Question

Utilizando o método de Cramer, determine a solução dos sistemas a seguir

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Answer 1

Resposta:

a)

$X=\dfrac{D_{X}}{D} =\dfrac{15}{1} = 15\\\\Y=\dfrac{D_{Y}}{D} =\dfrac{-11}{1} = -11\\\\Z=\dfrac{D_{Z}}{D} =\dfrac{-8}{1} = -8$

b)

$X=\dfrac{D_{X}}{D} =\dfrac{-13}{-3} = \dfrac{13}{3}\\\\Y=\dfrac{D_{Y}}{D} =\dfrac{-9}{-3} = 3\\\\Z=\dfrac{D_{Z}}{D} =\dfrac{-1}{-3} = \dfrac{1}{3}$

Explicação passo-a-passo:

Vamos calcular o Determinante e os Determinantes x, y e z

a)

$D = \left[\begin{array}{ccc}\ 2&1&\ \ 1\\\ \ 1&2&-2\\-1&3&-6\end{array}\right] \begin{array}{cc}\ 2&1\\\ \ 1&2\\-1&3\end{array}\right] \\\\Diagonais\ principais = 2.2.-6 + 1.-2.-1 + 1.1.3 = -24 + 2 + 3\\Diagonais\ secund\'{a}rias = - (-1).2.1 - 3.-2-2 - (-6).1.1 = +2 +12 + 6\\D = -24 + 2 + 3 +2 -12 + 6 = 1\\\\D_{X} = \left[\begin{array}{ccc}\ 11&1&\ \ 1\\\ \ 9&2&-2\\\ \ 0&3&-6\end{array}\right] \begin{array}{cc}\ 11&1\\\ \ 9&2\\ \ \ 0&3\end{array}\right]$

$Diagonais\ principais = 11.2.-6 + 1.-2.0 + 1.9.3 = -132 + 0 + 27\\Diagonais\ secund\'{a}rias = -0.2.1 - 3.-2.11 - (-6).9.1 = +0 +66 + 54\\\\D_{X} = -132 + 0 + 27 +0 +66 + 54 = 15\\\\D_{Y} = \left[\begin{array}{ccc}\ 2&11&\ \ 1\\\ \ 1&\ \ 9&-2\\-1&\ \ 0&-6\end{array}\right] \begin{array}{cc}\ 2&11\\\ \ 1&\ \ 9\\-1&\ \ 0\end{array}\right]\\Diagonais\ principais = 2.9.-6 + 11.-2.-1 + 1.1.0 = -108 + 22 + 0\\Diagonais\ secund\'{a}rias = - (-1).9.1 - 0.-2.2 - (-6).1.11 = +9 +0 + 66$

$D_{Y} = -108 + 22 + 0 +9 +0 + 66 = -11\\\\D_{Z} = \left[\begin{array}{ccc}\ 2&1&\ 11\\\ \ 1&2&\ \ 9\\-1&3&\ \ 0\end{array}\right] \begin{array}{cc}\ 2&1\\\ \ 1&2\\-1&3\end{array}\right]\\\\Diagonais\ principais = 2.2.0 + 1.9.-1 + 11.1.3 = 0 - 9 + 33\\Diagonais\ secund\'{a}rias = - (-1).2.11 - 3.9.2 - 0.1.1 = +22 -54 + 0\\D_{Z}= 0-9+33+22-54+0=-8\\\\X=\dfrac{D_{X}}{D} =\dfrac{15}{1} = 15\\\\Y=\dfrac{D_{Y}}{D} =\dfrac{-11}{1} = -11\\\\Z=\dfrac{D_{Z}}{D} =\dfrac{-8}{1} = -8$

b)

$D = \left[\begin{array}{ccc}1&\ \ 1&\ \ 2\\1&-1&-1\\2&-1&\ \ 1\end{array}\right] \begin{array}{cc}1&\ \ 1\\1&-1\\2&-1\end{array}\right] \\\\Diagonais\ principais = 1.-1.1 + 1.-1.2 + 2.1.-1 = -1 -2 -2\\Diagonais\ secund\'{a}rias = - 2.-1.2 - (-1).-1.1 - 1.1.1 = +4 -1 -1\\D = -1 - 2 -2 + 4 -1 -1 = -3\\\\D_{X} =\left[\begin{array}{ccc}8&\ \ 1&\ \ 2\\1&-1&-1\\6&-1&\ \ 1\end{array}\right] \begin{array}{cc}8&\ \ 1\\1&-1\\6&-1\end{array}\right]$

$Diagonais\ principais = 8.-1.1 + 1.-1.6 + 2.1.-1 = -8 -6 -2\\Diagonais\ secund\'{a}rias = - 6.-1.2 - (-1).-1.8 - 1.1.1 = +12 -8 -1\\D_{X} = -8 - 6 -2 + 12 -8 -1 = -13\\\\D_{Y} =\left[\begin{array}{ccc}1&\ \ 8&\ \ 2\\1&\ \ 1&-1\\2&\ \ 6&\ \ 1\end{array}\right] \begin{array}{cc}1&\ \ 8\\1&\ \ 1\\2&\ \ 6\end{array}\right]\\\\Diagonais\ principais = 1.1.1 + 8.-1.2 + 2.1.6 = 1 -16 +12\\Diagonais\ secund\'{a}rias = - 2.1.2 - 6.-1.1 - 1.1.8 = -4 +6 -8\\D_{Y} = 1-16+12-4+6-8 = -9$

$D_{Z} = \left[\begin{array}{ccc}1&\ \ 1&\ \ 8\\1&-1&\ \ 1\\2&-1&\ \ 6\end{array}\right] \begin{array}{cc}1&\ \ 1\\1&-1\\2&-1\end{array}\right] \\\\Diagonais\ principais = 1.-1.6 + 1.1.2 + 8.1.-1 = -6 +2 -8\\Diagonais\ secund\'{a}rias = - 2.-1.8 - (-1).-1.1 - 6.1.1 = +16 +1 -6\\D_{Z} = -6+2-8+16+1-6= -1\\\\\\X=\dfrac{D_{X}}{D} =\dfrac{-13}{-3} = \dfrac{13}{3}\\\\Y=\dfrac{D_{Y}}{D} =\dfrac{-9}{-3} = +3\\\\Z=\dfrac{D_{Z}}{D} =\dfrac{-1}{-3} = \dfrac{1}{3}$

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