January 14th, 2019, 10:10 AM  #1 
Newbie Joined: Jan 2019 From: UK Posts: 3 Thanks: 0  Matrix
Hi all, can someone please tell me how they got this equation from this matrix. I seem to get two products with X2 instead of one. Please see attachment. Thank you! 
January 14th, 2019, 01:50 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,680 Thanks: 658 
The right side should be a 2vector. You should end up with two equations.

January 14th, 2019, 03:21 PM  #3 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,035 Thanks: 810 Math Focus: Wibbly wobbly timeywimey stuff. 
This becomes: $\displaystyle \left ( \begin{matrix} 106.67 \times 10^3  129.5439 \times 300 & 53.34 \times 10^3 \\ 53.34 \times 10^3 & 106.67 \times 10^3 129.5439 \times 500 \end{matrix} \right ) \left ( \begin{matrix} 1 \\ x_2 \end{matrix} \right ) = \left ( \begin{matrix} 0 \\ 0 \end{matrix} \right )$ Does writing it like this help? Dan 
January 15th, 2019, 10:36 AM  #4 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 
It's easier to outline with letters. $\displaystyle [ \begin{bmatrix} a &b \\ b & a \end{bmatrix}+\begin{bmatrix} c & 0\\ 0& d \end{bmatrix}] \begin{bmatrix} 1\\ x \end{bmatrix}=\begin{bmatrix} a+c &b \\ b &a+d \end{bmatrix}\begin{bmatrix} 1\\x \end{bmatrix}=0$ $\displaystyle (a+c)+bx=0$ $\displaystyle b+(a+d)x=0$ $\displaystyle \frac{a+c}{b}=\frac{b}{a+d}$ This is the test for a solution. Now substitute numbers and if this equation is satisfied, you can solve for x which is probably the second equation of OP. 

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