Matemática, perguntado por ryuuyuuki2908, 8 meses atrás

Resolva a equação matricial:

Anexos:

Soluções para a tarefa

Respondido por CyberKirito
1

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\boxed{\begin{array}{l}\begin{bmatrix}\sf3&\sf1\\\sf-1&\sf2\end{bmatrix}\cdot\begin{bmatrix}\sf x&\sf y\\\sf z&\sf t\end{bmatrix}=\begin{bmatrix}\sf6&\sf5\\\sf5&\sf-4\end{bmatrix}\\\begin{bmatrix}\sf3x+z&\sf3y+t\\\sf-x+2z&\sf-y+2t\end{bmatrix}=\begin{bmatrix}\sf6&\sf5\\\sf5&\sf-4\end{bmatrix}\end{array}}

\boxed{\begin{array}{l}\begin{cases}\sf3x+z=6\\\sf-x+2z=5\end{cases}\\\begin{bmatrix}\sf3&\sf1\\\sf-1&\sf2\end{bmatrix}\cdot\begin{bmatrix}\sf x\\\sf z\end{bmatrix}=\begin{bmatrix}\sf6\\\sf5\end{bmatrix}\\\sf A=\begin{bmatrix}\sf3&\sf1\\\sf-1&\sf2\end{bmatrix}\\\sf det~A=6+1=7\\\sf A_x=\begin{bmatrix}\sf6&\sf1\\\sf5&\sf2\end{bmatrix}\\\sf det~A_x=12-5=7\\\sf A_z=\begin{bmatrix}\sf3&\sf6\\\sf-1&\sf5\end{bmatrix}\\\sf det~A_z=15+6=21\\\sf x=\dfrac{det~A_x}{det~A}=\dfrac{7}{7}=1\end{array}}

\boxed{\begin{array}{l}\sf z=\dfrac{det~A_z}{det~A}\\\\\sf z=\dfrac{21}{7}=3\end{array}}

\boxed{\begin{array}{l}\begin{cases}\sf3y+t=5\\\sf-y+2t=-4\end{cases}\\\begin{bmatrix}\sf3&\sf1\\\sf-1&\sf2\end{bmatrix}\cdot\begin{bmatrix}\sf y\\\sf t\end{bmatrix}=\begin{bmatrix}\sf 5\\\sf-4\end{bmatrix}\\\sf M=\begin{bmatrix}\sf3&\sf1\\\sf-1&\sf2\end{bmatrix}\\\sf det~M=6+1=7\\\sf M_y=\begin{bmatrix}\sf5&\sf1\\\sf-4&\sf2\end{bmatrix}\\\sf det~M_y=10+4=14\\\sf M_t=\begin{bmatrix}\sf3&\sf5\\\sf-1&\sf-4\end{bmatrix}\\\sf det~M_t=-12+5=-7\end{array}}

\boxed{\begin{array}{l}\sf y=\dfrac{det~M_y}{det~M}=\dfrac{14}{7}=2\\\\\sf t=\dfrac{det~M_t}{det~M}\\\\\sf t=\dfrac{-7}{7}\\\sf t=-1\end{array}}

\Large\boxed{\begin{array}{l}\huge\boxed{\boxed{\boxed{\boxed{\sf x=1}}}}\\\huge\boxed{\boxed{\boxed{\boxed{\sf z=3}}}}\\\huge\boxed{\boxed{\boxed{\boxed{\sf y=2}}}}\\\huge\boxed{\boxed{\boxed{\boxed{\sf t=-1}}}}\end{array}}

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