seja A=(aij) uma matriz quadrada de ordem 2 tal que aij=2i -3j e seja B= (1 0) (-1 1). calcule a matriz X tal que X+2A=B
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Primeiramente, temos que encontrar a matriz A.
A lei de criação da matriz A é dada por aij=2i-3j
a matriz genérica de A é dada por
![A= \left[\begin{array}{ccc}a_{11}&a_{12}\\a_{21}&a_{22}\\\end{array}\right] \\\\\\\\\text{Determinando a matriz A}\\\\a_{11}=2\cdot 1-3\cdot1~\longrightarrow a_{11}=2-3\longrightarrow \boxed{a_{11}=-1}\\\\a_{12}=2\cdot 1-3\cdot2~\longrightarrow a_{12}=2-6\longrightarrow \boxed{a_{12}=-4}\\\\a_{21}=2\cdot 2-3\cdot1~\longrightarrow a_{21}=4-3\longrightarrow \boxed{a_{21}=1}\\\\a_{22}=2\cdot 2-3\cdot2~\longrightarrow a_{22}=4-6\longrightarrow \boxed{a_{22}=-2}\\\\\\\text{a matriz A ficou assim:} A= \left[\begin{array}{ccc}a_{11}&a_{12}\\a_{21}&a_{22}\\\end{array}\right] \\\\\\\\\text{Determinando a matriz A}\\\\a_{11}=2\cdot 1-3\cdot1~\longrightarrow a_{11}=2-3\longrightarrow \boxed{a_{11}=-1}\\\\a_{12}=2\cdot 1-3\cdot2~\longrightarrow a_{12}=2-6\longrightarrow \boxed{a_{12}=-4}\\\\a_{21}=2\cdot 2-3\cdot1~\longrightarrow a_{21}=4-3\longrightarrow \boxed{a_{21}=1}\\\\a_{22}=2\cdot 2-3\cdot2~\longrightarrow a_{22}=4-6\longrightarrow \boxed{a_{22}=-2}\\\\\\\text{a matriz A ficou assim:}](https://tex.z-dn.net/?f=A%3D++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da_%7B11%7D%26amp%3Ba_%7B12%7D%5C%5Ca_%7B21%7D%26amp%3Ba_%7B22%7D%5C%5C%5Cend%7Barray%7D%5Cright%5D+%5C%5C%5C%5C%5C%5C%5C%5C%5Ctext%7BDeterminando+a+matriz+A%7D%5C%5C%5C%5Ca_%7B11%7D%3D2%5Ccdot+1-3%5Ccdot1%7E%5Clongrightarrow+a_%7B11%7D%3D2-3%5Clongrightarrow+%5Cboxed%7Ba_%7B11%7D%3D-1%7D%5C%5C%5C%5Ca_%7B12%7D%3D2%5Ccdot+1-3%5Ccdot2%7E%5Clongrightarrow+a_%7B12%7D%3D2-6%5Clongrightarrow+%5Cboxed%7Ba_%7B12%7D%3D-4%7D%5C%5C%5C%5Ca_%7B21%7D%3D2%5Ccdot+2-3%5Ccdot1%7E%5Clongrightarrow+a_%7B21%7D%3D4-3%5Clongrightarrow+%5Cboxed%7Ba_%7B21%7D%3D1%7D%5C%5C%5C%5Ca_%7B22%7D%3D2%5Ccdot+2-3%5Ccdot2%7E%5Clongrightarrow+a_%7B22%7D%3D4-6%5Clongrightarrow+%5Cboxed%7Ba_%7B22%7D%3D-2%7D%5C%5C%5C%5C%5C%5C%5Ctext%7Ba+matriz+A+ficou+assim%3A%7D)
![A= \left[\begin{array}{ccc}-1&-4\\1&-2\\\end{array}\right] \\\\\\\text{Agora temos que resolver a equacao matricial dada pelo exercicio}\\\\X+2A=B\\\\\boxed{X=B-2A}\\\\\\\text{Ja temos a matriz 'B', mas nao temos a matriz '2A', porem, temos}\\\text{a matriz 'A', entao, para encontrar a matriz '2A', basta multiplicar}\\\text{Cada elemento da matriz 'A' por 2}\\\\2A= \left[\begin{array}{ccc}2\cdot(-1)~~&~~2\cdot(-4)\\2\cdot(1)~~&~~2\cdot(-2)\\\end{array}\right]\\\\\\ A= \left[\begin{array}{ccc}-1&-4\\1&-2\\\end{array}\right] \\\\\\\text{Agora temos que resolver a equacao matricial dada pelo exercicio}\\\\X+2A=B\\\\\boxed{X=B-2A}\\\\\\\text{Ja temos a matriz 'B', mas nao temos a matriz '2A', porem, temos}\\\text{a matriz 'A', entao, para encontrar a matriz '2A', basta multiplicar}\\\text{Cada elemento da matriz 'A' por 