Question

resolva o sistema usando a regra de cremer​

Answer 1

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$\tt a)\\\begin{cases}\sf x-y=1\\\sf2x-y=3\end{cases}\\\tt A=\begin{bmatrix}\sf1&\sf-1\\\sf2&\sf-1\end{bmatrix}\\\sf det~A=1\cdot(-1)-(-1)\cdot2=-1+2=1\\\tt A_x=\begin{bmatrix}\sf1&\sf-1\\\sf3&\sf-1\end{bmatrix}\\\sf det~A_x=-1+3=2\\\tt A_y=\begin{bmatrix}\sf1&\sf1\\\sf2&\sf3\end{bmatrix}\\\sf det~A_y=3-2=1\\\sf x=\dfrac{det~A_x}{det~A}=\dfrac{2}{1}=2\\\sf y=\dfrac{det~A_y}{det~A}=\dfrac{1}{1}=1\\\sf s=\left\{2,1\right\}\checkmark$

$\tt b)~\begin{cases}\sf x-y+z=1\\\sf2x-y+z=3\\\sf x-2y+z=5\end{cases}\\\tt A=\begin{bmatrix}\sf1&\sf-1&~~\sf1\\\sf2&\sf-1&~~\sf1\\\sf1&\sf-2&~~\sf1\end{bmatrix}\\\sf det~A=1\cdot(-1+2)+1\cdot(2-1)+1\cdot(-4+1)\\\sf det~A=1+1-3=-1\\\tt A_x=\begin{bmatrix}\sf1&\sf-1&~~\sf1\\\sf3&\sf-1&~~\sf1\\\sf5&\sf-2&~~\sf1\end{bmatrix}\\\sf det~A_x=1\cdot(-1+2)+1\cdot(3-5)+1\cdot(-6+5)\\\sf det~A_x=\diagup\!\!\!1-2-\diagup\!\!\!1=-2\\\tt A_y=\begin{bmatrix}\sf1&\sf1&\sf1\\\sf2&\sf3&\sf1\\\sf1&\sf5&\sf1\end{bmatrix}$

$\sf det~A_y=1\cdot(3-5)-1\cdot(2-1)+1\cdot(10-3)\\\sf det~A_y=-2-1+7=4\\\tt A_z=\begin{bmatrix}\sf1&\sf-1&~~\sf1\\\sf2&\sf-1&~~\sf3\\\sf1&\sf-2&~~\sf5\end{bmatrix}\\\sf det~A_z=1\cdot(-5+6)+1\cdot(10-3)+1\cdot(-4+1)\\\sf det~A_z=1+7-3=5$

$\sf x=\dfrac{det~A_x}{det~A}\\\sf x=\dfrac{-2}{-1}\\\huge\boxed{\boxed{\boxed{\boxed{\sf x=2\checkmark}}}}\\\sf y=\dfrac{det~A_y}{det~A}\\\sf y=\dfrac{4}{-1}\\\huge\boxed{\boxed{\boxed{\boxed{\sf y=-4\checkmark}}}}\\\sf z=\dfrac{det~A_z}{det~A}\\\sf z=\dfrac{5}{-1}\\\huge\boxed{\boxed{\boxed{\boxed{\sf z=-5\checkmark}}}}$

$\Huge\boxed{\boxed{\boxed{\boxed{\sf s=\{2,-4,-5\}}}}}$