If A = [abb2−a2−ab]\begin{bmatrix} ab & b^2 \ -a^2 & -ab \end{bmatrix}[ab−a2b2−ab], show that A2 = 0.
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Solving A2,
⇒A2=[abb2−a2−ab]×[abb2−a2−ab]⇒[ab×ab+b2×(−a2)ab×b2+b2×(−ab)(−a2)×ab+(−ab)×(−a2)(−a2)×b2+(−ab)×(−ab)]⇒[a2b2−a2b2ab3−ab3−a3b+a3b−a2b2+a2b2]⇒[0000].\Rightarrow A^2 = \begin{bmatrix} ab & b^2 \ -a^2 & -ab \end{bmatrix} \times \begin{bmatrix} ab & b^2 \ -a^2 & -ab \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} ab \times ab + b^2 \times (-a^2) & ab \times b^2 + b^2 \times (-ab) \ (-a^2) \times ab + (-ab) \times (-a^2) & (-a^2) \times b^2 + (-ab) \times (-ab) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} a^2 b^2 -a^2 b^2 & a b^3 -a b^3 \ -a^3 b +a^3 b & -a^2 b^2 +a^2 b^2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}.⇒A2=[ab−a2b2−ab]×[ab−a2b2−ab]⇒[ab×ab+b2×(−a2)(−a2)×ab+(−ab)×(−a2)ab×b2+b2×(−ab)(−a2)×b2+(−ab)×(−ab)]⇒[a2b2−a2b2−a3b+a3bab3−ab3−a2b2+a2b2]⇒[0000].
Hence, proved that A2 = 0.
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