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a.caregnato

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About a.caregnato

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  1. You're right, I've got it now. [math]t1,t2,t3[/math] are row vectors (1x3), thats the only way We'll have a matrix T-1 with 3x3 dimensions. Thank you for your help, ajb.
  2. Yes, is the inverse. I'm trying to find the terms of the inverse transformation matrix (t1,t2,t3) which will "turn" F into A (It's easy to figure out T knowing T-1) . But that last equation doens't make any sense to me. Thank you for your answer.
  3. Hello everybody. I'm having a little bit of trouble understanding a passage of my textbook regarding a linear transformation and matrix multiplication, I wonder if you could help me out. So, I have this equation: [math] \dot x = \textbf{Fx} + \textbf{G}u [/math] Where F is some 3x3 matrix and x a 3x1 array. For now, these are the important variables. So, my objective is putting F in a specific format called control canonical form (A), which is: [math] A = \left| \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ a & b & c \\ \end{array} \right|.[/math] For that, the book shows a Linear Transformation in the variable x: [math] \textbf{x} = \textbf{Tz} [/math] Which leads to (see first equation): [math] \dot z =T^{-1} \textbf{FTz} + T^{-1}\textbf{G}u [/math] The equation for A is: [math]\textbf{A} = T^{-1} \textbf{FT} [/math] Where T-1is defined as: [math] T^{-1} = \left| \begin{array}{ccc} t1 \\ t2 \\ t3 \\ \end{array} \right|.[/math] Writing everything in therms of T-1: [math]\textbf{A} T^{-1} = T^{-1} \textbf{F} [/math] Now, the problem: [math]\left| \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ a & b & c \\ \end{array} \right| \left| \begin{array}{ccc} t1 \\ t2 \\ t3 \\ \end{array} \right| = \left| \begin{array}{ccc} t1 \textbf{F} \\ t2 \textbf{F} \\ t3 \textbf{F} \\ \end{array} \right| [/math] I don't understant the right part of the equation. How can I multiply T-1, which is a 3x1 array, with the 3x3 F matrix? Why the book shows a array with every single term of T-1 multiplying F? I apologize if this is some stupid question but linear algebra isn't my strong suit. Thanks!