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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?? Pic Last edited by skipjack; November 7th, 2016 at 12:37 AM. 
November 6th, 2016, 09:15 PM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 2,427 Thanks: 1314 
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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