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
Hello, mathlover700!
I hope I interpreted the problem correctly . . .
$\displaystyle \text{Given: }\:A = \begin{vmatrix} a&b&c \\ d&e&f \\ g&h&i \end{vmatrix},\;\text{det}(A) = 5$
$\displaystyle B =\begin{vmatrix}a&u&c \\ d&v&f \\ g&w&i\end{vmatrix},\;\text{det}(B) = -3$
$\displaystyle \text{Find the determinant of: }\:\begin{vmatrix}-3a & c & 2b-u \\ -3d & f & 2e-v \\ -3g & i & 2h-w \end{vmatrix}$
$\displaystyle 2A \;=\;2\cdot\begin{vmatrix}a&b&c\\d&e&f\\g&h&y\end{ vmatrix} \;=\;\begin{vmatrix}2a&2b&2c\\2d&2e&2f\\2g&2h&2i \end{vmatrix}$
. . $\displaystyle \text{det}(2A)\,=\,2(5)\,=\,10$
$\displaystyle 2A-B \;=\;\begin{vmatrix}2a&2b&2c\\2d&2e&2f \\ 2g&2h&2y\end{vmatrix} - \begin{vmatrix}a&u&c\\d&v&f \\ g&w&i\end{vmatrix} \;=\;\begin{vmatrix}a&2b-y&c\\d&2e-v&f \\ g&2h-w&i \end{vmatrix}$
. . $\displaystyle \text{det}(2A-B)\,=\,10 - (\text{-}3)\,=\,13$
Multiply -3 times column-1: .$\displaystyle \begin{vmatrix}\text{-}3a&2b-u&c \\ \text{-}3d&2e-v&f \\ \text{-}3g&2h-w&i\end{vmatrix}$
. . $\displaystyle \text{det}\,=\,-3(13)\,=\,-39$
Switch column-2 and column-3: .$\displaystyle \begin{vmatrix}\text{-}3a & c & 2b-u \\ \text{-}3d & f & 2e-v \\ \text{-}3g & i & 2h-w \end{vmatrix}$
. . $\displaystyle \text{det} \,=\,-1(-39) \,=\,+39$
Hey, I agree with Mathguy25 !
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. $\displaystyle \text{det}(2A)=2^n\text{det}A$, 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, $\displaystyle \text{det}(2A)=8\text{det}A=40$.
Also, $\displaystyle \text{det}(2A-B)=\text{det}2A-\text{det}B$ in this case*, but not in general.
What's happening here is that $\displaystyle \text{det}[A_i(2c_i+v)]=\text{det}[A_i(2c_i)]+\text{det}[A_i(v)]$, where the notation $\displaystyle A_i(v)$ means the matrix that's the same as A except the ith column is replaced with the column vector v, and $\displaystyle c_i$ represents the ith column of A, so $\displaystyle \text{det}(2A-B)$ 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.