Category: linear Algebra -Matrices

How do we use matrices to solve linear system equations?

  • 9- How to Use a matrix for the quadratic polynomial function?

    9- How to Use a matrix for the quadratic polynomial function?

    How to Use a matrix for the interpolating polynomial function?

    We discussed the quadratic function or the quadratic interpolation function in two posts. The first post is an easy introduction to quadratic interpolation, and the second post solves problems for quadratic interpolation.

    In this post, we will solve three solved problems. The main idea of these solved problems is to use the matrix for the quadratic function to get the values of its parameters or simply get the quadratic matrix.

    The first solved problem for the matrix for the interpolating polynomial function

    Use row echelon form to get a0,a1, and a2 values.

    We need to find the interpolating polynomial function P(t)= a0+a1t+a2t2 for the given data points, which are (1,12),(2,15), and (3,16).

    We start by writing the three equations for P(t) for the given three points as follows:

    We plug in the value of t=1, and P(1) equals 12. The first equation is P(1)=a0+a1(1)+a2(1)^2=12. We plug in the value of t=2, and P(1) equals 15. The second equation is P(2)=a0+a1(2)+a2(2)^2=15. We plug in the value of t=3, and P(1) equals 16. The second equation is P(3)=a0+a1(3)+a2(3)^2=16.

    Our unknowns are a0, a1, and a2. Using the augmented matrix, we will find the first row as (1 1 1 12). The second row is (1 2 4 15). Finally, the last row is (1 3 9 16). We will convert the augmented matrix to the row echelon form.

    Solved problem for a matrix for the quadratic function.

    The first row has a leading item equal to 1; to let a21 and a32 equal zero, we will perform the elementary row operations as we can see in the next slide image. Multiply the first row by -1 and add the value to the second row, while for the third row, we will multiply the first row by minus one and add the value to the third row.

    Next, we will move to the second leading item in the second row, equal to 1; we will let the element a22 equal to zero by multiplying the second row by -2 and adding the value to the third row. As we can see, we have the row echelon form for the quadratic function. We have a0=7. We can use back substitution to get the value of a1 and a2.

    first Solved problem, values of a0,a1 and a2

    Use a reduced row echelon form-interpolating polynomial function.

    If we wish to use the reduced row echelon form, for the matrix for the quadratic function, we want to have zeros above the pivots for the third and second rows. We consider a33 as a pivot, then multiply R2*-3 and add the value to R2; we will get a zero for the element a23. We will multiply the third row by -1 and add the value to the first row.

    Next, we will move to the second pivot column. Multiply R2*-1 and add the value to R4; we will get a zero for element a12. We have a0=7& a1=6, and a2=-1. We can rewrite the interpolating polynomial as P(t)=1+6*t-t^2.

    Use RRef for a0,a1 and a2- first Practice problem

    Check the interpolating function for the given three points’ data. The details can be seen in the next slide image.

    Practice problem 1-verify answer.

    This is a graph for the interpolating Quadratic polynomial.

    Graph for the interpolationg polynomial-practice problem -1.
    Solved problem-1-graph

    The second solved problem for the matrix for the quadratic polynomial function.

    The second solved problem is a practice problem, for the matrix for the quadratic polynomial function. It is required to find the quadratic polynomial function P(t)=a0+a1x+a2x^2 that satisfies the condition P(0)=f(0), p'(0)=f'(0) and p”(0)=f”(0).

    In this solved problem, instead of giving three points. One point is given, and the slope at that point shall equal the slope of the original function, e^2x. The second derivative P”(0) is also equal to the second derivation of the original function, which is e^2x.

    The interpolating quadratic polynomial y value at point zero is estimated by plugging in the function e^2x and setting x=0; the value at point zero equals e^0=1.

    Practice problem-2-interpolating polynominal

    The three polynomial equations are written in terms of x. Please refer to the next slide image.

    Practice problem-2-Polynomial conditions

    The first and second derivatives of the original function are estimated at x=0, as shown in the next slide image. We start by writing the three equations for P(x) for the given data: We plug in the value of x=0, and P(0) equals 1.

