q_{3}&0&0&0&1\\ 0&0&0&2p_{1}q_{1}&0&0&0&0&0&0&q_{1}&1&0&0\\ 0&0&0&0&0&0&0&0&0&0&q_{2}&0&1&0\\ 0&0&0&0&0&2p_{3}q_{3}&0&0&0&0&q_{3}&0&0&1\\ 0&0&0&0&0&0&2p_{1}q_{1}&0&0&0&0&1&0&0\\ 0&0&0&0&0&0&0&2p_{2}q_{2}&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&1\\ q_{1}&q_{2}&q_{3}&0&0&0&0&0&0&0&0&0&0&0\\ 0&0&0&q_{1}&q_{2}&q_{3}&0&0&0&0&0&0&0&0\\ 1&0&0&1&0&0&1&0&0&0&0&0&0&0\\ 0&1&0&0&1&0&0&1&0&0&0&0&0&0\\ 0&0&1&0&0&1&0&0&1&0&0&0&0&0 \end{array}\right]$$

This is an example for $n = 3,$ i.e. it is an $n^2+2n-1$-dimensional matrix, with all its non-negative entries strictly positive. We also have that $p_i+q_i = 1,$ and $0 < p_i,q_i.$ I entered this matrix to maxima CAS and it successfully finds its determinant. Unfortunately, I could not generalize from there, neither could I get any hypothesis. How to prove this matrix is always non-singular?

EDIT: all the numbers are reals, and the symbols are positive reals. As to symbolic computations, I have the following maxima-targeted format:

[0,0,0,0,0,0,0,0,0,q1,0,1,0,0],
[0,2*p2*q2,0,0,0,0,0,0,0,q2,0,0,1,0],
[0,0,2*p3*q3,0,0,0,0,0,0,q3,0,0,0,1],
[0,0,0,2*p1*q1,0,0,0,0,0,0,q1,1,0,0],
[0,0,0,0,0,0,0,0,0,0,q2,0,1,0],
[0,0,0,0,0,2*p3*q3,0,0,0,0,q3,0,0,1],
[0,0,0,0,0,0,2*p1*q1,0,0,0,0,1,0,0],
[0,0,0,0,0,0,0,2*p2*q2,0,0,0,0,1,0],
[0,0,0,0,0,0,0,0,0,0,0,0,0,1],
[q1,q2,q3,0,0,0,0,0,0,0,0,0,0,0],
[0,0,0,q1,q2,q3,0,0,0,0,0,0,0,0],
[1,0,0,1,0,0,1,0,0,0,0,0,0,0],
[0,1,0,0,1,0,0,1,0,0,0,0,0,0],
[0,0,1,0,0,1,0,0,1,0,0,0,0,0]

EDIT: For a concrete example of $q_1 = \ldots = q_n = 1/2,$ one would have $$ \left[ \begin{array}{cccccccccccccc} 0&0&0&0&0&0&0&0&0&{{1}\over{2}}&0&1&0&0\\ 0&{{1}\over{2}}&0&0&0&0&0&0&0&{{1}\over{2}}&0&0&1&0\\\ 0&0&{{1}\over{2}}&0&0&0&0&0&0&{{1}\over{2}}&0&0&0&1\\ 0&0&0&{{1}\over{2}}&0&0&0&0&0&0&{{1}\over{2}}&1&0&0\\ 0&0&0&0&0&0&0&0&0&0&{{1}\over{2}}&0&1&0\\ 0&0&0&0&0&{{1}\over{2}}&0&0&0&0&{{1}\over{2}}&0&0&1\\ 0&0&0&0&0&0&{{1}\over{2}}&0&0&0&0&1&0&0\\ 0&0&0&0&0&0&0&{{1}\over{2}}&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&1\\ {{1}\over{2}}&{{1}\over{2}}&{{1}\over{2}}&0&0&0&0&0&0&0&0&0&0&0\\ 0&0&0&{{1}\over{2}}&{{1}\over{2}}&{{1}\over{2}}&0&0&0&0&0&0&0&0\\ 1&0&0&1&0&0&1&0&0&0&0&0&0&0\\ 0&1&0&0&1&0&0&1&0&0&0&0&0&0\\ 0&0&1&0&0&1&0&0&1&0&0&0&0&0\\ \end{array} \right] $$

EDIT: The Wikipedia page says the matrix is positive definite if and only if all its eigenvalues are positive. I ran maxima's eigenvalues-function for the above matrix, and I got the following: $$\left[ \left[ \left({{-1}\over{2}}-{{\sqrt{3}\,i}\over{2}}\right)\, \left({{\sqrt{469}\,i}\over{4\,3^{{{3}\over{2}}}}}+{{1}\over{216}} \right)^{{{1}\over{3}}}+{{37\,\left({{\sqrt{3}\,i}\over{2}}+{{-1 }\over{2}}\right)}\over{36\,\left({{\sqrt{469}\,i}\over{4\,3^{{{3 }\over{2}}}}}+{{1}\over{216}}\right)^{{{1}\over{3}}}}}+{{1}\over{6}} , \left({{\sqrt{3}\,i}\over{2}}+{{-1}\over{2}}\right)\,\left({{ \sqrt{469}\,i}\over{4\,3^{{{3}\over{2}}}}}+{{1}\over{216}}\right)^{ {{1}\over{3}}}+{{37\,\left({{-1}\over{2}}-{{\sqrt{3}\,i}\over{2}} \right)}\over{36\,\left({{\sqrt{469}\,i}\over{4\,3^{{{3}\over{2}}}}} +{{1}\over{216}}\right)^{{{1}\over{3}}}}}+{{1}\over{6}} , \left({{ \sqrt{469}\,i}\over{4\,3^{{{3}\over{2}}}}}+{{1}\over{216}}\right)^{ {{1}\over{3}}}+{{37}\over{36\,\left({{\sqrt{469}\,i}\over{4\,3^{{{3 }\over{2}}}}}+{{1}\over{216}}\right)^{{{1}\over{3}}}}}+{{1}\over{6}} , {{1}\over{2}} \right] , \left[ 1 , 1 , 1 , 1 \right] \right] $$

Conceptually (if that helps and no-so-obvious from the above representation), this matrix can be seen as a block matrix $$B = \left[\begin{array}{cc}P&V^{T}\\V&\mathbf{0}\end{array}\right]$$