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Considere as matrizes A= 3 2 -1 2 e B= 1 0 1 -2. Sabendo que AxB-1 = C, calcule A-B+C.

T

Thaemy

04/10/24

Guru IA answer

Para resolver essa questão, vamos seguir os passos:

  1. Calcular B1B^{-1} (a inversa de BB).
  2. Calcular A×B1A \times B^{-1} para encontrar CC.
  3. Finalmente, calcular AB+CA - B + C.

Dadas as matrizes:

A=(3212),B=(1012) A = \begin{pmatrix} 3 & 2 \\ -1 & 2 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 0 \\ 1 & -2 \end{pmatrix}

Passo 1: Calcular B1B^{-1}

Para calcular a inversa de uma matriz 2×22 \times 2, usamos a fórmula:

B1=1det(B)(dbca) B^{-1} = \frac{1}{\text{det}(B)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

onde det(B)=adbc\text{det}(B) = ad - bc e B=(abcd)B = \begin{pmatrix} a & b \\ c & d \end{pmatrix}.

Para BB, temos:

a=1,b=0,c=1,d=2 a = 1, \quad b = 0, \quad c = 1, \quad d = -2

Portanto,

det(B)=(1)(2)(1)(0)=2 \text{det}(B) = (1)(-2) - (1)(0) = -2

E assim,

B1=12(2011)=(100.50.5) B^{-1} = \frac{1}{-2} \begin{pmatrix} -2 & 0 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0.5 & -0.5 \end{pmatrix}

Passo 2: Calcular A×B1=CA \times B^{-1} = C

A×B1=(3212)×(100.50.5) A \times B^{-1} = \begin{pmatrix} 3 & 2 \\ -1 & 2 \end{pmatrix} \times \begin{pmatrix} 1 & 0 \\ 0.5 & -0.5 \end{pmatrix}

Multiplicando as matrizes:

C=(3(1)+2(0.5)3(0)+2(0.5)1(1)+2(0.5)1(0)+2(0.5))=(4101) C = \begin{pmatrix} 3(1) + 2(0.5) & 3(0) + 2(-0.5) \\ -1(1) + 2(0.5) & -1(0) + 2(-0.5) \end{pmatrix} = \begin{pmatrix} 4 & -1 \\ 0 & -1 \end{pmatrix}

Passo 3: Calcular AB+CA - B + C

AB+C=(3212)(1012)+(4101) A - B + C = \begin{pmatrix} 3 & 2 \\ -1 & 2 \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 1 & -2 \end{pmatrix} + \begin{pmatrix} 4 & -1 \\ 0 & -1 \end{pmatrix}

Somando as matrizes:

AB+C=(31+420111+02+21)=(6123) A - B + C = \begin{pmatrix} 3-1+4 & 2-0-1 \\ -1-1+0 & 2+2-1 \end{pmatrix} = \begin{pmatrix} 6 & 1 \\ -2 & 3 \end{pmatrix}

Portanto, o resultado de AB+CA - B + C é (6123)\begin{pmatrix} 6 & 1 \\ -2 & 3 \end{pmatrix}.

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