Matemática, perguntado por 152baixinha, 11 meses atrás

Construa a matriz C = (Cij) de ordem 5×4, em que Cij = 3i - 2j

Soluções para a tarefa

Respondido por Júnior

Olá, sabemos que a ordem dessa matriz é 5x4, ou seja, 5 linhas e 4 colunas.

$C_{5\times4\:=}\begin{bmatrix}C_{11}&\:C_{12}&\:C_{13}& \:C_{14}\\ C_{21}&\:C_{22}&\:C_{23}& \:C_{24}\\ C_{31}&\:C_{32}&\:C_{33}& \:C_{34}\\ C_{41}&\:C_{42}&\:C_{43}& \:C_{44}\\ C_{51}&\:C_{52}&\:C_{53}& \:C_{54}\end{bmatrix}$

Sabemos também que Cij= 3i -2j. Observe:

$C_{ij} = \begin{bmatrix}(3\cdot1)-(2\cdot1)& \:(3\cdot1)-(2\cdot2)& \:(3\cdot1)-(2\cdot3)& \:(3\cdot1)-(2\cdot4)\\ (3\cdot2)-(2\cdot1)& \:(3\cdot2)-(2\cdot2)& \:(3\cdot2)-(2\cdot3)& \:(3\cdot2)-(2\cdot4)\\ (3\cdot3)-(2\cdot1)& \:(3\cdot3)-(2\cdot2)& \:(3\cdot3)-(2\cdot3)& \:(3\cdot3)-(2\cdot4)\\ (3\cdot4)-(2\cdot1)& \:(3\cdot4)-(2\cdot2)& \:(3\cdot4)-(2\cdot3)& \:(3\cdot4)-(2\cdot4)\\ (3\cdot5)-(2\cdot1)& \:(3\cdot5)-(2\cdot2)& \:(3\cdot5)-(2\cdot3)& \:(3\cdot5)-(2\cdot4)\end{bmatrix}$

Resolvendo as multiplicações:

$C_{ij}=\begin{bmatrix}(3-2)& \:(3-4)& \:(3-6)& \:(3-8)\\ (6-2)& \:(6-4)& \:(6-6)& \:(6-8)\\ (9-2)& \:(9-4)& \:(9-6)& \:(9-8)\\ (12-2)& \:(12-4)& \:(12-6)& \:(12-8)\\(15-2)& \:(15-4)& \:(15-6)& \:(15-8)\end{bmatrix}$

Portanto, essa é a matriz Cij de ordem 5×4:

$C_{ij}=\begin{bmatrix}1& \:-1&\:-3& \:-5\\ 4& \:\:\:\:2& \:\:\:\:0& \:-2\\ 7& \:\:\:\:5& \:\:\:\:3& \:\:\:\:1\\ 10& \:\:\:\:8& \:\:\:\:6& \:\:\:\:4\\13& \:\:\:\:11& \:\:\:\:9& \:\:\:\:7\end{bmatrix}$

Espero ter ajudado! Aprenda mais sobre matrizes em:

https://brainly.com.br/tarefa/45687736

Anexos:

Respondido por Skoy

A resposta é:

$\large\begin{array}{lr}\left[\begin{array}{cccc}1&-1&-3}&-5\\4&2&0&-2\\7&5&3&1\\10&8&6&4\\13&11&9}&7\end{array}\right] \end{array}$

Para que possamos calcular sua questão devemos primeiramente criar uma matriz genérica 5x4 e calcular cada termo da mesma.

$\large\begin{array}{lr}\left[\begin{array}{cccc}a_{11}&a_{12}&a_{13}&a_{14}\\a_{21}&a_{22}&a_{23}&a_{24}\\a_{31}&a_{32}&a_{33}&a_{34}\\a_{41}&a_{42}&a_{43}&a_{44}\\a_{51}&a_{52}&a_{53}&a_{54}\end{array}\right] \end{array}$

Agora devemos calcular cada termo da matriz genérica seguindo a lei "3i - 2j"

Primeira Linha:

$\large\begin{array}{lr} \sf a11= 3i - 2j =3*1-2*1 \rightarrow \underline{\boxed{\red{\sf 1}}}\\\\\sf a12= 3i - 2j =3*1-2*2 \rightarrow \underline{\boxed{\red{\sf -1}}}\\\\\sf a13= 3i - 2j =3*1-2*3 \rightarrow \underline{\boxed{\red{\sf -3}}}\\\\\sf a14= 3i - 2j =3*1-2*4 \rightarrow \underline{\boxed{\red{\sf -5}}}\\\end{array}$

Segunda Linha:

$\large\begin{array}{lr} \sf a21= 3i - 2j =3*2-2*1 \rightarrow \underline{\boxed{\red{\sf 4}}}\\\\\sf a22= 3i - 2j =3*2-2*2 \rightarrow \underline{\boxed{\red{\sf 2}}}\\\\\sf a23= 3i - 2j =3*2-2*3 \rightarrow \underline{\boxed{\red{\sf 0}}}\\\\\sf a24= 3i - 2j =3*2-2*4 \rightarrow \underline{\boxed{\red{\sf -2}}}\end{array}$

Terceira Linha:

$\large\begin{array}{lr} \sf a31= 3i - 2j =3*3-2*1 \rightarrow \underline{\boxed{\red{\sf 7}}}\\\\\sf a32= 3i - 2j =3*3-2*2 \rightarrow \underline{\boxed{\red{\sf 5}}}\\\\\sf a33= 3i - 2j =3*3-2*3 \rightarrow \underline{\boxed{\red{\sf 3}}}\\\\\sf a34= 3i - 2j =3*3-2*4 \rightarrow \underline{\boxed{\red{\sf 1}}}\end{array}$

Quarta Linha:

$\large\begin{array}{lr} \sf a41= 3i - 2j =34-21 \rightarrow \underline{\boxed{\red{\sf 10}}}\\\\\sf a42= 3i - 2j =34-22 \rightarrow \underline{\boxed{\red{\sf 8}}}\\\\\sf a43= 3i - 2j =34-23 \rightarrow \underline{\boxed{\red{\sf 6}}}\\\\\sf a44= 3i - 2j =34-24 \rightarrow \underline{\boxed{\red{\sf 4}}}\end{array}$

Quinta Linha:

$\large\begin{array}{lr} \sf a51= 3i - 2j =3*5-2*1 \rightarrow \underline{\boxed{\red{\sf 13}}}\\\\\sf a52= 3i - 2j =3*5-2*2 \rightarrow \underline{\boxed{\red{\sf 11}}}\\\\\sf a53= 3i - 2j =3*5-2*3 \rightarrow \underline{\boxed{\red{\sf 9}}}\\\\\sf a54= 3i - 2j =3*5-2*4 \rightarrow \underline{\boxed{\red{\sf 7}}}\end{array}$