Question

Dada a matriz
A= [-2 3 1]
......[-1 0 4]
......[ 1 -2 3]
determine:
a) cof (a12) c) cof (a22) e) cof (a23)
b) cof (a31) d) cof (a13) f) cof (a33)

Answer 1

Explicação passo a passo:

Temos a matriz:

$\begin{bmatrix} -2&3&1 \\ -1&0&4 \\ 1&-2&3 \end{bmatrix}$

Para encontrar o cofator, devemos eliminar a linha e a coluna do elemento

Lembrando que i = linha, j = coluna

a)

$\sf A_{12}$ <= linha 1, coluna 2. Logo devemos eliminar a linha e a coluna do elemento 3, dessa forma formando uma outra matriz:

$\begin{bmatrix} -1&4 \\ 1&3 \end{bmatrix}$

Calcule seu determinante

$\sf D = (- 1*3) - (4*1)$

$\sf D = - 3 - 4$

$\red{\sf D = - 7}$

Agora vamos calcular o cofator

$\sf C_{ij} = (-1)^{i+j} * D$

$\sf C_{12} = (-1)^{1+2} * (-7)$

$\sf C_{12} = (-1)^{3} * (-7)$

$\sf C_{12} = (-1) * (-7)$

$\boxed{\sf C_{12} = 7}$

Vamos fazer a mesma coisa nas outras alternativas:

b)

$\sf A_{31}$ <= eliminar linha 3 e coluna 1

$\begin{bmatrix} 3&1 \\ 0&4 \end{bmatrix}$

$\sf D = (3*4) - (1*0)$

$\sf D = 12 - 0$

$\red{\sf D = 12}$

__________________

$\sf C_{ij} = (-1)^{i+j} * D$

$\sf C_{31} = (-1)^{3+1} * 12$

$\sf C_{31} = (-1)^{4} * 12$

$\sf C_{31} = 1 * 12$

$\boxed{\sf C_{31} = 12}$

__________________

c)

$\sf A_{22}$ <= eliminar linha 2 e coluna 2

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

$\sf D = (-2*3) - (1*1)$

$\sf D = - 6 - 1$

$\red{\sf D = - 7}$

__________________

$\sf C_{ij} = (-1)^{i+j} * D$

$\sf C_{22} = (-1)^{2+2} * (-7)$

$\sf C_{22} = (-1)^{4} * (-7)$

$\sf C_{22} = 1 * (-7)$

$\boxed{\sf C_{22} = - 7}$

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d)

$\sf A_{13}$ <= eliminar linha 1 e coluna 3

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

$\sf D = (-1*(-2)) - (0*1)$

$\sf D = 2 - 0$

$\red{\sf D = 2}$

__________________

$\sf C_{ij} = (-1)^{i+j} * D$

$\sf C_{13} = (-1)^{1+3} * 2$

$\sf C_{13} = (-1)^{4} * 2$

$\sf C_{13} = 1 * 2$

$\boxed{\sf C_{13} = 2}$

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e)

$\sf A_{23}$ <= eliminar linha 2 e coluna 3

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

$\sf D = (-2*(-2)) - (3*1)$

$\sf D = 4 - 3$

$\red{\sf D = 1}$

__________________

$\sf C_{ij} = (-1)^{i+j} * D$

$\sf C_{23} = (-1)^{2+3} * 1$

$\sf C_{23} = (-1)^{5} * 1$

$\sf C_{23} = (-1) * 1$

$\boxed{\sf C_{23} = -1}$

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f)

$\sf A_{33}$ <= eliminar linha 3 e coluna 3

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

$\sf D = (-2*0) - (3*(-1))$

$\sf D = 0 + 3$

$\red{\sf D = 3}$

__________________

$\sf C_{ij} = (-1)^{i+j} * D$

$\sf C_{33} = (-1)^{3+3} * 3$

$\sf C_{33} = (-1)^{6} * 3$

$\sf C_{33} = 1 * 3$

$\boxed{\sf C_{33} = 3}$

Answer 2

Explicação passo-a-passo:

