Question

Considerando que A, B, C e X - 1/2AB = 2C, o valor do determinante de X é:

a) 285/2
b) 325/2
c) 335/2
d) 245/2
e) 315/2​

Answer 1

$X-\frac{1}{2}AB~=~2C\\\\\\X~=~2C+\frac{1}{2}AB\\\\\\\left[\begin{array}{ccc}x_{11}&x_{12}\\x_{21}&x_{22}\end{array}\right]~=~2\,.\,\left[\begin{array}{ccc}7&0\\-3&6\end{array}\right]+\frac{1}{2}\,.\,\left[\begin{array}{ccc}1&-2\\4&3\end{array}\right] \,.\,\left[\begin{array}{ccc}2&-3\\-4&1\end{array}\right]$

$\left[\begin{array}{ccc}x_{11}&x_{12}\\x_{21}&x_{22}\end{array}\right]~=~2\,.\,\left[\begin{array}{ccc}7&0\\-3&6\end{array}\right]+\frac{1}{2}\,.\,\left[\begin{array}{ccc}1\,.\,2+(-2)\,.\,(-4)&1\,.\,(-3)+(-2)\,.\,1\\4\,.\,2+3\,.\,(-4)&4\,.\,(-3)+3\,.\,1\end{array}\right]$

$\left[\begin{array}{ccc}x_{11}&x_{12}\\x_{21}&x_{22}\end{array}\right]~=~2\,.\,\left[\begin{array}{ccc}7&0\\-3&6\end{array}\right]+\frac{1}{2}\,.\,\left[\begin{array}{ccc}10&-5\\-4&-9\end{array}\right]\\\\\\\\\left[\begin{array}{ccc}x_{11}&x_{12}\\x_{21}&x_{22}\end{array}\right]~=~\left[\begin{array}{ccc}2\,.\,7&2\,.\,0\\2\,.\,(-3)&2\,.\,6\end{array}\right]+\left[\begin{array}{ccc}\frac{1}{2}\,.\,10&\frac{1}{2}\,.\,(-5)\\\frac{1}{2}\,.\,(-4)&\frac{1}{2}\,.\,(-9)\end{array}\right]$

$\left[\begin{array}{ccc}x_{11}&x_{12}\\x_{21}&x_{22}\end{array}\right]~=~\left[\begin{array}{ccc}14&0\\-6&12\end{array}\right]+\left[\begin{array}{ccc}5&-\frac{5}{2}\\-2&-\frac{9}{2}\end{array}\right]\\\\\\\\\left[\begin{array}{ccc}x_{11}&x_{12}\\x_{21}&x_{22}\end{array}\right]~=~\left[\begin{array}{ccc}14+5&0+\left(-\frac{5}{2}\right)\\-6+(-2)&12+\left(-\frac{9}{2}\right)\end{array}\right]$

$\left[\begin{array}{ccc}x_{11}&x_{12}\\x_{21}&x_{22}\end{array}\right]~=~\left[\begin{array}{ccc}19&-\frac{5}{2}\\-8&\frac{15}{2}\end{array}\right]$

$det(X)~=~\left|\begin{array}{ccc}19&-\frac{5}{2}\\-8&\frac{15}{2}\end{array}\right|\\\\\\det(X)~=~\left(19~.~\frac{15}{2}\right)~-~\left(-\frac{5}{2}~.~(-8)\right)\\\\\\det(X)~=~\left(\frac{19~.~15}{2}\right)~-~\left(\frac{-5~.~(-8)}{2}\right)\\\\\\det(X)~=~\left(\frac{285}{2}\right)~-~\left(\frac{40}{2}\right)\\\\\\det(X)~=~\frac{285-40}{2}\\\\\\\boxed{det(X)~=~\frac{245}{2}}$