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Area of a Triangle
We have already learnt in the previous class that the area of triangle whose vertices are (x 1 , y 1 ), (x 2 , y 2 ), (x 3 , y 3 ) is given by Hence area of a triangle having vertices at (x 1 , y 1 ), (x 2 , y 2 ) and (x 3 , y 3 ) is given by..
We have already learnt in the previous class that the area of triangle whose vertices are (x 1 , y 1 ), (x 2 , y 2 ), (x 3 , y 3 ) is given by Hence area of a triangle having vertices at (x 1 , y 1 ), (x 2 , y 2 ) and (x 3 , y 3 ) is given by..Cramer's rule for the solution of a system of equations in 2 variables
We recall from our earlier classes that a system of linear equation with two variables is given by This system of linear equation may have either one solution or infinitely many solutions or no solutio..
We recall from our earlier classes that a system of linear equation with two variables is given by This system of linear equation may have either one solution or infinitely many solutions or no solutio..Note:
In the above method note that To obtain D 1 , replace a 1 , a 2 , a 3 by d 1 , d 2 , d 3 in D To obtain D 2 , replace b 1 , b 2 , b 3 by d 1 , d 2 , d 3 in D To obtain D 3 , replace c 1 , c 2 , c 3 by d 1 , d 2 , d 3 in..
In the above method note that To obtain D 1 , replace a 1 , a 2 , a 3 by d 1 , d 2 , d 3 in D To obtain D 2 , replace b 1 , b 2 , b 3 by d 1 , d 2 , d 3 in D To obtain D 3 , replace c 1 , c 2 , c 3 by d 1 , d 2 , d 3 in..Suggested answer:
The determinant of coefficients ..
The determinant of coefficients ..Case II:
If D = 0 and D 1 , D 2 and D 3 are not all zero, then the system is inconsistent, that is the system has no solutio..
Non Homogenous Equations (Solution by the Matrix Method)
Consider the non-homogeneous equations a 1 x + b 1 y + c 1 z = d 1 a 2 x + b 2 y + c 2 z = d 2 a 3 x + b 3 y + c 3 z = d 3 This can be written as |A| may or may not be ze..
Consider the non-homogeneous equations a 1 x + b 1 y + c 1 z = d 1 a 2 x + b 2 y + c 2 z = d 2 a 3 x + b 3 y + c 3 z = d 3 This can be written as |A| may or may not be ze..Case II:
A - 1 does not exist But if (adj A) B = 0, then the system is consistent with infinite number of solutions or has no solution. the system is inconsistent i.e., it has no solutio..
A - 1 does not exist But if (adj A) B = 0, then the system is consistent with infinite number of solutions or has no solution. the system is inconsistent i.e., it has no solutio..Suggested answer:
The given equations are 2x - y + z = -3 3x - 0.y - z = - 8 2x + 6y + 0.z= 2 = 2(6) +1(2) + 1(18) = 12 +2 + 18 = 32 The system has a unique solutions. A 1 1 = (0 + 6) = 6, A 1 2 = -(0 + 2) = -2, A 1 3 = 18 A 2 1 = 6, A 2 2 = -2, A 2 3 = -14 A 3 1 = 1, A 3 2 = 5, A 3 3 = 3 ..
The given equations are 2x - y + z = -3 3x - 0.y - z = - 8 2x + 6y + 0.z= 2 = 2(6) +1(2) + 1(18) = 12 +2 + 18 = 32 The system has a unique solutions. A 1 1 = (0 + 6) = 6, A 1 2 = -(0 + 2) = -2, A 1 3 = 18 A 2 1 = 6, A 2 2 = -2, A 2 3 = -14 A 3 1 = 1, A 3 2 = 5, A 3 3 = 3 ..Suggested answer:
= (14 - 12) - (7 - 3) + (4 - 2) = 2 - 4 + 2 = 0 The system may have infinite number of solutions or no solution. Put x = k in (1) and (2) and solve y + z = 6 - k 2y + 3z = 14 - k. Solving the above two equations, we have z = k + 2 and y = 4 - 2k When x = k, substituting these values of x, y ..
= (14 - 12) - (7 - 3) + (4 - 2) = 2 - 4 + 2 = 0 The system may have infinite number of solutions or no solution. Put x = k in (1) and (2) and solve y + z = 6 - k 2y + 3z = 14 - k. Solving the above two equations, we have z = k + 2 and y = 4 - 2k When x = k, substituting these values of x, y ..See what our Users say :
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