We should start by expanding the minors of the matrix. Take each element in the first row, then multiply it by the minor. The minor of an element is whats left of the matrix when the row and column of the element are removed.
#((color(red)(4),~cancel(10),cancel(3),cancel(2)),(cancel(2),1,4,5),(cancel(1),~1,2,2),(cancel(2),~14,~1,~2)) , ((cancel(4),color(red)(~10),cancel(3),cancel(2)),(2,cancel(1),4,5),(1,~cancel(1),2,2),(2,~cancel(14),~1,~2)), ...#
Alternate adding and subtracting the product of each element and its minor. In other words;
#4 |(1,4,5),(~1,2,2),(~14, ~1, ~2) | -(~10) |(2,4,5),(1,2,2),(2,~1,~2) | +3 |(2,1,5),(1,~1,2),(2,~14,~2) | -2 |(2,1,4),(1,~1,2),(2,~14,~1) |#
We can expand the minors again to continue. For convenience in double checking, I'll work each element in its own row.
#4(1|(2,2),(~1,~2)| - 4|(~1,2),(~14,~2)|+5|(~1,2),(~14,~1)|)#
#+10(2|(2,2),(~1,~2)| -4|(1,2),(2,~2)|+5|(1,2),(2,~1)|)#
#+3(2|(~1,2),(~14,~2)|-1|(1,2),(2,~2)|+5|(1,~1),(2,~14)|)#
#-2(2|(~1,2),(~14,~1)|-1|(1,2),(2,~1)|+4|(1,~1),(2,~14)|)#
To find the determinant of each of the #2xx2# matrices, multiply the top left element by the bottom right element and subtract the product of the other two elements.
#det{((A,B),(C,D))} = A*D-B*C#
Solving each determinant yields;
#4((2)(~2)-(~1)(2)) - 16((~1)(~2) - (~14)(2)) +20((~1)(~1)-(~14)(2))#
#+20((2)(~2)-(~1)(2))-40((1)(~2)-(2)(2)) + 50((1)(~1)-(2)(2))#
#+6((~1)(~2)-(~14)(2)) -3((1)(~2)-(2)(2))+5((1)(~14)-(2)(~1))#
#-4((~1)(~1)-(~14)(2))+2((1)(~1)-(2)(2))-8((1)(~14)-(2)(~1))#
The remainder of the problem is arithmetic.
#4(~4+2)-16(2+28)+20(1+28)#
#+20(~4+2)-40(~2-4)+50(~1-4)#
#+6(2+28)-3(~2-4)+5(~14+2)#
#~4(1+28)+2(~1-4)-8(~14+2)#
Solving inside the parenthesis;
#4(~2)-16(30)+20(29)#
#+20(~2)-40(~6)+50(~5)#
# +6(30)-3(~6)+5(~12)#
#~4(29)+2(4)-8(~28)#
Take each product;
#~8-480+580#
#-40+240-250#
#+180+18-60#
#-116+8+224#
So the determinant of our matrix is;
#=296#
