What is the Binomial Formula in Matrices without Evaluating Determinants?

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SUMMARY

The discussion centers on demonstrating the equality of two matrices without evaluating their determinants, specifically using the binomial formula. The matrices in question involve variables a1, b1, a2, b2, a3, and b3, and the goal is to show that the left matrix equals twice the right matrix. The solution hints at utilizing the properties of matrix operations, particularly the rule that states λ*A = λ*every value in the matrix, to simplify the problem without direct determinant evaluation.

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  • Understanding of matrix operations, including addition and scalar multiplication.
  • Familiarity with the binomial formula as it applies to matrices.
  • Knowledge of matrix equality and properties of determinants.
  • Basic algebraic manipulation skills in the context of matrices.
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  • Study the properties of matrix addition and how they apply to column operations.
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  • Explore techniques for proving matrix equalities without determinants.
  • Investigate scalar multiplication in matrices and its implications for matrix transformations.
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Students studying linear algebra, particularly those focusing on matrix theory and operations, as well as educators looking for teaching strategies related to matrix equality proofs.

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Homework Statement



I'm sorry this doesn't look too nice but it is supposed to be two matricces.

Show:

|1 a1-b1 a1+b1| |1 a1 b1|
|1 a2-b2 a2+b2|=2*|1 a2 b2|
|1 a3-b3 a3+b3| |1 a3 b3|

without evaluating the determinants.

Homework Equations



The Attempt at a Solution



It pretty obviously has got something to do with the 3. binomial formula and the rule that λ*A = λ*every value in the matrix.

I really don't know where to start on this, since I can't evaluate the determinants.
I'm pretty sure there is an easy way to do this, but I just can't see it.

Help is very much appreciated!

Thank you
 
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Welcome to PF!

Hi Crution! Welcome to PF! :smile:

Do you know any rules for adding one column to another column? :wink:
 
thanks that was enough help :-)
 

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