```Date: Mar 2, 2013 3:44 AM
Author: Bill Rowe
Subject: Re: An unknown Greek matrix

On 2/28/13 at 9:26 PM, marshfeldman@gmail.com wrote:>Forgive me if this is obvious, but I'm a Mathematica newbie and have>given up trying to figure this out any other way.>I have the following defined matrices:>A = {{5, 0}, {0, 10}} B = {{6, 3}, {1, 12}}>and a row vector:>L = {{1, 1}}.>I don't know how to use Greek here, so I'll use "Lambda" as the name>of a row vector whose name is really the Greek letter lambda>capitalized and "lambda" for as the name of the elements of Lambda>subscripted, with the elements really being lower-case versions of>the Greek letter lambda and subscripts indicated by appending _n,>where n is the subscript (e.g. lambda_1 is lowercase lambda>subscripted with 1). In other words,You can enter Greek letters in one of two ways. Either by usingthe full name, i.e.,\[Lambda]or by using the esc key as followsesc l escSubscripts can also be entered using simple key strokes. But Iam not going to do that here. Mathematica can be made to usesubscripted variables as you would see in a text book. But IMO,usage of subscripted variables is more of an advancedMathematica technique and more trouble than it is worth.>Lambda = {{lambda_1, lambda_2}}.>Now given the following equation, solve for Lambda:>Lambda B = Lambda A + L.>Also, display the elements of Lambda as>lambda_1 = -1 lambda_2 = 2.>Can one do this in Mathematica? How?So, I will show a solution without the subscripted variables.Also, it is highly recommended not to use a single upper caseletter as a variable name. There are a several built-infunctions which are named with a single upper case letter.Avoiding the use of single upper case letters as variable namesensures you will not have conflicts with built-in symbols andsave you a lot of grief in the long run.So, here are your matricesIn[1]:= a = {{5, 0}, {0, 10}};b = {{6, 3}, {1, 12}};In[3]:= l = {{1, 1}};In[4]:= lambda = {{x, y}};Now to set up the equationIn[5]:= eq = lambda.b == lambda.a + lOut[5]= {{6*x + y, 3*x + 12*y}} == {{5*x + 1, 10*y + 1}}note the use of a double = which defines an equation inMathemtica and the usage of '.' to tell Mathematica I want toperform a matrix multiplication rather than and element byelement multiplicationand the solution can be found withIn[6]:= Solve[eq, {x, y}]Out[6]= {{x -> -1, y -> 2}}
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