Question

) Sendo A = (
2 1
3 2
) e B = (
1 5
2 −2
), determine:

Answer 1

$A = \left[\begin{array}{cc}2&1\\3&2\\\end{array}\right] \\ \\ B = \left[\begin{array}{cc}1&5\\2&-2\\\end{array}\right]$

a)

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

b)

$5A = 5.\left[\begin{array}{cc}2&1\\3&2\\\end{array}\right] \\ \\ 5A = \left[\begin{array}{cc}5.2&5.1\\5.3&5.2\\\end{array}\right] \\ \\ 5A = \left[\begin{array}{cc}10&5\\15&10\\\end{array}\right]$

c)

$B^t$ = (Transposta da matriz B, onde o que é linha se transforma em coluna)

$B = \left[\begin{array}{cc}1&5\\2&-2\\\end{array}\right] \\ \\ B^t = \left[\begin{array}{cc}1&2\\5&-2\\\end{array}\right]$

d)

$A = \left[\begin{array}{cc}2&1\\3&2\\\end{array}\right] \\ \\ A^t = \left[\begin{array}{cc}2&3\\1&2\\\end{array}\right] \\ \\ A^t + B = \left[\begin{array}{cc}2&3\\1&2\\\end{array}\right] + \left[\begin{array}{cc}1&5\\2&-2\\\end{array}\right] \\ \\ A^t + B = \left[\begin{array}{cc}2 + 1&3+5\\1+2&2-2\\\end{array}\right] \\ \\ A^t + B = \left[\begin{array}{cc}3&8\\3&0\\\end{array}\right]$

e)

$3.A^t = 3 . \left[\begin{array}{cc}2&3\\1&2\\\end{array}\right] \\ \\ 3.A^t = \left[\begin{array}{cc}3.2&3.3\\3.1&3.2\\\end{array}\right] \\ \\ 3.A^t = \left[\begin{array}{cc}6&9\\3&6\\\end{array}\right]$

f)

$5A - B = \left[\begin{array}{cc}10&5\\15&10\\\end{array}\right]-\left[\begin{array}{cc}1&5\\2&-2\\\end{array}\right] \\ \\5A - B = \left[\begin{array}{cc}10-1&5-5\\15-2&10-(-2)\\\end{array}\right] \\ \\5A - B = \left[\begin{array}{cc}9&0\\13&12\\\end{array}\right] \\ \\(5A - B)^t = \left[\begin{array}{cc}9&13\\0&12\\\end{array}\right]$

g)

$3A = 3 . \left[\begin{array}{cc}2&1\\3&2\\\end{array}\right] \\ \\ 3A = \left[\begin{array}{cc}3.2&3.1\\3.3&3.2\\\end{array}\right] \\ \\ 3A = \left[\begin{array}{cc}6&3\\9&6\\\end{array}\right]$
$(3A)^t = \left[\begin{array}{cc}6&9\\3&6\\\end{array}\right] \\ \\ (3A)^t - 3.A^t = \left[\begin{array}{cc}6&9\\3&6\\\end{array}\right] - \left[\begin{array}{cc}6&9\\3&6\\\end{array}\right] \\ \\ (3A)^t - 3.A^t = \left[\begin{array}{cc}6-6&9-9\\3-3&6-6\\\end{array}\right]\\ \\ (3A)^t - 3.A^t = \left[\begin{array}{cc}0&0\\0&0\\\end{array}\right]$

h)

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

Answer 2

Vamos definir as propriedades das operações com matrizes antes de iniciar os cálculos:

• Soma e subtração de matrizes: somar ou subtrair cada elemento por seu respectivo na outra matriz;
• Multiplicação por escalar: multiplicar todos os elementos pelo escalar;
• Matriz transposta: trocar as linhas por colunas;

a) Subtração de matrizes:

$A - B = \left[\begin{array}{cc}2&1\\3&2\end{array}\right] - \left[\begin{array}{cc}1&5\\2&-2\end{array}\right] = \left[\begin{array}{cc}1&-4\\1&4\end{array}\right]$

b) Multiplicação por escalar:

$5A = \left[\begin{array}{cc}10&5\\15&10\end{array}\right]$

c) Transposta:

$B^t = \left[\begin{array}{cc}1&2\\5&-2\end{array}\right]$

d) Transposta e soma de matrizes:

$A^t + B = \left[\begin{array}{cc}2&3\\1&2\end{array}\right] + \left[\begin{array}{cc}1&5\\2&-2\end{array}\right] = \left[\begin{array}{cc}3&8\\3&0\end{array}\right]$

e) Multiplicação por escalar e transposta:

$3A^t = \left[\begin{array}{cc}6&9\\3&6\end{array}\right]$

f) Multiplicação por escalar, subtração de matrizes e transposta:

$(5A-B)^t = \left[\begin{array}{cc}10&5\\15&10\end{array}\right] - \left[\begin{array}{cc}1&5\\2&-2\end{array}\right] = \left[\begin{array}{cc}9&13\\0&12\end{array}\right]$

g) Multiplicação por escalar, subtração de matrizes e transposta:

$(3A)^t-3A^t = \left[\begin{array}{cc}6&9\\3&6\end{array}\right] - \left[\begin{array}{cc}6&9\\3&6\end{array}\right] = \left[\begin{array}{cc}0&0\\0&0\end{array}\right]$

h) Soma de matrizes e transposta:

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

Leia mais em:

https://brainly.com.br/tarefa/15788616