Question

Estou estudando Matrizes e determinantes, no exercício me mostra isso
1 0
2 4
E tá pedindo pra calcular A-1
Esse -1 tá canto aonde fica essa bolinha assim: A°

Answer 1

A⁻¹ é a matriz inversa de A com a propriedade:

A x A⁻¹ = I    (I é a matriz identidade)

Logo:

$\left[\begin{array}{ccc}1&0\\2&4\end{array}\right] * \left[\begin{array}{ccc}a&b\\c&d\end{array}\right] = \left[\begin{array}{ccc}1&0\\0&1\end{array}\right] \\ \\ \\ \left[\begin{array}{ccc}a&b\\2a+4c&2b+4c\end{array}\right] =\left[\begin{array}{ccc}1&0\\0&1\end{array}\right] \\ \\ \\ a=1\\ b=0\\ c=-\frac{1}{2}\\ d=-\frac{1}{4}\\ \\ A^{-1}= \left[\begin{array}{ccc}1&0\\-\frac{1}{2}&-\frac{1}{4}\end{array}\right] \\ \\ det(A^{-1})=-\frac{-1}{4}-0=-\frac{1}{4}$

Answer 2

Seja $A^{-1}=\left(\begin{array}{cc}a_1&a_2\\a_3&a_4\end{array}\right)$.

Temos que, $A\cdot\ A^{-1}=I_n$, ou seja:

$\left(\begin{array}{cc}1&0\\2&4\end{array}\right)\cdot\left(\begin{array}{cc}a_1&a_2\\a_3&a_4\end{array}\right)=\left(\begin{array}{cc}1&0\\0&1\end{array}\right)$

Assim:

$\left(\begin{array}{cc}1\cdot\ a_1+0\cdot\ a_3&1\cdot\ a_2+0\cdot\ a_3\\2\cdot\ a_1+4\cdot\ a_3&2\cdot\ a_2+4\cdot\ a_4\end{array}\right)=\left(\begin{array}{cc}1\cdot\ a_1&1\cdot\ a_2\\2\cdot\ a_1+4\cdot\ a_3&2\cdot\ a_2+4\cdot\ a_4\end{array}\right)$

$=\left(\begin{array}{cc}1&0\\0&1\end{array}\right)$

Com isso, 

$\boxed{a_1=1}$

$\boxed{a_2=0}$

$2\cdot1+4a_3=0$

$2+4a_3=0~~\Rightarrow~~4a_3=-2~~\Rightarrow~~a_3=-\dfrac{2}{4}~~\Rightarrow~~\boxed{a_3=-\dfrac{1}{2}}$

$2\cdot0+4a_4=1$

$0+4a_4=1~~\Rightarrow~~\boxed{a_4=\dfrac{1}{4}}$.

Assim, $\boxed{A^{-1}=\left(\begin{array}{cc}1&0\\-\frac{1}{2}&\frac{1}{4}\end{array}\right)}$.

Dada uma matriz quadrada $M$ de ordem $2$:

$M=\left[\begin{array}{cc}a_1&a_2\\a_3&a_4\end{array}\right]$

Teremos que, $\text{Det}_M=a_1\cdot\ a_4-a_2\cdot\ a_3$.

Deste modo, sendo $A^{-1}=\left(\begin{array}{cc}1&0\\-\frac{1}{2}&\frac{1}{4}\end{array}\right)$, temos:

$\boxed{\text{Det}_{A^{-1}}=1\cdot\dfrac{1}{4}-0\cdot\dfrac{-1}{2}=\dfrac{1}{4}}$.