1. ## determinant question 5

if $\begin{vmatrix}
a & b &c \\
d & e &f \\
g&h &k
\end{vmatrix}$
=2

then
$\begin{vmatrix}
f & k-4c &2k+f \\
d & g-4a &2g+d \\
e&h-4b &2h+e
\end{vmatrix}$
=-16

is it true?

2. Originally Posted by transgalactic
if $\begin{vmatrix}
a & b &c \\
d & e &f \\
g&h &k
\end{vmatrix}$
=2

then
$\begin{vmatrix}
f & k-4c &2k+f \\
d & g-4a &2g+d \\
e&h-4b &2h+e
\end{vmatrix}$
=-16

is it true?
I don't know, is it?

3. i dont know how to break the big determinant into smaller packages of the original determinant

4. 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.

5. Originally Posted by Drexel28
I don't know, is it?

Remember that determinant is a multilineal functions, so:

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

So....??

Tonio

6. Originally Posted by tonio
Remember that determinant is a multilineal functions, so:

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

So....??

Tonio
Why did you quote me? I know how to do this.

7. Originally Posted by Drexel28
Why did you quote me? I know how to do this.

I'm a selective reader: only got "I don't know" and thought it was the OP. Hopefully I won't confuse him/her.

Tonio

8. I'd do column reduction and then factor out the 2 and -4, so 2(2)(-4)=-16