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Re: An unknown Greek matrix

On 2/28/13 at 9:26 PM, marshfeldman at 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 using
the full name, i.e.,


or by using the esc key as follows

esc l esc

Subscripts can also be entered using simple key strokes. But I
am not going to do that here. Mathematica can be made to use
subscripted variables as you would see in a text book. But IMO,
usage of subscripted variables is more of an advanced
Mathematica 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 case
letter as a variable name. There are a several built-in
functions which are named with a single upper case letter.
Avoiding the use of single upper case letters as variable names
ensures you will not have conflicts with built-in symbols and
save you a lot of grief in the long run.

So, here are your matrices

In[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 equation

In[5]:= eq = lambda.b == lambda.a + l

Out[5]= {{6*x + y, 3*x + 12*y}} == {{5*x + 1, 10*y + 1}}

note the use of a double = which defines an equation in
Mathemtica and the usage of '.' to tell Mathematica I want to
perform a matrix multiplication rather than and element by
element multiplication

and the solution can be found with

In[6]:= Solve[eq, {x, y}]

Out[6]= {{x -> -1, y -> 2}}

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