Question 1201590
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The template P if and only if Q breaks into two pieces<ul><li>If P, then Q</li><li>If Q, then P</li></ul>
For this problem,
AB = BA if and only if (A-B)(A+B) = A^2-B^2
breaks into<ul><li>If AB = BA, then (A-B)(A+B) = A^2-B^2</li><li>If (A-B)(A+B) = A^2-B^2, then AB = BA</li></ul>


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Part 1)
If AB = BA, then (A-B)(A+B) = A^2-B^2


We'll start with (A-B)(A+B) and try to reach A^2-B^2 through use of AB = BA


(A-B)(A+B) = A(A+B)-B(A+B)
(A-B)(A+B) = (A^2+AB)+(-BA-B^2)
(A-B)(A+B) = (A^2+AB)+(-AB-B^2) ... use AB = BA
(A-B)(A+B) = A^2+(AB-AB)-B^2
(A-B)(A+B) = A^2+0*AB-B^2
(A-B)(A+B) = A^2+0-B^2
(A-B)(A+B) = A^2-B^2


We have proven that if AB=BA, then (A-B)(A+B) leads to A^2-B^2


In other words, if AB=BA, then (A-B)(A+B) = A^2-B^2


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Part 2)


If (A-B)(A+B) = A^2-B^2, then AB = BA



(A-B)(A+B) = A^2-B^2
A(A+B)-B(A+B) = A^2-B^2
(A^2+AB)+(-BA-B^2) = A^2-B^2
A^2+(AB-BA)-B^2 = A^2-B^2
(AB-BA)-B^2 = -B^2
AB-BA = 0
AB = BA


The second portion of the "if and only if" statement has been confirmed.


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Therefore, overall we can say AB = BA if and only if (A-B)(A+B) = A^2-B^2
Matrices A and B must be square, and the same size.
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