March 28th, 2018, 08:36 PM  #1 
Member Joined: Nov 2016 From: Ireland Posts: 84 Thanks: 3  Reversing a matrix
Ok I think this should be easy, if I can explain it properly. Let's say I had a 2*2 matrix = matrix A And I was multiplying it 2*1 matrix = matrix B And it would produce a 2*1 matrix = matrix C Right? But let's say  I want to go back ways, reverse it... I have Matrix C and I have Matrix A but I need to find Matrix B. I'd better explain this as well, as I'm not sure the rules are exactly the same. I'm using a nominal Pi model to analyse a transmission line The layout is like this Vs  A B  Vr Is  C D  Ir Where Vs = A*Vr + B*Ir And Is = C*Vr + D*Ir But now I need to find Vr and Ir I have ABCD and I have Vs and Is At first I thought well I'll just rearrange the equation to find Vr. But 2 of the variables are missing, when I do that... Any help would be great guys. Thx 
March 29th, 2018, 05:05 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
So basically, you want to solve a matrix equation of the form AB= C for B? If A is invertible then $\displaystyle B= A^{1}C$. If A is not invertible then there is no solution. Write $\displaystyle A= \begin{bmatrix}a & b \\ c & d \end{bmatrix}$, $\displaystyle B= \begin{bmatrix}x \\ y \end{bmatrix}$, and $\displaystyle C= \begin{bmatrix}p \\ q \end{bmatrix}$. A is invertible if and only if its determinant, ad bc is not 0 and, in that case, its inverse is $\displaystyle A^{1}= \frac{1}{ad bc}\begin{bmatrix}d & b \\ c & a\end{bmatrix}$ and $\displaystyle B= A^{1}C= \frac{1}{ad bc}\begin{bmatrix}d & b \\ c & a\end{bmatrix}\begin{bmatrix}p \\ q \end{bmatrix}= \frac{1}{ad bc}\begin{bmatrix}dp bq \\ aq cp\end{bmatrix}$. Or you could simply treat the matrix equation as a system of two equations in two unkowns. We have $\displaystyle \begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}ax+ by \\ cx+ dy\end{bmatrix}= \begin{bmatrix}p \\ q\end{bmatrix}$. That is equivalent to the two equations ax+ by= p and cx+ dy= q. Multiply the first equation by d to get adx+ bdy= pd and multiply the second equation by b to get bcx+ bdy= bq. The coefficents of y are now the same, bd, so subtracting one equation from the other eliminates y: (ad bc)x= pd bq. If ad bc is not 0, divide both sides by ad bc to get $\displaystyle x= \frac{pd bq}{ad bc}$ as before. If ad bc= 0 and pd bq is not 0, there is no x that makes that equation true. If both ad bc= 0 and pd bq= 0 then any x works. Much the same thing is true of $\displaystyle V_s = AV_r + BI_r$ and $\displaystyle I_s = CV_r + DI_r$. Multiply the first equation by D to get $\displaystyle DV_s= ADV_r+ BDI_r$ and multiply the second equation by B to get $\displaystyle BI_s= BCV_r+ BDI_r$. Now $\displaystyle I_r$ has the same coefficient in both equations so subtracting one equation from the other eliminates $\displaystyle I_r$: $\displaystyle DV_s BI_s= (AD BC)V_r$ and $\displaystyle V_r= \frac{DV_s BI_s}{AD BC}$, assuming, of course, that AD BC is nonzero. Then $\displaystyle V_s= AV_r+ BI_r= \frac{ADV_s ABI_s}{AD BC}+ BI_r$ so $\displaystyle BI_r= V_s \frac{ADV_S ABI_s}{AD BC}= \frac{ADV_s BCV_s ADV_s+ ABI_s}{AD BC}= \frac{ABI_s BCV_s}{AD BC}$. 
March 29th, 2018, 06:59 AM  #3 
Senior Member Joined: Oct 2009 Posts: 784 Thanks: 280  Why wouldn't there be a solution if A is not invertible? Sure, there isn't always a solution for all A and C. But some choices certainly will have (multiple) solutions!