    The expression is P(x)=ax^2+bx+c. The first equation is P(0)=a(0)^2+b(0)+C=1. We plug in the value of P'(0), equal to 2. The expression is P'(x)=2ax+b. The second equation is P'(0)=2a(0)+b=2.

    We plug in the value of P”(0), equal to 4. The expression is P”(x)=2a. The third equation is P”(0)=2a=4. We can write the three equations in the form of two-matrix multiplication. The product is equal to (1 2 4).

    The second solved problem for quadratic interpolation.

    We can write the augmented matrix. We will swap the first and third rows in the row echelon form and convert it to a reduced echelon form. The final values for a,b, and c can be obtained as a=2, b=2, and c=1.

    The polynomial can be written as p(x)=2x^2+2x+1.

    The following slide image shows the derivatives of the polynomial function and the original functions.

    Practice problem-2-Polynomial expressions

    The third solved the problem of the matrix for the quadratic polynomial function.

    For the third solved problem for a quadratic matrix.It is required to find the quadratic polynomial function P(t)=a0+a1x+a2x^2 that satisfies the condition P(1)=f(1),p'(1)=f'(1) and p”(1)=f”(1).

    One point is given, and the slope at that point shall equal the slope of the original function, which is xe^(x-1). The second derivative P”(0) is also equal to the second derivation of the original function, which is xe^(x-1). The interpolating quadratic polynomial y value at point zero is estimated by plugging in the function xe^(x-1) and setting x=1; the value at point zero equals 1*e^(1-1)=1.

    The first and second derivatives of the original function are estimated at x=1. The next slide image shows how to estimate these values.

    Practice problem-3-interpolating polynominal
    Practice problem-3-interpolating polynominal

    We start by writing the three equations for P(x) for the given data: We plug in the value of x=1, and P(1) equals 1. The expression is P(x)=ax^2+bx+c.The first equation is P(1)=a(1)^2+b(1)+C=1. We plug in the value of P'(1), equal to 2. The expression is P'(x)=2ax+b. The second equation is P'(1)=2a(1)+b=2. We plug in the value of P”(1), equal to 3. The expression is P”(x)=2a. The third equation is P”(1)=2a=3.

    Practice problem-3-writting equations

    We can write the augmented matrix. We will swap the first and third rows in the row echelon form. We have a row echelon form for the third solved problem. We will convert it to the reduced echelon form.

    We can write the augmented matrix. We will swap the first and third rows in the row echelon form and convert it to a reduced echelon form.

    Practice problem-3-matrix operations

    The final values for a,b, and c can be obtained as a=3/2, b=-1, and c=1/2. The final value for the polynomial is shown in the next slide image. The following slide image shows the derivatives of the polynomial function and the original functions.

    Practice problem-3-Polynomial expressions

    For a useful external link, math is fun for the matrix part.

Practice problem for REF-The first part.

Practice problem for reduced echelon form-REF.

In the first part of the post, we will introduce Practice problems for reduced echelon form. or abbreviated as REF. There are several parts a,b,c, and d.

For the first practice problem, it is required to check which matrices are in reduced echelon form and justify the reason for

Part A practice problem for reduced echelon form.

In the first part, part a, of the Practice problem for reduced echelon form, we have three pivots, each equaling 1. The three are located in the first, second, and third rows. We have zeros below the first pivot.

There are zeros above and below the pivot in the second column. Meanwhile, the third column has two zeros above the pivot. These conditions satisfy the requirement of the reduced row echelon form shown in the next slide image.

Part A-excercise 1-reduced echelon form

Part B practice problem for reduced echelon form-REF.

In the second part, part b, we have only two pivots, each equaling 1, which can be considered a non-zero element. The two ones are located in the first and second rows. We have zeros below the first pivot.

There are zeros above and below the pivot in the second column.

These conditions satisfy the requirement of the reduced row echelon. For the third row, there is no leading entry. These conditions satisfy the requirement of the reduced row echelon form shown in the next slide image.

