Question

Determine, se existem, os produtos:​

Answer 1

O produto de matrizes existe se e somente se o número de colunas da primeira matriz for igual ao número de linhas da 2ª matriz.

a)

$\begin{bmatrix}1&2\\3&4\end{bmatrix} \begin{bmatrix}2&3\\-2&1\end{bmatrix}= \\ \begin{bmatrix}1.2+2.(-2)&1.3+2.1\\3.2+4.(-2)&3.3+4.1\end{bmatrix} \\ = \begin{bmatrix}-2&5\\-2&13\end{bmatrix}$

b)

$\begin{bmatrix}1&-2\\3&4\end{bmatrix}.\begin{bmatrix}-2&3&2&-1\\-1&0&0&-4\end{bmatrix}\\=\begin{bmatrix}1.-2-2.(-1)&1.3-2.0&1.2-2.0&1.(-1)-2.(-4)\\3.(-2)+4.(-1)&3.3+4.0&3.2+4.0&3.(-1)+4.(-4)\end{bmatrix}\\= \begin{bmatrix}0&3&2&7\\-10&9&6&-19\end{bmatrix}$

c)

O produto é impossível pois o número de colunas da 1ª matriz é diferente do número de linhas da 2ª matriz ferindo a definição de produto de matrizes.

d)

$\begin{bmatrix}3&4&1\\5&6&1\\7&8&1\end{bmatrix}.\begin{bmatrix}2\\-3\\4\end{bmatrix}\\=\begin{bmatrix}3.2+4.(-3)+1.4\\5.2+6.(-3)+1.4\\7.2+8.(-3)+1.4\end{bmatrix}\\= \begin{bmatrix}-2\\-4\\-6\end{bmatrix}$

e)

$\begin{bmatrix}-5&0\\-1&3\\1&1\\2&2\end{bmatrix}.\begin{bmatrix}3&-2\\1&5\end{bmatrix}$

$\begin{bmatrix}-5.3+0.1&-5.(-2)+0.5\\-1.3+3.1&-1(-2)+3.5\\1.3+1.1&1.(-2)+1.5\\2.3+2.1&2.(-2)+2.5\end{bmatrix}\\=\begin{vmatrix}-15&10\\0&17\\4&3\\8&6\end{vmatrix}</p><p>$

Answer 2

a)  $\left[\begin{array}{ccc}-2&5\\-2&13\end{array}\right]$                                          b) $\left[\begin{array}{cccc}0&3&2&7\\-10&9&6&-19\end{array}\right]$

c) não é matriz

d)  $\left[\begin{array}{ccc}-2\\-4\\-6\end{array}\right]$                                                 e)  $\left[\begin{array}{ccc}-15&10\\-6&17\\4&3\\9&6\end{array}\right]$

f) $\left[\begin{array}{ccc}8&20&29\\1&16&29\\-2&4&12\end{array}\right]$                                     g)  $\left[\begin{array}{ccc}12&-4&16\\18&-6&24\\30&-10&40\end{array}\right]$

h) não é matriz

i) a -  $\left[\begin{array}{ccc}11&4\\4&3\\10&3\end{array}\right]$                                         b - não é matriz

c -  $\left[\begin{array}{ccc}1\\8\\-8\end{array}\right]$                                                 d -  $\left[\begin{array}{ccc}1\\8\\-8\end{array}\right]$

e - $\left[\begin{array}{ccc}5&4&2\\6&6&1\end{array}\right]$

Multiplicação de matriz e matriz transposta

Na multiplicação de matriz, o número de colunas da primeira matriz, sempre deve ser igual ao número de linhas da segunda matriz.

Na matriz transposta, a primeira linha vira a primeira coluna e a segunda linha, vira a segunda coluna.

a) $\left[\begin{array}{ccc}1&2\\3&4\end{array}\right]$ . $\left[\begin{array}{ccc}2&3\\-2&1\end{array}\right]$ = $\left[\begin{array}{ccc}1.2+2.(-2)&1.3+2.1\\3.2+4.(-2)&3.3+4.1\end{array}\right]$ = $\left[\begin{array}{ccc}-2&5\\-2&13\end{array}\right]$

b) $\left[\begin{array}{ccc}1&-2\\-2&4\end{array}\right]$ . $\left[\begin{array}{cccc}-2&3&2&-1\\-1&0&0&-4\end{array}\right]$ =

