Algebra Pre-Algebra and Basic Algebra Math Forum

 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
Math Team

Joined: May 2013
From: The Astral plane

Posts: 2,230
Thanks: 908

Math Focus: Wibbly wobbly timey-wimey stuff.
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
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

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:

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

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

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

$\displaystyle \qquad X \;=\;\begin{bmatrix}\text{-}2&\text{-}12 \\ 2&7 \end{bmatrix}$ Tags inversion, matrix Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post ktalebian Linear Algebra 2 September 2nd, 2013 09:27 AM milenko76 Linear Algebra 2 November 4th, 2011 02:49 AM maximus101 Algebra 1 March 8th, 2011 07:25 AM yiorgos Applied Math 0 November 23rd, 2010 04:33 AM callingearthlings Linear Algebra 1 February 8th, 2009 09:11 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      