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November 6th, 2016, 07:22 PM   #1
Newbie

Joined: Nov 2016
From: HOuston, TX

Posts: 3
Thanks: 0

Can't figure out how to solve

I don't know how to even start...

Suppose | a b c | $\hspace{62px}$ |a -g 2d|
$\hspace{60px}$| d e f | = -3. Find |b -h 2e|
$\hspace{60px}$| g h i | $\hspace{66px}$ |c $\,$-i$\,$ 2f|

Ok, so -3 is the determinant of the 1st matrix; how do I use it to solve for the determinant of the second one??

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Last edited by skipjack; November 7th, 2016 at 12:37 AM.

 November 6th, 2016, 08:18 PM #2 Newbie   Joined: Nov 2016 From: HOuston, TX Posts: 3 Thanks: 0 Does this look right???? So I thought about it hard... Could this possibly be how you solve this? Does it look right??? Please help Picimage.jpg
 November 6th, 2016, 09:15 PM #3 Senior Member     Joined: Sep 2015 From: USA Posts: 2,530 Thanks: 1390 let $M=\begin{pmatrix}a &b &c \\ d &e &f \\ g &h &i \end{pmatrix}$ $\begin{pmatrix}a &-g &2d \\ b &-h &2e \\ c &-i &2f \end{pmatrix} = \left(\begin{pmatrix}1 &0 &0 \\ 0 &0 &-1 \\ 0 &2 &0\end{pmatrix} M\right)^T$ $\left |\begin{pmatrix}1 &0 &0 \\ 0 &0 &-1 \\ 0 &2 &0\end{pmatrix}\right|$ is easily evaluated as $(1)(0 + 2) = 2$ for any square matrix $\left |A^T \right| = \left |A\right|$ $\left |\begin{pmatrix}a &-g &2d \\ b &-h &2e \\ c &-i &2f \end{pmatrix} \right | =\left |\begin{pmatrix}1 &0 &0 \\ 0 &0 &-1 \\ 0 &2 &0\end{pmatrix}\right| \left | M \right | = (2)(-3) = -6$ Well done!

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