$\left[\begin{array}{cccc}1.(-2)+(-2).(-1)&1.3+(-2).0&1.2+(-2).0&1.(-1)+(-2).(-4)\\3.(-2)+4.(-1)&3.3+4.0&3.3+4.0&3.(-1)+4.(-4)\end{array}\right]$ =

$\left[\begin{array}{cccc}0&3&2&7\\-10&9&6&-19\end{array}\right]$

d) $\left[\begin{array}{ccc}3&4&1\\5&6&1\\7&8&1\end{array}\right]$ . $\left[\begin{array}{ccc}2\\-3\\4\end{array}\right]$ = $\left[\begin{array}{ccc}3.2+4.(-3)+1.4\\5.2+6.(-3)+1.4\\7.2+8.(-3)+1.4\end{array}\right]$ = $\left[\begin{array}{ccc}-2\\-4\\-6\end{array}\right]$

e) $\left[\begin{array}{ccc}-5&0\\-1&3\\1&1\\2&2\end{array}\right]$ . $\left[\begin{array}{ccc}3&-2\\1&5\end{array}\right]$ = $\left[\begin{array}{ccc}-5.3+0.1&-5.(-2)+0.5\\-1.3+3.1&-1.(-2)+3.5\\1.3+1.1&1.(-2)+1.5\\2.3+2.1&2.(-2)+2.5\end{array}\right]$ = $\left[\begin{array}{ccc}-15&10\\-6&17\\4&3\\9&6\end{array}\right]$

f) $\left[\begin{array}{ccc}8&2&-1\\1&7&5\\-2&4&6\end{array}\right]$ . $\left[\begin{array}{ccc}1&2&3\\0&2&3\\0&0&1\end{array}\right]$ =

$\left[\begin{array}{ccc}8.1+2.0+(-1).0&8.2+2.2+(-1).0&8.3+2.3+(-1).1\\1.1+7.0+5.0&1.2+7.2+5.0&1.3+7.3+5.1\\(-2).1+4.0+6.0&(-2).2+4.2+6.0&(-2).3+4.3+6.1\end{array}\right]$ =

$\left[\begin{array}{ccc}8&20&29\\1&16&29\\-2&4&12\end{array}\right]$

g) $\left[\begin{array}{ccc}2\\3\\5\end{array}\right]$ . $\left[\begin{array}{ccc}6&-2&8\end{array}\right]$ = $\left[\begin{array}{ccc}2.6&2.(-2)&2.8\\3.6&3.(-2)&3.8\\5.6&5.(-2)&5.8\end{array}\right]$ = $\left[\begin{array}{ccc}12&-4&16\\18&-6&24\\30&-10&40\end{array}\right]$

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

Determine:

a) A.B ⇒ $\left[\begin{array}{ccc}1.2+3.3&1.1+3.1\\2.2+0.3&2.1+0.1\\-1.2+4.3&-1.1+4.1\end{array}\right]$ = $\left[\begin{array}{ccc}11&4\\4&3\\10&3\end{array}\right]$

c) A.C ⇒ $\left[\begin{array}{ccc}1.4+3.(-1)\\2.4+0.(-1)\\-1.4+4.(-1)\end{array}\right]$ = $\left[\begin{array}{ccc}1\\8\\-8\end{array}\right]$

d) Bt.C ⇒ $\left[\begin{array}{ccc}2&3\\1&1\end{array}\right]$ . $\left[\begin{array}{ccc}4\\-1\end{array}\right]$ = $\left[\begin{array}{ccc}2.4+3.(-1)\\1.4+1.(-1)\end{array}\right]$ = $\left[\begin{array}{ccc}5\\3\end{array}\right]$

e) B.At ⇒ $\left[\begin{array}{ccc}2&1\\3&1\end{array}\right]$ . $\left[\begin{array}{ccc}1&2&-1\\3&0&4\end{array}\right]$ = $\left[\begin{array}{ccc}2.1+1.3&2.2+1.0&2.(-1)+1.4\\3.1+1.3&3.2+1.0&3.(-1)+1.4\end{array}\right]$ =

$\left[\begin{array}{ccc}5&4&2\\6&6&1\end{array}\right]$

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