Question

A matriz   A2X2  é definida por aij= $\left \{ {{13-6i, se, i=j} \atop {-j-1, se, i \neqj }} \right.$ .
O valor do determinante da matriz $( A^{-1}) {t}$ , é :
a) -2
b)-1
c)$\frac{1}{2}$
d)1
e)2

Answer 1

$\begin{vmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{vmatrix}\\\\ \texttt{Quando i = j fica :}\\\\ x_{11} = 13 - 6*1 = 7\\\\ x_{22} = 13 - 6*2 = 1\\\\ \texttt{Quando i e diferente de j fica :}\\\\ x_{12} = -2 - 1 = -3\\\\ x_{21} = -1 - 1 = -2\\\\ \begin{vmatrix} 7 & -3\\ -2&1 \end{vmatrix}\\\\ A^{-1} = \begin{vmatrix} a & c \\ b& d \end{vmatrix}\\\\\begin{vmatrix}a & c\\  b& d\end{vmatrix}= \begin{vmatrix}7 & -3\\  -2&1 \end{vmatrix} = \begin{vmatrix}1 &0 \\  0& 1\end{vmatrix}$
$\begin{cases} & \text{ } 7a - 2b = 1 \\ & \text{ } -3a + c = 0 \to c = 3a \end{cases}\\\\ 7a - 6a = 1\\\\ \boxed{a = 1; c = 3}\\\\ \begin{cases} & \text{ } 7b - 2d = 0 \\ & \text{ } -3b + d = 1 \to d = 1 + 3b \end{cases}\\\\ 7b - 2(1 + 3b) = 0\\\\ \boxed{b = 2;d = 7}\\\\ A^{-1} = \begin{vmatrix} 1 &3 \\ 2& 7 \end{vmatrix}\\\\ (A^{-1})^{t} = \begin{bmatrix} 1 &2 \\ 3& 7 \end{bmatrix}\\\\ D = -6 + 7\\\\ \boxed{\boxed{D = 1}}$