My Math Forum Matrix Inversion?

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 April 22nd, 2014, 08:48 AM #1 Newbie   Joined: Apr 2014 From: Jersey Posts: 2 Thanks: 0 Matrix Inversion? I only know how to do the inversion of one matrix, how does one do 2 like that?
April 22nd, 2014, 09:05 AM   #2
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Quote:
 Originally Posted by great username I only know how to do the inversion of one matrix, how does one do 2 like that?
LaTeX is messing with me for the moment.

Your equation is of the type:
AX = B

So, formally
A^{-1} A X = A^{-1} B

X = A^{-1} B

So all you need to do is invert A.

-Dan

PS This only works for the case where det(A) is not 0, of course.

Last edited by topsquark; April 22nd, 2014 at 09:08 AM.

 April 22nd, 2014, 10:27 AM #3 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Top is correct. Alternately, since this is multiple choice, you could multiply out each of the possible X values and see which work out.
April 22nd, 2014, 10:51 AM   #4
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Quote:
 $\displaystyle \text{Solve: }\;\begin{bmatrix}6&\text{-}3 \\ 7&9\end{bmatrix}\,X \;=\;\begin{bmatrix}\text{-}18 & \text{-}93 \\ 4 & \text{-}21 \end{bmatrix}$ $\displaystyle \;\;A.\;\begin{bmatrix}\text{-}3&9\\4&\text{-}2\end{bmatrix} \qquad B.\;\begin{bmatrix}\text{-}9&\text{-}7 \\ -11&3\end{bmatrix} \qquad C.\;\begin{bmatrix}\text{-}2&\text{-}12\\2&7\end{bmatrix}$ $\displaystyle \;\;D.\;\begin{bmatrix}1&\text{-}9\\12&11 \end{bmatrix} \qquad E.\;\begin{bmatrix}\text{-}10&1\\\text{-}9&6\end{bmatrix} \qquad F.\;\begin{bmatrix}12&\text{-}5\\3&\text{-}11\end{bmatrix}$ I only know how to do the inversion of one matrix. How does one do 2 like that? $\displaystyle \color{blue}{\text{We don't "do" two of them.}}$

Find the inverse of $\displaystyle \begin{bmatrix}6&\text{-}3 \\ 7&9\end{bmatrix}$ by your favorite method.
$\displaystyle \qquad$It turns out to be: $\displaystyle \;\frac{1}{75}\begin{bmatrix}9&3\\\text{-}7&6\end{bmatrix}$

Multiply both sides by the inverse:

$\displaystyle \quad \frac{1}{75}\begin{bmatrix}9&3\\\text{-}7&6\end{bmatrix}\begin{bmatrix}6&\text{-}3\\7&9\end{bmatrix}X \;=\;\frac{1}{75}\begin{bmatrix}9&3\\\text{-}7&6\end{bmatrix}\begin{bmatrix}\text{-}18&\text{-}93\\ 4 & \text{-}21\end{bmatrix}$

$\displaystyle \qquad\frac{1}{75}\begin{bmatrix}75&0\\0&75 \end{bmatrix}X \;=\;\frac{1}{75} \begin{bmatrix}\text{-}150&\text{-}900\\150 & 525 \end{bmatrix}$

$\displaystyle \qquad \begin{bmatrix}1&0\\0&1\end{bmatrix}X \;=\;\begin{bmatrix}\text{-}2&\text{-}12 \\ 2&7 \end{bmatrix}$

$\displaystyle \qquad X \;=\;\begin{bmatrix}\text{-}2&\text{-}12 \\ 2&7 \end{bmatrix}$

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