Question

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)

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

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

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

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

$\mathsf{a)}$

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

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

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

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

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

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

$\mathsf{b)}$

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

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

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

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

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

$\mathsf{c)}$

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

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

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

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

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

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

$\mathsf{d)}$

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

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

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

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

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

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

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