Matemática, perguntado por marcospaulo7991, 10 meses atrás

como resolver essa ​

Anexos:

Soluções para a tarefa

Respondido por jbsenajr
0

Resposta:

Explicação passo-a-passo:

a)

A+B=\left[\begin{array}{ccc}3&-1\\1&2\\4&0\end{array}\right]+\left[\begin{array}{ccc}2&3\\0&1\\1&2\end{array}\right]=\left[\begin{array}{ccc}3+2&-1+3\\1+0&2+1\\4+1&0+2\end{array}\right]=\left[\begin{array}{ccc}5&2\\1&3\\5&2\end{array}\right]

b)

B+C=\left[\begin{array}{ccc}2&3\\0&1\\1&2\end{array}\right]+\left[\begin{array}{ccc}1&0\\-1&3\\2&5\end{array}\right]=\left[\begin{array}{ccc}2+1&3+0\\0+(-1)&1+3\\1+2&2+5\end{array}\right]=\left[\begin{array}{ccc}3&3\\-1&4\\3&7\end{array}\right]

c)

B-A=\left[\begin{array}{ccc}2&3\\0&1\\1&2\end{array}\right]-\left[\begin{array}{ccc}3&-1\\1&2\\4&0\end{array}\right]=\left[\begin{array}{ccc}2-3&3-(-1)\\0-1&1-2\\1-4&2-0\end{array}\right]=\left[\begin{array}{ccc}-1&4\\-1&-1\\-3&2\end{array}\right]

d)

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

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