March 29th, 2018, 10:14 AM  #4  
Member Joined: Nov 2016 From: Ireland Posts: 84 Thanks: 3 
Beautiful Country Boy. I've written it all out, substituting in numbers on one side, with the equations on the other. I think I understand most of it now. This bit at the end however, seems to be off by a factor of 2. Quote:
$\displaystyle V_s = AV_r + BI_r$ = $\displaystyle 92 = 12*3 + 14*4$ and $\displaystyle I_s = CV_r + DI_r$ = $\displaystyle 85 = 11*3 + 13*4$ Therefore: $\displaystyle BI_r = 28 $ But: $\displaystyle (ABI_s  BCV_s) / (ADBC) $ $\displaystyle 112/2 = 56$ Last edited by Kevineamon; March 29th, 2018 at 10:21 AM.  
March 29th, 2018, 11:12 AM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902  Yes, I should have said right at the start that there is not a unique solution. If A is not invertible then either there is no solution or there are an infinite number of solutions. I did later say that "If ad bc= 0 and pd bq is not 0, there is no x that makes that equation true. If both ad bc= 0 and pd bq= 0 then any x works."

March 29th, 2018, 07:06 PM  #6 
Member Joined: Nov 2016 From: Ireland Posts: 84 Thanks: 3 
How about my example Country Boy... Those numbers don't meet any of you conditions. However: $\displaystyle BI_r ≠ (ABI_s−BCV_s)/(AD−BC)$ Although it is off by a factor of 2, which is the answer I'm looking for... Am I supposed to guess something here? My math is Ok, when I work on it. But it takes a lot of effort. I'm not naturally gifted with it, like your good self. My mathematically intuition = 0 Can you help me take this final step? 
March 29th, 2018, 09:07 PM  #7  
Senior Member Joined: Sep 2016 From: USA Posts: 609 Thanks: 378 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
The fact is that if $A$ has full rank, then $AB = C$ may have a unique solution even if $A$ is not invertible. This is easily seen by noticing that $A^TA$ is always invertible where $A^T$ is the conjugate transpose. Now suppose $Ax = b$ where $x,b$ are vectors, then you can easily see that a formula for $x$ is given by \[x = (A^TA)^{1} A^T b. \] In this case, $(A^TA)^{1}A^T$ is the weak inverse for $A$ which is also a left inverse (sometimes called a pseudoinverse). Notice that if $A$ is not square, then it doesn't even have a determinant, but that has nothing to do with the question of invertibility.  
March 30th, 2018, 06:31 AM  #8 
Member Joined: Nov 2016 From: Ireland Posts: 84 Thanks: 3 
While this high brow mathematically discussion is, I can guess... absolutely fascinating. There's a n00b engineer over here with a transmission line to analyse. Can one of you geniuses, climb down from the lofty tower and explain to lil old me, how to find $\displaystyle I_r$? Please...

March 30th, 2018, 06:45 AM  #9 
Member Joined: Nov 2016 From: Ireland Posts: 84 Thanks: 3 
Wait... Inspiration has come... I think...

March 30th, 2018, 07:10 AM  #10 
Member Joined: Nov 2016 From: Ireland Posts: 84 Thanks: 3 
$\displaystyle I_r= (CV_sAI_s)/(CBAD)$ Well I think your math is great, Country Boy. Never mind the haters. 

Tags 
matrix, reversing 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Help reversing an algorithm...  Vidofner  Applied Math  4  January 12th, 2014 08:47 AM 
Help reversing formula  oneup  Elementary Math  4  September 2nd, 2013 04:08 AM 
Logarithms and Reversing them  Liqwde  Computer Science  2  August 31st, 2010 08:59 PM 
Reversing the order of Integration  OSearcy4  Calculus  4  October 17th, 2009 03:32 PM 
Reversing Digits  Cat  Number Theory  6  January 7th, 2007 07:32 PM 