Let $A$ and $B$ be two invertible matrices in $M_2(\mathbb{R})$such that $(A+B)^3=A^3+3A^2B+3AB^2+B^3$ then prove or disprove that $ AB=BA$
My working:
$$(A+B)^3=A^3+3A^2B+3AB^2+B^3$$ $$\implies BA^2+B^2A+ABA+BAB =2A^2B+2AB^2$$
Now what should I do?
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I saw the counterexample $A = \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix}$ and $B = \begin{bmatrix}-2 & 0 \0 & 1\end{bmatrix}$. This gives $AB = \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix}$, $BA= \begin{bmatrix} 0 & -2 \ 0 & 0 \end{bmatrix}$, $(A+B)^3 = A^3+3A^2B + 3AB^2 + B^3 = \begin{bmatrix} -8 & 3 \ 0 & 1 \end{bmatrix}$, but why is it deleted? – GNUSupporter 8964民主女神 地下教會 Jan 22 '18 at 16:38
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@EwanDelanoy I got it thanks – GNUSupporter 8964民主女神 地下教會 Jan 22 '18 at 16:40
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I got one result $AB=BA\iff B=xA+yI_2$ where $x,y\in \mathbb{R}$ – Makar Jan 22 '18 at 17:01
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Counter-example :
$$ A= \left(\begin{array}{cc} 2 & 0 \\ & \\ 0 & 3 \\ \end{array}\right), B= \left(\begin{array}{cc} 1 & 0 \\ & \\ 1 & -10 \\ \end{array}\right). $$
Ewan Delanoy
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It should be more enlightening that you explain to the OP how you obtained the counterexample. :) – Pedro Jan 22 '18 at 16:43
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