Part b-exercise 1 for reduced echelon

Part C-practice problem for reduced echelon form.

In the third part, part c, of the Practice problem for reduced echelon form, we have only two pivots, each equaling 1, which can be considered a non-zero element. The two ones are located in the first and second rows. We have zeros below the first pivot.

In the second column, there are zeros above and below the pivot. In the third row, we have zeros. The pivot is not leading in the last row. The given matrix is not in the reduced echelon form since the row of the zeros should be placed as the last row.

.

Practice problem -part C-reduced echelon form.

Part D-practice problem for reduced echelon form.

In the fourth part, part d, of the Practice problem for reduced echelon form, we have only one leading pivot as one in the first row. In the second row, the leading pivot is not equal to 1; it is equal to 2, which does not comply with the rules of the reduced echelon form. There is no zero above the a21 element.

The matrix can be considered an echelon form but not a reduced one. The next slide image shows the details of the arrangement of the leading pivots.

Part d-excercise 2 reduced echelon form.

In the fourth part, part d, of exercise 2 of the Practice problem for reduced echelon form, we have a leading pivot as zero in the first row. This is not acceptable whether in the reduced form of the echelon form. The details of the arrangement of the leading pivots are shown in the next slide image.

Part d-excercise 2 reduced echelon form.

Traffic flow application-the second part.

The application for traffic flow is the second part of the post. We have a practice problem with the distribution of traffic at various intersections.

The traffic flows are expressed as linear equations based on the inflows and outflows for every joint.

The main principle is that the sum of inflows equals the sum of outlaws. Based on the given data at joint A, the sum of entering=x1+450 while the sum of outgoing is x2+610.

The equations for points B, C and D can be written accordingly. The next slide image shows the details of the given incoming and outgoing joints or intersections.

Traffice flow example use matrix form.

We can rearrange a set of linear equations for the four joints of intersections and find values of the unknowns x1&x2&x3 and x4.

A series of matrix operations can turn the augmented matrix into an echelon form. The next slide image shows the detail of the echelon form of the four equations or intersections.

A-Solving system of linear equations for traffic flow.

We can proceed to perform the reduced echelon form and arrange the pivots. We have basic variables and, accordingly three pivot columns. We have a free variable X4, which can take any value. The last row has the equation of zero equal to zero.

Row echelon form of the equations for traffic.

We can rearrange the relations between the four variables and write the general equations that show that x4 is a free variable. The variables X1&x2&X3 are the basic variables.

Solving system of linear equations

Here is a link to the post: 9- How to Use a Matrix for the quadratic polynomial function?

This is a link to the matrix calculator.

For a useful external link, math is fun for the matrix part.

  • 7b- Practice problems for Gauss Jordan elimination.

    7b- Practice problems for Gauss Jordan elimination.

    Practice problems for Gauss Jordan elimination.

    Practice problems for Gauss Jordan elimination-first practice problem.

    We will introduce a practice problem Gauss Jordan elimination, or the reduced echelon form. Practice problem number #33 is from Stewart’s book College Algebra.

    For the first practice problem #33 for Gauss Jordan elimination. Solve the following systems of linear equations. The detail of the given matrix for practice problem #33 is shown in the next slide image.

    Solvd problem 33 for gausss jordan elimination.

    Practice problems for Gauss Jordan elimination a11 as a pivot to eliminate a21 and a31.

    For the given system of linear equations, we will write these equations in augmented form.

    Practice problems for Gauss Jordan elimination

    We will introduce a vertical line between the third and fourth columns to show the augmented matrix. The next step is to let a21 or the second row/first column be equal to zero. Make sure that the element a11 is equal to 1, then add (-a21/a11)*R1 to row2 to get the zero value. In this process, we consider a11 as the pivot element.

    Similarly, let a31 or the third row/first column equal zero. Add (-a31/a11)*R1 to row 3 to get the zero value. Again, in this process, we consider a11 as the pivot element.

    Carry on the row operations. Make sure that element a22 is equal to 1, and this is done by multiplying row 2 by (-1/2).

