Question

Determine os valores para as incógnitas de cada situação, sabendo que:

Answer 1

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$\tt a)~\begin{pmatrix}\sf x\\\sf y\\\sf z\end{pmatrix}+\begin{pmatrix}\sf 3\\\sf-1\\\sf5\end{pmatrix}=\begin{pmatrix}\sf10\\\sf-4\\\sf5\end{pmatrix}\\\sf x+3=10\implies x=10-3\implies\boxed{\boxed{\boxed{\boxed{\sf x=7\checkmark}}}}}\\\sf y-1=-4\implies y=1-4\implies\boxed{\boxed{\boxed{\boxed{\sf y=-3\checkmark}}}}\\\sf z+5=5\implies z=5-5\implies \boxed{\boxed{\boxed{\boxed{\sf z=0\checkmark}}}}$

$\tt b)\begin{pmatrix}\sf x&\sf y\\\sf3&\sf z\end{pmatrix}+\begin{pmatrix}\sf x&\sf3\\\sf t&\sf z\end{pmatrix}=\begin{pmatrix}\sf10&\sf-1\\\sf1&\sf18\end{pmatrix}\\\sf 2x=10\\\sf x=\dfrac{10}{2}\\\huge\boxed{\boxed{\boxed{\boxed{\sf x=5}}}}\\\sf y+3=-1\\\sf y=-3-1\\\huge\boxed{\boxed{\boxed{\boxed{\sf y=-4}}}}\\\sf3+t=1\\\sf t=1-3\\\huge\boxed{\boxed{\boxed{\boxed{\sf t=-2}}}}\\\sf2z+z=18\\\sf3z=18\\\sf z=\dfrac{18}{3}\\\huge\boxed{\boxed{\boxed{\boxed{\sf z=6}}}}$

$\tt c)\begin{pmatrix}\sf x&\sf6\\\sf1&\sf2z\end{pmatrix}\cdot \begin{pmatrix}\sf -x&\sf4\\\sf-3&\sf z\end{pmatrix}=\begin{pmatrix}\sf12&\sf y\\\sf4&\sf -1\end{pmatrix}\\\begin{pmatrix}\sf-x^2-18&\sf4x+6z\\\sf-x-6z&\sf4+2z^2\end{pmatrix}=\begin{pmatrix}\sf12&\sf y\\\sf4&\sf-1\end{pmatrix}$

$\sf 4+2z^2=-1\\\sf 2z^2=-4-1\\\sf 2z^2=-5\implies \not\exists~z\in\mathbb{R}$

$\sf\not\exists~x,y\in\mathbb{R}$

$\tt d)\begin{pmatrix}\sf~a+b&\sf b+c\\\sf2b&\sf2a-3d\end{pmatrix}=\begin{pmatrix}\sf9&\sf-1\\\sf6&\sf18\end{pmatrix}\\\begin{cases}\sf a+b=9\\\sf b+c=-1\\\sf2b=6\\\sf2a-3d=18\end{cases}$

$\sf 2b=6\implies b=\dfrac{6}{2}=\boxed{\boxed{\boxed{\boxed{\sf b=3}}}}\\\sf a+b=9\\\sf a+3=9\\\sf a=9-3\\\sf\boxed{\boxed{\boxed{\boxed{\sf a=6}}}}\\\sf b+c=-1\\\sf 3+c=-1\\\sf c=-3-1\\\boxed{\boxed{\boxed{\boxed{\sf c=-4}}}}\\\sf 2a-3d=18\\\sf2\cdot6-3d=18\\\sf 12-3d=18\\\sf 3d=12-18\\\sf 3d=-6\\\sf d=-\dfrac{6}{3}\\\boxed{\boxed{\boxed{\boxed{\sf d=-2}}}}$