Matemática, perguntado por CristianFischer, 5 meses atrás

Se A = \left[\begin{array}{cc}1&7&\\2&6\end{array}\right] , B = \left[\begin{array}{cc}2&1&\\4&3\end{array}\right] e C = \left[\begin{array}{cc}0&2&\\2&0\end{array}\right] , determine X em cada uma das euqações abaixo:

a) 2X + A = 3B + C
b) X + A = \frac{1}{2}(B - C)
c) 3X + A = B - X
d) \frac{1}{2} (X -A - B) = \frac{1}{3} (X - C)

Anexos:

Soluções para a tarefa

Respondido por auditsys
1

Resposta:

\textsf{Leia abaixo}

Explicação passo a passo:

\textsf{A =} \begin{bmatrix}\cancel1&\cancel7\\\cancel2&\cancel6\end{bmatrix}

\textsf{B =} \begin{bmatrix}\cancel2&\cancel1\\\cancel4&\cancel3\end{bmatrix}

\textsf{C =} \begin{bmatrix}\cancel0&\cancel2\\\cancel2&\cancel0\end{bmatrix}

\mathsf{a)}

\mathsf{\:2X + A = 3B + C}

\mathsf{\:2X  = 3B + C - A}

\mathsf{\:X  = \dfrac{3B + C - A}{2}}

\textsf{X =} \begin{bmatrix}\cancel(3(2) + 0 - 1)/2 &\cancel(3(1) + 2 - 7)/2\\\cancel(3(4) + 2 - 2)/2&\cancel(3(3) + 0 - 6)/2\end{bmatrix}

\textsf{X =} \begin{bmatrix}\cancel 5/2 &\cancel -2/2\\\cancel12/2&\cancel 3/2\end{bmatrix}

\textsf{X =} \begin{bmatrix}\cancel 5/2 &\cancel -1\\\cancel6&\cancel 3/2\end{bmatrix}

\mathsf{b)}

\mathsf{X + A = \dfrac{1}{2}(B - C)}

\mathsf{X = \dfrac{B - C - 2A}{2}}

\textsf{X =} \begin{bmatrix}\cancel (2 - 0 - 2(1))/2 &\cancel (1 - 2 - 2(7))/2\\\cancel (4 - 2 - 2(2))/2&\cancel (3 - 0 - 2(6))/2\end{bmatrix}

\textsf{X =} \begin{bmatrix}\cancel 0/2 &\cancel -15/2\\\cancel -2/2&\cancel -9/2\end{bmatrix}

\textsf{X =} \begin{bmatrix}\cancel 0/2 &\cancel -15/2\\\cancel -1&\cancel -9/2\end{bmatrix}

\mathsf{c)}

\mathsf{3X + A = B - X}

\mathsf{4X = B - A}

\mathsf{X = \dfrac{B - A}{4}}

\textsf{X =} \begin{bmatrix}\cancel (2 - 1)/4 &\cancel (1 - 7)/4\\\cancel (4 - 2)/4&\cancel (3 - 6)/4\end{bmatrix}

\textsf{X =} \begin{bmatrix}\cancel 1/4 &\cancel -6/4\\\cancel 2/4&\cancel -3/4\end{bmatrix}

\textsf{X =} \begin{bmatrix}\cancel 1/4 &\cancel -3/2\\\cancel 1/2&\cancel -3/4\end{bmatrix}

\mathsf{d)}

\mathsf{\dfrac{1}{2}(X - A - B) = \dfrac{1}{3}(X - C)}

\mathsf{3(X - A - B) = 2(X - C)}

\mathsf{3x - 3A - 3B = 2x - 2C}

\mathsf{X = 3A + 3B - 2C}

\textsf{X =} \begin{bmatrix}\cancel 3(1) + 3(2) - 2(0) &\cancel 3(7) + 3(1) - 2(2)\\\cancel 3(2) + 3(4) - 2(2)&\cancel 3(6) + 3(3) - 2(0)\end{bmatrix}

\textsf{X =} \begin{bmatrix}\cancel 3 + 6 - 0 &\cancel 21 + 3 - 4\\\cancel 6 + 12 - 4&\cancel 18 + 9 - 0\end{bmatrix}

\textsf{X =} \begin{bmatrix}\cancel 9 &\cancel 20\\\cancel 14&\cancel 27\end{bmatrix}

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