if $\displaystyle \begin{vmatrix}

a & b &c \\

d & e &f \\

g&h &k

\end{vmatrix}$=2

then

$\displaystyle \begin{vmatrix}

f & k-4c &2k+f \\

d & g-4a &2g+d \\

e&h-4b &2h+e

\end{vmatrix}$=-16

is it true?

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- Nov 16th 2010, 12:35 PMtransgalacticdeterminant question 5
if $\displaystyle \begin{vmatrix}

a & b &c \\

d & e &f \\

g&h &k

\end{vmatrix}$=2

then

$\displaystyle \begin{vmatrix}

f & k-4c &2k+f \\

d & g-4a &2g+d \\

e&h-4b &2h+e

\end{vmatrix}$=-16

is it true? - Nov 16th 2010, 05:41 PMDrexel28
- Nov 16th 2010, 09:22 PMtransgalactic
i dont know how to break the big determinant into smaller packages of the original determinant

- Nov 18th 2010, 08:38 PMAnthonny
Have you tried computing the determinants for both and comparing them? Although I'm not completely sure if this is the best way, but it may work.

- Nov 18th 2010, 09:01 PMtonio

Remember that determinant is a multilineal functions, so:

$\displaystyle \begin{vmatrix}f & k-4c &2k+f \\ d & g-4a &2g+d \\ e&h-4b &2h+e \end{vmatrix}=\begin{vmatrix}f & k &2k \\ d & g &2g \\ e&h &2h \end{vmatrix}+\begin{vmatrix}f & k &f \\ d & g &d \\ e&h &e \end{vmatrix}+$$\displaystyle \begin{vmatrix}f & -4c &2k \\ d & -4a &2g \\ e&-4b &2h \end{vmatrix}+\begin{vmatrix}f & -4c &f \\ d & -4a &d \\ e&-4b &e \end{vmatrix}$ ,

So....??

Tonio - Nov 18th 2010, 09:11 PMDrexel28
- Nov 18th 2010, 09:14 PMtonio
- Nov 18th 2010, 10:50 PMmatheagle
I'd do column reduction and then factor out the 2 and -4, so 2(2)(-4)=-16