Question

Alguém pode me ajudar com esta questão de Matrix?

Answer 1

1)

$\mathbf{A}=\left( \begin{array}{cc} -1&2\\ 3&4\\ 0&1 \end{array} \right ),\;\;\;\mathbf{B}=\left( \begin{array}{ccc} 3&4&1\\ 0&0&2 \end{array} \right )$


a) $\mathbf{M}=\mathbf{A}\cdot\mathbf{B}$

$\mathbf{M}=\left( \begin{array}{cc} -1&2\\ 3&4\\ 0&1 \end{array} \right )\cdot\left( \begin{array}{ccc} 3&4&1\\ 0&0&2 \end{array} \right )$


A matriz $\mathbf{M}$ terá o número de linhas de $\mathbf{A}$ e o número de colunas de $\mathbf{B}$. Logo, $\mathbf{M}$ é uma matriz $3\times 3$:

$\mathbf{M}=\left( \begin{array}{cc} -1&2\\ 3&4\\ 0&1 \end{array} \right )\cdot\left( \begin{array}{ccc} 3&4&1\\ 0&0&2 \end{array} \right )\\ \\ \\ \mathbf{M}=\left( \begin{array}{rrr} -1\cdot 3+2\cdot 0\;&\;-1\cdot 4+2\cdot 0\;&\;-1\cdot 1+2\cdot 2\\ 3\cdot 3+4\cdot 0\;&\;3\cdot 4+4\cdot 0\;&\;3\cdot 1+4\cdot 2\\ 0\cdot 3+1\cdot 0\;&\;0\cdot 4+1\cdot 0\;&\;0\cdot 1+1\cdot 2 \end{array} \right )\\ \\ \\ \mathbf{M}=\left( \begin{array}{rrr} -3+0\;&\;-4+0\;&\;-1+4\\ 9+0\;&\;12+0\;&\;3+8\\ 0+0\;&\;0+0\;&\;0+2 \end{array} \right )$


$\mathbf{M}=\left( \begin{array}{ccc} -3&-4&3\\ 9&12&11\\ 0&0&2 \end{array} \right )$


b) $\mathrm{det\,}\mathbf{M}$

$=\mathrm{det}\left( \begin{array}{ccc} -3&-4&3\\ 9&12&11\\ 0&0&2 \end{array} \right )\\ \\ \\ =\begin{array}{lll} -3\cdot 12\cdot 2&-4\cdot 11\cdot 0&+3\cdot 9\cdot 0\\ -0\cdot 12\cdot 3&-0\cdot 11\cdot \left(-3 \right )&-2\cdot 9\cdot \left(-4 \right ) \end{array}\\ \\ \\ =\begin{array}{lll} -72&-0&+0\\ -0&-0&+72 \end{array}\\ \\ \\ =0\\ \\ \\ \mathrm{det\,}\mathbf{M}=0$


2) $\overset{\to}{\mathbf{x}}=\left(2,\,1 \right ),\;\;\;\overset{\to}{\mathbf{y}}=\left(-2,\,4 \right )$


a) $\overset{\to}{\mathbf{z}}=\overset{\to}{\mathbf{x}}-2\overset{\to}{\mathbf{y}}$

$\overset{\to}{\mathbf{z}}=\left(2,\,1 \right )-2\cdot \left(-2,\,4 \right )\\ \\ \overset{\to}{\mathbf{z}}=\left(2,\,1 \right )+\left(4,\,-8 \right )\\ \\ \overset{\to}{\mathbf{z}}=\left(2+4,\,1-8 \right )\\ \\ \overset{\to}{\mathbf{z}}=\left(6,\,-7 \right )$


b) $\overset{\to}{\mathbf{x}}\cdot \overset{\to}{\mathbf{y}}$

$=\left(2,\,1 \right )\cdot \left(-2,\,4 \right )\\ \\ =2\cdot \left(-2 \right )+1\cdot 4\\ \\ =-4+4\\ \\ =0$