Question

preciso urgente
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Answer 1

Resposta:

Explicação passo a passo:

Exercício 1:

$A=\left[\begin{array}{cc}-3&-1\\4&-1\end{array}\right]\quad\quad\quad B=\left[\begin{array}{cc}6&-5\\0&1\end{array}\right]$

a) $A+B$

$A + B= \left[\begin{array}{cc}-3&-1\\4&-1\end{array}\right]+\left[\begin{array}{cc}6&-5\\0&1\end{array}\right]\\A+B= \left[\begin{array}{cc}-3+6&-1-5\\4+0&-1+1\end{array}\right]\\A+B= \left[\begin{array}{cc}3&-6\\4&0\end{array}\right]$

b) $A + A^t$

$A + A^t= \left[\begin{array}{cc}-3&-1\\4&-1\end{array}\right]+\left[\begin{array}{cc}-3&4\\-1&-1\end{array}\right]\\A+B= \left[\begin{array}{cc}-3-3&-1+4\\4-1&-1-1\end{array}\right]\\A+B= \left[\begin{array}{cc}-6&3\\3&-2\end{array}\right]$

c) $A-B$

$A + (-B)= \left[\begin{array}{cc}-3&-1\\4&-1\end{array}\right]+\left[\begin{array}{cc}-6&5\\0&-1\end{array}\right]\\A+(-B)= \left[\begin{array}{cc}-3-6&-1+5\\4-0&-1-1\end{array}\right]\\A+(-B)= \left[\begin{array}{cc}-9&4\\4&-2\end{array}\right]$

d) $A \cdot B$

$A \cdot B= \left[\begin{array}{cc}-3&-1\\4&-1\end{array}\right]\cdot\left[\begin{array}{cc}6&-5\\0&1\end{array}\right]\\A\cdot B= \left[\begin{array}{cc}(-3)\cdot 6+(-1)\cdot0&(-3)\cdot(-5)+(-1)\cdot1\\4\cdot6+(-1)\cdot0&4\cdot(-5)+(-1)\cdot1\end{array}\right]\\A\cdot B= \left[\begin{array}{cc}-18+0&15-1\\24+0&-20-1\end{array}\right]\\A\cdot B= \left[\begin{array}{cc}-18&14\\24&-21\end{array}\right]$

Exercício 2:

$A=\left[\begin{array}{cc}1&-1\\2&-3\end{array}\right]\quad\quad\quad B=\left[\begin{array}{ccc}2&-5&0\\7&1&-2\end{array}\right]\quad\quad\quad C=A\cdot B$

$C=A \cdot B= \left[\begin{array}{cc}1&-1\\2&-3\end{array}\right]\cdot\left[\begin{array}{ccc}2&-5&0\\7&1&-2\end{array}\right]\\C=A\cdot B= \left[\begin{array}{ccc}1\cdot 2+(-1)\cdot7&1\cdot(-5)+(-1)\cdot1&1\cdot0+(-1)\cdot(-2)\\2\cdot2+(-3)\cdot7&2\cdot(-5)+(-3)\cdot1&2\cdot0+(-3)\cdot(-2)\end{array}\right]\\C=A\cdot B= \left[\begin{array}{ccc}2+(-7)&(-5)+(-1)&0+2\\4+(-21)&(-10)+(-3)&0+6\end{array}\right]\\C=A\cdot B= \left[\begin{array}{ccc}-5&-6&2\\-17&-13&6\end{array}\right]$

a) $c_{12} = -6$

b) $c_{21}=-17$

c) $c_{23}=6$