2}\\\\2A= \left[\begin{array}{ccc}2\cdot(-1)~~&~~2\cdot(-4)\\2\cdot(1)~~&~~2\cdot(-2)\\\end{array}\right]\\\\\\](https://tex.z-dn.net/?f=A%3D++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%26amp%3B-4%5C%5C1%26amp%3B-2%5C%5C%5Cend%7Barray%7D%5Cright%5D+%5C%5C%5C%5C%5C%5C%5Ctext%7BAgora+temos+que+resolver+a+equacao+matricial+dada+pelo+exercicio%7D%5C%5C%5C%5CX%2B2A%3DB%5C%5C%5C%5C%5Cboxed%7BX%3DB-2A%7D%5C%5C%5C%5C%5C%5C%5Ctext%7BJa+temos+a+matriz+%27B%27%2C+mas+nao+temos+a+matriz+%272A%27%2C+porem%2C+temos%7D%5C%5C%5Ctext%7Ba+matriz+%27A%27%2C+entao%2C+para+encontrar+a+matriz+%272A%27%2C+basta+multiplicar%7D%5C%5C%5Ctext%7BCada+elemento+da+matriz+%27A%27+por+2%7D%5C%5C%5C%5C2A%3D++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%5Ccdot%28-1%29%7E%7E%26amp%3B%7E%7E2%5Ccdot%28-4%29%5C%5C2%5Ccdot%281%29%7E%7E%26amp%3B%7E%7E2%5Ccdot%28-2%29%5C%5C%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5C+)
![2A= \left[\begin{array}{ccc}-2&-8\\2&-4\\\end{array}\right]\\\\\\\text{Como o enunciado deu a matriz 'B', Ja temos todas as }\\\text{informacoes necessarias, entao vamos encontrar a matriz X}\\\\\\X=B-2A\\\\\\X= \left[\begin{array}{ccc}1&0\\-1&1\\\end{array}\right]- \left[\begin{array}{ccc}-2&-8\\2&-4\\\end{array}\right]\\\\\\X= \left[\begin{array}{ccc}1-(-2)~~&~~0-(-8)\\-1-2&1-(-4)\\\end{array}\right] 2A= \left[\begin{array}{ccc}-2&-8\\2&-4\\\end{array}\right]\\\\\\\text{Como o enunciado deu a matriz 'B', Ja temos todas as }\\\text{informacoes necessarias, entao vamos encontrar a matriz X}\\\\\\X=B-2A\\\\\\X= \left[\begin{array}{ccc}1&0\\-1&1\\\end{array}\right]- \left[\begin{array}{ccc}-2&-8\\2&-4\\\end{array}\right]\\\\\\X= \left[\begin{array}{ccc}1-(-2)~~&~~0-(-8)\\-1-2&1-(-4)\\\end{array}\right]](https://tex.z-dn.net/?f=2A%3D++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%26amp%3B-8%5C%5C2%26amp%3B-4%5C%5C%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5C%5Ctext%7BComo++o+enunciado+deu+a+matriz+%27B%27%2C+Ja+temos+todas+as+++%7D%5C%5C%5Ctext%7Binformacoes+necessarias%2C+entao+vamos+encontrar+a+matriz+X%7D%5C%5C%5C%5C%5C%5CX%3DB-2A%5C%5C%5C%5C%5C%5CX%3D++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26amp%3B0%5C%5C-1%26amp%3B1%5C%5C%5Cend%7Barray%7D%5Cright%5D-++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%26amp%3B-8%5C%5C2%26amp%3B-4%5C%5C%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CX%3D++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1-%28-2%29%7E%7E%26amp%3B%7E%7E0-%28-8%29%5C%5C-1-2%26amp%3B1-%28-4%29%5C%5C%5Cend%7Barray%7D%5Cright%5D)
![\boxed{X= \left[\begin{array}{ccc}3&8\\-3&5\\\end{array}\right]}~~~~~~~~~\longleftarrow\text{Resposta} \boxed{X= \left[\begin{array}{ccc}3&8\\-3&5\\\end{array}\right]}~~~~~~~~~\longleftarrow\text{Resposta}](https://tex.z-dn.net/?f=%5Cboxed%7BX%3D++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26amp%3B8%5C%5C-3%26amp%3B5%5C%5C%5Cend%7Barray%7D%5Cright%5D%7D%7E%7E%7E%7E%7E%7E%7E%7E%7E%5Clongleftarrow%5Ctext%7BResposta%7D)
Primeiramente, temos que encontrar a matriz A.
A lei de criação da matriz A é dada por aij=2i-3j
a matriz genérica de A é dada por
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