given: A =
|a b c|
|d e f| det(A) = 5
|g h i|
B =
|a u c|
|d v f| det(B) = -3
|g w i|
FIND:
|-3a c 2b - u|
|-3d f 2e - v|
|-3g i 2h - w|
-it's pretty much "B" except with the column of A in it...not sure what to do about that
given: A =
|a b c|
|d e f| det(A) = 5
|g h i|
B =
|a u c|
|d v f| det(B) = -3
|g w i|
FIND:
|-3a c 2b - u|
|-3d f 2e - v|
|-3g i 2h - w|
-it's pretty much "B" except with the column of A in it...not sure what to do about that
Given
aei + bfg + cdh - ceg - bdi - afh = 5
avi + ufg + cdw - cvg - udi - afw = -3
Find
(-3a)(f)(2h-w) + c(2e-v)(-3g) + (2b-u)(-3d)(i) - (2b - u)f(-3g) - c(-3d)(2h - w) - (-3a)(2e - v)(i) = ?
-6afh + 3afw - 6ceg + 3cvg - 6bdi + 3udi + 6bfg - 3ufg + 6cdh - 3cdw + 6aei - 3avi
[6aei + 6bfg + 6cdh - 6ceg - 6bdi - 6afh] + [-3avi - 3ufg - 3cdw + 3cvg + 3udi + 3afw]
6(aei + bfg + cdh - ceg - bdi - afh) + (-3)(avi + ufg + cdw - cvg - udi - afw)
6(5) + (-3)(-3)
30 + 9
39
Soroban, wow. My newbieness is showing. I attempted to approach the problem that way using the properties of manipulating columns but found no clear solution. Then I resorted to the basic way I know: brute force. Well done, sir. I'm impressed.
I hope I'm interpreting your solution correctly, but I think there are two errors here that cancel each other out. , where n is the dimension of A. This is because each column is multiplied by 2, and each of those multiplications multiplies the determinant by 2. In this case, .
Also, in this case*, but not in general.
What's happening here is that , where the notation means the matrix that's the same as A except the ith column is replaced with the column vector v, and represents the ith column of A, so does indeed equal 13 and the rest of the proof works fine.
*edit: whoops, no it doesn't. It would if det(2A)=2det(A) is what I should have said.