Dissertativa 8 - 2ª Fase - Dia 2 - Unicamp 2026

a) 

I. Considerando $\mathrm{A} = \left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right]$, então $\left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right] · \left[\begin{array}{cc}\mathrm{a}& \mathrm{d}\\ \mathrm{c}& \mathrm{d}\end{array}\right] = \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ executando a multiplicação de matrizes pode-se construir o seguinte sistema: $\left\{\begin{array}{l}{\mathrm{a}}^2 + \mathrm{b} · \mathrm{c} = 1\\ \mathrm{a} · \mathrm{b} + \mathrm{bd} = 0\\ \mathrm{a} · \mathrm{c} + \mathrm{d} · \mathrm{c} = 0\\ \mathrm{b} · \mathrm{c} + {\mathrm{d}}^2 = 1\end{array}\right.$

II. Considerando $\mathrm{b} \ne 0$ pode-se concluir que $\mathrm{a} · \mathrm{b} + \mathrm{b} · \mathrm{d} = 0 \to \mathrm{a} · \mathrm{b} = -\mathrm{b} · \mathrm{d} \to \mathrm{a} = -\mathrm{d} \to \mathrm{a} = \mathrm{d} = 0 \to \mathrm{b} · \mathrm{c} =1 \to \mathrm{b} = \mathrm{c} = 1$

Assim, tem-se uma possibilidade, $\mathrm{A} = \left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$.

III. Considerando $\mathrm{b} = 0 \to {\mathrm{a}}^2 = 1 \to \mathrm{a} = 1 \to \mathrm{c} = 0 \to \mathrm{d} = 1$, assim tem-se a segunda possibilidade,  $\boxed{\mathrm{A} = \left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]}$.

b) Sendo, $\mathrm{B} = \left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right]$, tem-se:

I. Do texto vem $\left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right] · \left[\begin{array}{c}1\\ 2\end{array}\right] = \left[\begin{array}{c}0\\ 1\end{array}\right] \to \left\{\begin{array}{l}\mathrm{a} + 2 · \mathrm{b} = 0\\ \mathrm{c} + 2 · \mathrm{d} = 1\end{array}\right.$

II. Do texto vem $\left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right] · \left[\begin{array}{c}3\\ 1\end{array}\right] = \left[\begin{array}{c}1\\ 0\end{array}\right] \to \left\{\begin{array}{l}3 · \mathrm{a} + \mathrm{b} = 1\\ 3\mathrm{c} +\mathrm{d} = 0\end{array}\right.$

Resolvendo os dois sistemas lineares é possível encontrar os elementos de B, que são:

$\mathrm{a} = \frac{2}{5}$; $\mathrm{b} =\frac{1}{-5}$; $\mathrm{c} = \frac{1}{-5}$ e $\mathrm{d} = \frac{3}{5}$, logo $\mathrm{B} = \left[\begin{array}{cc}\frac{2}{5}& \frac{1}{-5}\\ \frac{1}{-5}& \frac{3}{5}\end{array}\right]$

III. Cálculo de ${\mathrm{B}}^{-1}$:

Da definição de matriz inversa podemos encontrá-la através da seguinte equação matricial, $\mathrm{B} · \mathrm{B} -1 = \mathrm{Id}$.

Sendo ${\mathrm{B}}^{-1} = \left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right]$, vem:

$\left[\begin{array}{cc}\frac{2}{5}& \frac{1}{-5}\\ \frac{1}{-5}& \frac{3}{5}\end{array}\right] · \left[\begin{array}{cc}\mathrm{x}& \mathrm{y}\\ \mathrm{z}& \mathrm{w}\end{array}\right] = \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \to \left\{\begin{array}{l}2 · \mathrm{x} - \mathrm{z} = 5\\ -\mathrm{x} + 3 · \mathrm{z} = 0\\ 2 · \mathrm{y} - \mathrm{w} = 0\\ -\mathrm{y} + 3 · \mathrm{w} = 5\end{array}\right. \to \mathrm{x} = 3; \mathrm{y} = 1, \mathrm{z} = 1 \mathrm{e} \mathrm{w} = 2$, 

Assim, $\mathrm{B} = \left[\begin{array}{cc}3& 1\\ 1& 2\end{array}\right]$

IV. ${\mathrm{B}}^{-1} · \left[\begin{array}{c}2\\ 6\end{array}\right] = \left[\begin{array}{cc}3& 1\\ 1& 2\end{array}\right] · \left[\begin{array}{c}2\\ 6\end{array}\right] = \boxed{\left[\begin{array}{c}12\\ 14\end{array}\right]}$