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January 14th, 2019, 09:10 AM   #1
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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.

Thank you!
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 January 14th, 2019, 12:50 PM #2 Global Moderator   Joined: May 2007 Posts: 6,823 Thanks: 723 The right side should be a 2-vector. You should end up with two equations. Thanks from topsquark and Johncena12
 January 14th, 2019, 02:21 PM #3 Math Team     Joined: May 2013 From: The Astral plane Posts: 2,273 Thanks: 943 Math Focus: Wibbly wobbly timey-wimey 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 Thanks from Johncena12
 January 15th, 2019, 09:36 AM #4 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 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. Thanks from topsquark and Johncena12

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