MTH603 Quiz 1 Solution and Discussion
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The statement, 7265 instead of 7269 lies in the category of:
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@zaasmi said in MTH603 Quiz 1 Solution and Discussion:
The statement, 7265 instead of 7269 lies in the category of:
With regard to the dangers from lead opinion , if you removed the loose paint off the hands poisoning , it is pretty hard on … Do you not think your statement is rather a 7269. … That 7265. Of course they will not need soap if they can is where sand - papering comes in . rub the paint off dry from their hands P - I said you 7293.
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If there are three equations in two variables, then which of the following is true?
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@zaasmi said in MTH603 Quiz 1 Solution and Discussion:
If there are three equations in two variables, then which of the following is true?
A system of equations in three variables involves two or more equations, each of … Plug in these values to each of the equations to see that the solution satisfies all … when graphing each equation in the system and then finding the intersection point of … A three dimensional box with three slanted planes crossing through it.
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@zaasmi said in MTH603 Quiz 1 Solution and Discussion:
@zaasmi said in MTH603 Quiz 1 Solution and Discussion:
If there are three equations in two variables, then which of the following is true?
A system of equations in three variables involves two or more equations, each of … Plug in these values to each of the equations to see that the solution satisfies all … when graphing each equation in the system and then finding the intersection point of … A three dimensional box with three slanted planes crossing through it.
Explanation:
The simple answer is yes.If you have two consistent equations and they are linearly independent, then you will have to assign arbitrary values to one of the variables. If you have two consistent equations and they are linearly dependent, then you will have to assign arbitrary values to two of the variables, both cases lead to an infinite number of solutions. If they inconsistent, then obviously there is no solution.
Graphically two consistent linearly independent equations will form 2 planes in
R
3
that intersect and all solutions will lie on the line of intersection leading to an infinite number of solutions.Two consistent linearly dependent equations will just be a single line in
R
3
and all solutions will lie on this line, leading to an infinite number of solutions.If the two equations are inconsistent then you will have 2 parallel lines in
R
3
, and no point of intersection, hence no solutions. -
@zaasmi said in MTH603 Quiz 1 Solution and Discussion:
If there are three equations in two variables, then which of the following is true?
Dependent Systems of Equations with Three Variables
We know from working with systems of equations in two variables that a dependent system of equations has an infinite number of solutions. The same is true for dependent systems of equations in three variables. An infinite number of solutions can result from several situations. The three planes could be the same, so that a solution to one equation will be the solution to the other two equations. All three equations could be different but they intersect on a line, which has infinite solutions (see below for a graphical representation). Or two of the equations could be the same and intersect the third on a line (see the example problem for a graphical representation).
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The first row of the augmented matrix of the system of linear equations is: 2x+z=4 x-y+z=-3 -y+z=-5
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@zaasmi said in MTH603 Quiz 1 Solution and Discussion:
The first row of the augmented matrix of the system of linear equations is: 2x+z=4 x-y+z=-3 -y+z=-5
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The root of the equation 3x−ex=0 is bounded in the interval.
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A square matrix is said to be diagonally dominant if the magnitude of the diagonal element in a row is greater than or equal to the sum of the magnitudes of all the other non-diagonal elements in that row for each row of the matrix.
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Simplifying
0x + 0y = 2Anything times zero is zero.
0x + 0y = 2Anything times zero is zero.
0 + 0y = 2Combine like terms: 0 + 0 = 0
0 = 2Solving
0 = 2Couldn’t find a variable to solve for.
This equation is invalid, the left and right sides are not equal, therefore there is no solution.
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In numerical linear algebra, the Jacobi method (a.k.a. the Jacobi iteration method) is an iterative algorithm for determining the solutions of a strictly diagonally dominant system of linear equations. Each diagonal element is solved for, and an approximate value is plugged in.
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while using relaxation method, which of th folowing is the
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Solution to the linear equation 2x + 0y = 0
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the linear equation 2x-0y-2=0 has