    Use a22 as a pivot to eliminate a32.

    Use a22 as a pivot by adding R3 to -(a23/a21)*R2. Perform the row operation to get the reduced echelon form or simply an upper matrix with three ones as diagonal.

    Refer to the slide image to check the necessary row operations for the reduced row echelon form of Gauss elimination.

    Practice problems for Gauss Jordan elimination

    Use a33 as a pivot to eliminate a23& a13.

    We aim to convert the matrix to a diagonal matrix of 1s. We will use the element a33 to eliminate a23. This is done by adding(-a23/a33)*R3 to R2.

    Similarly, we use the element a33 to eliminate a13. This is done by adding(-a13/a33)*R3 to R1 to get the zero value. Again, in this process, we consider a33 as the pivot element.

    Use a22 as a pivot to eliminate a12.

    We will use the element a22, to eliminate a12, this is done by adding(-a12/a22)*R2 to R1. Now we have obtained the required diagonal matrix with 1 at the diagonals; please refer to the next slide image for clarity.

    Gauss Jordan elimination steps for matrix.

    We can then get the values of Z=1&Y=0 and X=-1. In this method, we do not need to make back substitutions. However, it is important to check the Values of X, Y, and Z in any of the three systems of equations to ensure that the solution is accurate.

    Check the values of x&y and z by substitution for problem -38

    Practice problems for Gauss Jordan elimination-second practice problem.

    1-For the second practice problem #38 for Gauss Jordan elimination. Solve the following systems of linear equations. The next slide image shows the details of the matrix given for practice problem #38.

    Solvd problem 38 for gausss jordan elimination.

    For the given system of linear equations, we will write these equations in augmented form.

    How to write matrix in augmented form?

    We will introduce a vertical line between the third and fourth columns to show the augmented matrix. The next step is to let a21 or the second row/first column be equal to zero.

    Make sure that element a11 is equal to 1. First, we will divide the first row by 10. We will add (-a21/a11)*R1 to row 2 to get the zero value. In this process, we consider a11 as the pivot element.

    Similarly, let a31 or the third row/first column equal zero. Add (-a31/a11)*R1 to row 3 to get the zero value. Again, in this process, we consider a11 as the pivot element.

    Use a22 as a pivot to eliminate a32.

    First, divide R2 by 5 to let the pivot a22 equal 1; also, divide r3/5 to minimize the numbers.

    Use a22 as a pivot by adding R3 to -(a23/a21)*R2. Divide r3 by 7 to let a33=1.

    Perform the row operation to get the reduced echelon form or simply an upper matrix with three ones as diagonal.

    Use a33 as a pivot to eliminate a23&a13.

    We aim to convert the matrix to a diagonal matrix of 1’s. We will use the element a33, to eliminate a23, this is done by adding(-a23/a33)*R3 to R2.

    Similarly, we use the element a33 to eliminate a13; this is done by adding(-a13/a33)*R3 to R1 to get the zero value. Again, in this process, we consider a33 as the pivot element. Now we have obtained the required diagonal matrix with 1 at the diagonals; please refer to the next slide image for clarity.

    The values of x &y and z for the practice problem-38.

    We can then get the values of Z=-2&Y=1 and X=+1. In this method, we do not need to make back substitutions. It is important to check the Values of X& Y, and Z in any of the three equation systems to ensure that the solution is accurate.

    Check the values of x&y and z by substitution for problem -38

    The next post will discuss the Practice problems for reduced echelon form.

    This is a link to the matrix calculator.

    For a useful external link, math is fun for the matrix part.

  • 7a- Practice problems for back substitution.

    7a- Practice problems for back substitution.

    Back substitution practice problems.

    First practice problem#25.

    We will introduce practice problems for back substitution from Stewart’s book College Algebra: Two Practice Problems, number 25&27.

    1-For the first practice problem, #25 for back substitution.

    A matrix is given in the row-echelon form. a) write the system of equations for which the given matrix is the augmented matrix. b) use back substitution to solve the system. The following slide image shows the details of the matrix given for practice problem #25.

