How find this matrix determinant value

Question

Find the value $$ \det\left| \begin{array}{c&c&c&c&c} 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & AB^2 & AC^2 & AP^2 \\ 1 & AB^2 & 0 & BC^2 & BP^2 \\ 1 & AC^2 & BC^2 & 0 & CP^2 \\ 1 & AP^2 & BP^2 & CP^2 & 0 \end{array} \right|\,. $$ where $$AB=a,AC=c,AP=8,BC=\sqrt{a^2-ac+c^2},BP=8,CP=6$$ when I deal a math problem, and I feel very ugly by hand, can you someone can use computer to help me find this determinant? Thank you

Answer 1

I would find the determinant by computer rather than by hand, but finding the determinant manually is actually not that difficult. By permutation, the determinant is equal to \begin{align} &\left|\begin{array}{ccc|cc} 0&a^2&c^2&8^2&1\\ a^2&0&a^2-ac+c^2&8^2&1\\ c^2&a^2-ac+c^2&0&6^2&1\\ \hline 8^2&8^2&6^2&0&1\\ 1&1&1&1&0 \end{array}\right|\\ =& \left|\begin{array}{c|c} X&Y\\ \hline Y^T&Z \end{array}\right| =\det(Z)\det(X-YZ^{-1}Y^T)\\ =& -\det\left(\begin{bmatrix} 0&a^2&c^2\\ a^2&0&a^2-ac+c^2\\ c^2&a^2-ac+c^2&0\\ \end{bmatrix} - \begin{bmatrix} 8^2&1\\ 8^2&1\\ 6^2&1 \end{bmatrix} \begin{bmatrix} 0&1\\ 1&0 \end{bmatrix} \begin{bmatrix} 8^2&8^2&6^2\\ 1&1&1 \end{bmatrix} \right)\\ =& -\det\begin{bmatrix} -128&a^2-128&c^2-100\\ a^2-128&-128&a^2-ac+c^2-100\\ c^2-100&a^2-ac+c^2-100&-72\\ \end{bmatrix}. \end{align} So, you only need to evaluate the determinant of a $3\times3$ matrix.

Answer 2

If you just need a computer-generated answer, here's one from Wolfram Alpha. Looks like it's not surprising your hand calculations became a mess.

By the way, it's worth taking the time to learn simple Mathematica syntax so that you can use Wolfram Alpha, which is a great, free tool for simple quick & dirty symbolic computations.