Introduction to types of linear systems-Two variables.
There are three types of linear systems-Two variables, the detailed view can be viewed as per the next slide image:
1-The first type is the consistent independent, the system is a system of linear equations that is consistent independent when it has exactly one solution. When this is the case, the graphs of the lines in the system cross at exactly one point.
2-The second type is the consistent dependent. A system of linear equations is consistent -dependent if it has an infinite number of solutions. When this is the case, the graphs of the lines in the system are the same, meaning the equations in the system represent the same line.
3-The third type is inconsistent. A system of linear equations is inconsistent if it has no solutions.
When this is the case, the graphs of the lines in the system do not intersect, meaning they are parallel.
Matrix Row operations.
Prior to solving types of linear systems-Two variables, it is necessary to have a review of the matrix row operations.
1-The first operation is the switch between any two rows, for the example included in the slide image, we have two rows of(2×3) matrix. The first row is ( 2 5 3) while the second row is ( 3 4 6). it is possible to switch row 1 to row 2 and also the reverse can be done.
2-The second operation is the multiplication of any row by a non-zero constant. For instance, for the same matrix, previously indicated, we can multiply the first row by 3, then a new row is formed by the value of the product of 3 by the first row.
3-The third operation is the addition of two rows and the output is placed as a new second row.
The Use of row operations to solve a system of two/three equations.
By using row operations, we can convert a system of linear equations with two unknowns to a row echelon form, for which the diagonal will contain a series of 1’s, while the second row/the first column is zero.
By using row operations, we can convert a system of linear equations with three unknowns to an echelon form, for which the diagonal will contain a series of 1’s, while the second row/the first column is zero and also the Third row/the first column is zero.
The Gaussian elimination method is an algorithm that uses elementary row operations to solve a system of linear equations. The goal of this method is to rewrite an augmented matrix in row echelon form. Please refer to College Algebra and Trigonometry chapter -10 from Richard N. Aufmann ‘s book. This is quoted from Algebra for college students by Margret Lial. This is the Amazon link for the book.
Solved problems for types of linear systems-Two variables.
The linear systems-Two variables can be expressed as two equations. The first equation is a1*x+b1*y=c1. The second equation is a2*x+b2*y=c2, where a1, a2,c1, and c2 are non-zeros.
1-The first type is the consistent independent, two lines intersecting with each other, for example, we have the first equation as y=x+2, while the second equation is (-4x-3y =-20).
Solving these two equations together, we get an intersecting point as (2,4). The intersecting point will satisfy the two equations.
for the inconsistent, A system of linear equations is inconsistent if it has no solutions, we have two equations, for example, the first line equation is 2x+y=4. While the second line equation is 2x+2y=6. The system of equations is parallel to each other, we can get an equal slope as m1=m2.
The consistent dependent is a system of linear equations is consistent, where we have multiple solutions. We have two equations.(x+y=1) and (2x+2y=2).
For the given equations we will use the following steps to figure out the types of linear systems-Two variables.
Steps used to find Types of linear systems-Two variables.
Case of consistent independence.
We have the first equation as -x+y=+2, while the second equation is (-4x-3y =-20). We will follow the next steps to find Types of linear systems
Step 1:We write the two equations in augmented form.
Step 2: make sure that we have 1 at the first column/first row, otherwise, we will do row necessary row operation.
Step 3:use row operation to let the element of the 2nd/first column=0, this is done by multiplying the first row by(-4) and adding to row 2, the result will be placed in the second row.
We have used the multiplication and addition row operations. Please notice that the elements of the first row have not been changed.
The next step is to divide the second row by (-7), to let the diagonal of at 2nd row/the second column.
The augmented matrix will lead us to two new equations. The first equation is -x+y=2, and the second equation is y=4. We will substitute by y=4 in the -x+y=2. We get a value of x=2. Thus we have an intersecting point (2&4).
Case of a consistent dependent.
We have the first equation as x+y=1, while the second equation is (2x+2y=2). We will follow the next steps to find Types of linear systems
Step 1:We write the two equations in augmented form.
Step 2: make sure that we have 1 at the first column/first row, otherwise, we will do row necessary row operation.
Step 3:use the row operation to let the element of the 2nd/first column=0, this is done by multiplying the first row by(-2) and adding to row 2, the result will be placed in the second row.
The augmented matrix will lead us to two new equations. The first equation is -x+y=2, the second equation is 0*x+0*y=0
There are multi-solution for these two equations. This is a typical case of consistent dependent equations.
A system of linear equations is inconsistent if it has no solutions.
We have the first equation as 2x+y=4, while the second equation is (4x+2y=+6). We will follow the next steps to find Types of linear systems
Step 1:We write the two equations in augmented form.
Step 2:use row operation to let the element of the 2nd/first column=0, this is done by multiplying the first row by(-2) and adding to row 2, the result will be placed in the second row.
The augmented matrix will lead us to two new equations. The first equation is 2x+y=1, the second equation is 0*x+0*y=-2
since for any value of x,y the left-hand side is zero, while the right-hand side is equal to -2, this is nonlogic since zero does not equal -2, and this will lead to a contradiction.
This is a typical case of inconsistent equations.
This is a link to the next post: Easy introduction to row echelon form.
This is a link to the matrix calculator.
For a useful external link, math is fun for the matrix part.