    Practice problems for back substitution.

    For part a of the first practice problem, we will introduce a vertical line between the third and fourth columns to show the augmented matrix.

    The next step is to write the system of linear equations; we have three rows, which means that we will write three systems of equations that Can be written as follows: A—assume that we have a vertical column with three rows and one column that is (X Y Z).

    The first row (1 -2 4) will be multiplied by the column vector (X Y Z). The first system of linear equations is
    X-2y+4Z=3.

    The second row (0 1 2) will be multiplied by the column vector (X Y Z). The second system of linear equation is 0X+Y+2Z=7.

    The third row (0 0 1) is to be multiplied by the column vector (X Y Z). The third system of linear equations is 0X+0Y+1Z=2.

    The next slide image shows the triangular form for the three systems of equations.

    Practice problem#25 for back substitution.

    We want to get the values of X, Y, and Z. We start with the last row, for which we have a Z value equal to 2.

    We move back to the second system of linear equations and substitute the value of Z in that equation to get the value of Y, which is equal to 7-4=3.

    We move back to the first system of linear equations and substitute the value of Y&Z in that equation to get the value of X equal to 3-8+6=1. It is essential to check the Values of X& Y, and Z in any of the three systems of equations to ensure the solution is accurate.

    Solving the system of linear equations by back substitution.

    Second practice problem #27.

    1-For the second practice problem #27 for back substitution. A matrix is given in the row-echelon form.

    a) write the system of equations for which the given matrix is the augmented matrix. b) use back substitution to solve the system. The next slide image shows the details of the matrix given for practice problem #27.

    practice problem 27 for back substitution.

    We will introduce a vertical line between the fourth and fifth columns to show the augmented matrix.

    The next step is to write the system of linear equations; we have four rows, which means that we will write four systems of equations that Can be written as follows: A- assume that we have a vertical column that has three rows with one column that is (X Y Z W)

    The first row (11 2 3 -1) is to be multiplied by the column vector (X Y Z W). The first system of linear equations is
    X+2Y+3Z-W=7.

    The second row (0 1 -2 0) is to be multiplied by the column vector (X Y Z W). The second system of linear equation is 0X+Y-2Z+0W=5.

    The third row (0 0 1 2) is to be multiplied by the column vector (X Y Z W). The third system of linear equation is 0X+0Y+1Z+2W=5.

    The fourth row (0 0 0 1) is to be multiplied by the column vector (X Y Z W). The fourth system of linear equation is 0X+0Y+0Z+1W=3. Two systems of linear equations are shown in the next slide image.

    Practice problem#27 for back substitution.

    To get the values of X, Y, Z, and W, we start with the last row, for which we have a W value equal to 3.

    We proceed to the third system of linear equations and substitute the value of W in that equation to get the value of Z equal to 5-6=-1. These are the arrangements for the 3×4 matrix to have a row echelon form.

    The values of W& Z and y for practice problem #27.

    We proceed to the second system of linear equations and substitute the values of Z&W in that equation to get the value of Y equal to 3.

    The values of W& Z and y for practice problem #27.

    Finally, we go to the first system of linear equations and substitute the values of Y & Z, and W in that equation to get the X value that is equal to 7.

    The value of X in practice problem #27.

    Check whether the estimated x, y, and z are correct.

    It is important to check the Values of X, Y, and Z in any of the four systems of equations to ensure that the solution is accurate. The next slide shows the steps I use to check the x,y,z, and w values for all the given equations.

    Check that all the unkowns will satisfy the system of equations.

    In the next post, we will solve practice problems for Gauss Jordan elimination.

    This is a link to the matrix calculator.

    For a valid external link, math is fun for the matrix part.

  • 7- Easy introduction to row echelon form.

    7- Easy introduction to row echelon form.

    Introduction to row echelon form or REF.

    Our subject, as of today, is the row echelon form. and also the reduced row echelon form. We are using the row matrix operation to create a new form which is the RE form. The following steps are explained:

    1-For the first row of the first item, the first row/first column is called the leading item, which must equal 1.

