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I become very bad in linear algebra when matrices are involved in the calculation. Here is one equation:

$H = (A_2 - a_2p)(A_1 - a_1p)^{-1}$

where:

$H$, $A_1$ and $A_2$ are $3\times3$ matrices, $a_1$ and $a_2$ are $3\times 1$ matrices $p$ is a $1\times3$ matrix.

$p$ is the only unknown I want to find. How can I do please?

Thanks.

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    Thank you again. At least this will help me to go further next time!2012-01-08

1 Answers 1

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Multiplying on the right by $A_1-a_1p$, one gets $ K=bp\quad\text{with}\quad K=A_2-HA_1\quad\text{and}\quad b=a_2-Ha_1. $ Recall that $K=(K_{ij})_{1\leqslant i,j\leqslant3}$ is a $3\times3$ matrix, $b=(b_i)_{1\leqslant i\leqslant3}$ is a $3\times1$ (column) vector and one looks for a $1\times3$ (line) vector $p=(p_j)_{1\leqslant j\leqslant 3}$.

That is, $K=bp$ is a shorthand for the condition that $K_{ij}=b_ip_j$ for every $(i,j)$. Thus, $K=bp$ is solved by $ p_j=K_{ij}/b_i $ for every $(i,j)$ if $K_{ij}/b_i$ does not depend on $i$, and has no solution otherwise. Recall finally that, by definition, $ K_{ij}=(A_2)_{ij}-\sum\limits_{k=1}^3H_{ik}(A_1)_{kj}\quad\text{and}\quad b_i=(a_2)_i-\sum\limits_{k=1}^3H_{ik}(a_1)_k. $ Edit (This is due to @David Mitra in a comment.) Drawing the matrices $K$ and $bp$ may help to see what is going on: $ \underbrace{ \left[ \matrix {K_{11} & K_{12}& K_{13}\cr K_{21}& K_{22}& K_{23}\cr K_{31}& K_{32}& K_{33}\cr } \right]}_{K=A_2-HA_1}= \underbrace{ \left[ \matrix { b_1p_1& b_1p_2& b_1p_3 \cr b_2p_1& b_2p_2& b_2p_3 \cr b_3p_1& b_3p_2& b_3p_3 \cr } \right]}_{bp=(a_2-Ha_1)p }. $

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    @DavidMitra: Included your (helpful) comment in my post. Hope you do not mind.2012-01-08