Given-A-1-2-1-2-a-b-If-A-3-A-2-then-2a-3b-

Question Number 166377 by cortano1 last updated on 19/Feb/22

Given A = (\begin{matrix} \frac{1}{2} \frac{1}{2} \\ a b \end{matrix}) . If A^{3} = A^{2}

then 2 a - 3 b = ?

Answered by bobhans last updated on 19/Feb/22

A^{2} = (\begin{matrix} \frac{1}{2} \frac{1}{2} \\ a b \end{matrix}) (\begin{matrix} \frac{1}{2} \frac{1}{2} \\ a b \end{matrix}) = (\begin{matrix} \frac{1}{4} + \frac{1}{2} a \frac{1}{4} + \frac{1}{2} b \\ \frac{1}{2} a + ab \frac{1}{2} a + b^{2} \end{matrix})

A^{3} = (\begin{matrix} \frac{1}{4} + \frac{1}{2} a \frac{1}{4} + \frac{1}{2} b \\ \frac{1}{2} a + ab \frac{1}{2} a + b^{2} \end{matrix}) (\begin{matrix} \frac{1}{2} \frac{1}{2} \\ a b \end{matrix})

= (\begin{matrix} \frac{1}{8} + \frac{1}{2} a + \frac{1}{2} ab \frac{1}{8} + \frac{1}{4} a + \frac{1}{4} b + \frac{1}{2} b^{2} \\ \frac{1}{4} a + \frac{1}{2} ab + \frac{1}{2} a^{2} + {ab}^{2} \frac{1}{4} a + \frac{1}{2} ab + \frac{1}{2} a^{2} + b^{3} \end{matrix})

{\begin{cases} \frac{1}{4} + \frac{1}{2} a = \frac{1}{8} + \frac{1}{2} a + \frac{1}{2} ab \Rightarrow ab = \frac{1}{4} \\ \frac{1}{4} + \frac{1}{2} a = \frac{1}{8} + \frac{1}{4} a + \frac{1}{4} b + \frac{1}{2} b^{2} \Rightarrow \frac{1}{8} + \frac{1}{4} a = \frac{1}{4} b + \frac{1}{2} b^{2} \end{cases}

\Rightarrow 1 + 2 a = 2 b + {4 b}^{2} \land a = \frac{1}{4 b}

\Rightarrow 1 + \frac{1}{2 b} = 2 b + {4 b}^{2}; {8 b}^{3} + {4 b}^{2} - 2 b - 1 = 0

\Rightarrow (2 b + 1) ({4 b}^{2} - 1) = 0

\Rightarrow {(2 b + 1)}^{2} (2 b - 1) = 0 \to {\begin{cases} b = - \frac{1}{2} \Rightarrow a = - \frac{1}{2} \\ b = \frac{1}{2} \Rightarrow a = \frac{1}{2} \end{cases}

∴ 2 a - 3 b = {\begin{cases} - 1 + \frac{3}{2} = \frac{1}{2} \\ 1 - \frac{3}{2} = - \frac{1}{2} \end{cases}

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