I want to know how to determine det(B) if det(A) = 3? 1 a 1 4 2a+1 2+b A = 2 1...

Expandinfg thruogh the 1st row, we get
:

Det (A) =  |[(1  a  1),(2  1  b), (3  c  0)]| =
3......(1)

Det B = |[(4  2a+1  2+b ) , (15  5c  0) , (2 1 
b)]|.

Det B = R1-R3 = |[(4-2  2a+1-1  2+b-b) , (15  5c  0),
(2  1  b)]|.

DetB = |[(2  2a  2), (15   5c  0) , (2  1  
b)]|

Det B = 2 *5 |[(1  a  1), (2  c  0) , (2  1  b)]| ,
where 2 and 5 are factored out from 1st and 2nd rows.

Now
we inter change R2 and R3  which results

Det B = -10 |[( 1 
a  1), (2   1   b) , ( 2   c   0)]|.....(2)

Now we notice
from the right side of  eq (1) and  left eq (2) that:

Det
(B) = -10 det (A). But det(A) = 3.

Therefore det(B) = -10*3
= -30.

Therefore det(B) =
-30