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Re: An unknown Greek matrix
*To*: mathgroup at smc.vnet.net
*Subject*: [mg129982] Re: An unknown Greek matrix
*From*: Bob Hanlon <hanlonr357 at gmail.com>
*Date*: Sat, 2 Mar 2013 03:44:11 -0500 (EST)
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*References*: <20130301022642.797786882@smc.vnet.net>
A = {{5, 0}, {0, 10}};
B = {{6, 3}, {1, 12}};
L = {1, 1};
Lambda = {Subscript[lambda, 1],
Subscript[lambda, 2]};
Solve[Lambda.B == Lambda.A + L, Lambda]
{{Subscript[lambda, 1] -> -1,
Subscript[lambda, 2] -> 2}}
Bob Hanlon
On Thu, Feb 28, 2013 at 9:26 PM, <marshfeldman at 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,
>
> 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?
>
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