$\sf A=\Big[\begin{array}{ccc} \sf -2 & \sf 3 & \sf 1 \\ \sf -1 & \sf 0 & \sf 4 \\ \sf 1 & \sf -2 & \sf 3 \end{array}\Big]$

$\sf cof~(a_{ij})=(-1)^{1+j}\cdot D_{ij}$

a)

$\sf D_{12}=\Big|\begin{array}{cc} \sf -1 & 4 \\ \sf 1 & \sf 3 \end{array}\Big|$

$\sf D_{12}=(-1)\cdot3-1\cdot4$

$\sf D_{12}=-3-4$

$\sf D_{12}=-7$

Assim:

$\sf cof~(a_{12})=(-1)^{1+2}\cdot D_{12}$

$\sf cof~(a_{12})=(-1)^{3}\cdot(-7)$

$\sf cof~(a_{12})=(-1)\cdot(-7)$

$\sf \red{cof~(a_{12})=7}$

b)

$\sf D_{31}=\Big|\begin{array}{cc} \sf 3 & 1 \\ \sf 0 & \sf 4 \end{array}\Big|$

$\sf D_{31}=3\cdot4-0\cdot1$

$\sf D_{31}=12-0$

$\sf D_{31}=12$

Assim:

$\sf cof~(a_{31})=(-1)^{3+1}\cdot D_{31}$

$\sf cof~(a_{31})=(-1)^{4}\cdot12$

$\sf cof~(a_{31})=1\cdot12$

$\sf \red{cof~(a_{31})=12}$

c)

$\sf D_{22}=\Big|\begin{array}{cc} \sf -2 & 1 \\ \sf 1 & \sf 3 \end{array}\Big|$

$\sf D_{22}=(-2)\cdot3-1\cdot1$

$\sf D_{22}=-6-1$

$\sf D_{22}=-7$

Assim:

$\sf cof~(a_{22})=(-1)^{2+2}\cdot D_{22}$

$\sf cof~(a_{22})=(-1)^{4}\cdot(-7)$

$\sf cof~(a_{22})=1\cdot(-7)$

$\sf \red{cof~(a_{22})=-7}$

d)

$\sf D_{13}=\Big|\begin{array}{cc} \sf -1 & 0 \\ \sf 1 & \sf -2 \end{array}\Big|$

$\sf D_{13}=(-1)\cdot(-2)-1\cdot0$

$\sf D_{13}=2-0$

$\sf D_{13}=2$

Assim:

$\sf cof~(a_{13})=(-1)^{1+3}\cdot D_{13}$

$\sf cof~(a_{13})=(-1)^{4}\cdot2$

$\sf cof~(a_{13})=1\cdot2$

$\sf \red{cof~(a_{13})=2}$

e)

$\sf D_{23}=\Big|\begin{array}{cc} \sf -2 & 3 \\ \sf 1 & \sf -2 \end{array}\Big|$

$\sf D_{23}=(-2)\cdot(-2)-1\cdot3$

$\sf D_{23}=4-3$

$\sf D_{23}=1$

Assim:

$\sf cof~(a_{23})=(-1)^{2+3}\cdot D_{23}$

$\sf cof~(a_{23})=(-1)^{5}\cdot1$

$\sf cof~(a_{23})=(-1)\cdot1$

$\sf \red{cof~(a_{23})=-1}$

f)

$\sf D_{33}=\Big|\begin{array}{cc} \sf -2 & 3 \\ \sf -1 & \sf 0 \end{array}\Big|$

$\sf D_{33}=(-2)\cdot0-(-1)\cdot3$

$\sf D_{33}=0+3$

$\sf D_{33}=3$

Assim:

$\sf cof~(a_{33})=(-1)^{3+3}\cdot D_{33}$

$\sf cof~(a_{33})=(-1)^{6}\cdot3$

$\sf cof~(a_{33})=1\cdot3$

$\sf \red{cof~(a_{33})=3}$