Matemática, perguntado por kekelzinhaestronda, 11 meses atrás


Calcule:
а) А - B
b) B- C
C) A-C
D) C-A+B
E) C-(B-A)​

Anexos:

Soluções para a tarefa

Respondido por antoniosbarroso2011
6

Resposta:

Explicação passo-a-passo:

a)

A-B=\left[\begin{array}{cc}1-(-2)&5-(-3)\\2-1&4-0\\-1-4&3-2\end{array}\right]=\left[\begin{array}{cc}3&8\\1&4\\-5&1\end{array}\right]

b)

B-C=\left[\begin{array}{cc}-2-6&-3-(-1)\\1-3&0-(-2)\\4-0&2-1\end{array}\right]=\left[\begin{array}{cc}-8&-2\\-2&2\\4&1\end{array}\right]

c)

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

d)

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

e)

C-(B-A)=>C-B+A=\left[\begin{array}{cc}6-(-2)+1&-1-(-3)+5\\3-1+2&-2-0+4\\0-4+(-1)&1-2+3\end{array}\right]=\left[\begin{array}{cc}9&7\\4&2\\-5&2\end{array}\right]

Respondido por cleitonromes
0

Explicação passo-a-passo:

Resposta:

Explicação passo-a-passo:

a)

\begin{gathered}A-B=\left[\begin{array}{cc}1-(-2)&5-(-3)\\2-1&4-0\\-1-4&3-2\end{array}\right]=\left[\begin{array}{cc}3&8\\1&4\\-5&1\end{array}\right]\end{gathered}A−B=⎣⎢⎡1−(−2)2−1−1−45−(−3)4−03−2⎦⎥⎤=⎣⎢⎡31−5841⎦⎥⎤

b)

\begin{gathered}B-C=\left[\begin{array}{cc}-2-6&-3-(-1)\\1-3&0-(-2)\\4-0&2-1\end{array}\right]=\left[\begin{array}{cc}-8&-2\\-2&2\\4&1\end{array}\right]\end{gathered}B−C=⎣⎢⎡−2−61−34−0−3−(−1)0−(−2)2−1⎦⎥⎤=⎣⎢⎡−8−24−221⎦⎥⎤

c)

\begin{gathered}A-C=\left[\begin{array}{cc}1-6&5-(-1)\\2-3&4-(-2)\\-1-0&3-1\end{array}\right]=\left[\begin{array}{cc}-5&6\\-1&6\\-1&2\end{array}\right]\end{gathered}A−C=⎣⎢⎡1−62−3−1−05−(−1)4−(−2)3−1⎦⎥⎤=⎣⎢⎡−5−1−1662⎦⎥⎤

d)

\begin{gathered}C-A+B=\left[\begin{array}{cc}6-1+(-2)&-1-5+(-3)\\3-2+1&-2-4+0\\0-(-1)+4&1-3+2\end{array}\right]=\left[\begin{array}{cc}3&-9\\0&-6\\5&0\end{array}\right]\end{gathered}C−A+B=⎣⎢⎡6−1+(−2)3−2+10−(−1)+4−1−5+(−3)−2−4+01−3+2⎦⎥⎤=⎣⎢⎡305−9−60⎦⎥⎤

e)

\begin{gathered}C-(B-A)= > C-B+A=\left[\begin{array}{cc}6-(-2)+1&-1-(-3)+5\\3-1+2&-2-0+4\\0-4+(-1)&1-2+3\end{array}\right]=\left[\begin{array}{cc}9&7\\4&2\\-5&2\end{array}\right]\end{gathered}C−(B−A)=>C−B+A=⎣⎢⎡6−(−2)+13−1+20−4+(−1)−1−(−3)+5−2−0+41−2+3⎦⎥⎤=⎣⎢⎡94−5722⎦⎥⎤

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