Answer
12
Work Step by Step
Let \[A=\left|\begin{array}{ccccc}
4&8&8&8&5\\
0&1&0&0&0\\
6&8&8&8&7\\
0&8&8&3&0\\
0&8&2&0&0\end{array}\right|\]
Expanding determinant $A$ along 2nd row
\[A=(-1)^{2+2}
\left|\begin{array}{cccc}4&8&8&5\\
6&8&8&7\\
0&8&3&0\\
0&2&0&0\end{array}\right|\]
\[A=
\left|\begin{array}{cccc}4&8&8&5\\
6&8&8&7\\
0&8&3&0\\
0&2&0&0\end{array}\right|\]
Again expanding along 4th row
\[A=(-1)^{4+2}(2)
\left|\begin{array}{ccc}4&8&5\\
6&8&7\\
0&3&0\\\end{array}\right|\]
\[A=(2)
\left|\begin{array}{ccc}4&8&5\\
6&8&7\\
0&3&0\\\end{array}\right|\]
Again expanding along 3th row
\[A=(2)(-3)[(4)(7)-(5)(6)]\]
\[A=(-6)[28-30]\]
\[A=(-6)(-2)=12\]
Hence $A=12$.