    2-For the first column/2nd row and third row, all the elements are zeros.

    2-For the second row, the leading item which is 1 will be below and to the right of the previous row’s leading item, as we can see the number 1 is the second row/2nd column position. Put a zero below the leading item in the third row.

    For the third row, we expect to have a leading I as in the third column. This is one definition for the RE form Quoted from Steven Leon’s book Linear Algebra with Applications.

    But other authors, state that the leading item is to be non-zero items that conclude that the number could be 1 or any positive value.

    Please refer to the wiki for the RE form for more information. The details of the RE form and how to create it are shown in the next slide image, and the two definitions are indicated.

    Row Echelon Form expression.

    These are the three items for the REF, which stands for the free echelon form based on the leading item being nonzeros. All rows of zeros should be the last row at the bottom.

    Definition of the row echelon form.

    What is the RREF-reduced row echelon form?

    The RREF stands for the reduced row echelon form, this form has the three previous items included in the RE form but all leading entries must be 1s.

    The column that has the first leading one should have zeros below it. At the same time, the columns of the leading ones should include zeros above and below the leading ones.

    The definition of reduced row echelon.

    This is the case of a matrix with a 3×3 dimension in the REF. We have a diagonal that contains three ones; the leading items are nonzero values. The first leading one has two zeros below the leading item.

    A matrix in the row echelon form.

    This is the case of a matrix of a dimension of 3×4 in the augmented form. We have a diagonal of nonzeros items, shown as ones. The solid boxes include any possible values.

    The pivot column in matrix A is the location that corresponds to the leading items. In the shown matrix, we have three leading ones, meaning we have three pivot columns.

    3x4 natrix with REF and pivot column expression.

    The use of the RREF to solve a system of linear equations.

    The RREF echelon form is the method used by Gauss-Jordan to get the solution of linear equations to avoid the back substitution method.

    Step 1: Create an RE form using row operations.
    Step 2: Use the leading term in the third row to create a zero for the elements above it.
    Step 3 Use the leading term in the second row to create a zero for the element above it.

    Based on these operations, the values of the fourth column will be changed, and at the end, we can get the value of the unknowns directly.

    For instance, if we have the following system of linear equations as x+y+z=6 & 2x-y+z=5 and 3x+y-2z=9.

    Perform a row operation to create an RREF, and we will get the arranged augmented matrix.
    This new arrangement will give us the values of x=3, y = 2, and z = 1. There is no need to use the back substitution to get these values.

    A solved problem for Gauss jordan.

    Another example is for the REF.

    In the given matrix A of a dimension 4×5, it is required to perform row reduction to the matrix and locate the pivot columns.

    The first row has a leading term of zero, which can’t be accepted, so we will replace the first row with row number 4, which has a leading term of 1 and is accepted as the non-zero item.

    We will perform a series of row operations to let the items below the first leading item be zeros. The steps used to have the reduced echelon form for a 4×5 matrix. The shape of the reduced echelon form is shown in the next slide image.

    Example 2 for the row echlon form of a matrix.

    We have the third row containing five zeros, which we cannot accept, and we will swap the third row to the fourth row to match the condition of the REF. The places of the leading items and the pivot columns are shown in the next slide image. the leading terms are 1&2&-5 which are a11 & a22 and a34.

    Example 2-continued- for the row echlon form of a matrix.

    We have used the REF to find the places of leading terms and the pivot column, so we will return back to the original matrix A, with the information that we have in the next slide image.

    page 9- matrix 7-The location of the pivot columns and leading terms in Matrix A.

    In the next post, we will look at practice problems for back substitution.

    This is a link to the matrix calculator.

    For a useful external link, math is fun for the matrix part.

  • 6- Introduction to types of linear systems-Two variables.

    6- Introduction to types of linear systems-Two variables.

    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.

    Types of linear systems-Two variables

    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, and 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.

    Matrix row operations.

    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.

    The Use of row operations to solve a system of two/three equations.

    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).

    page 5- matrix 6-System of two equations for two unkowns.

    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.

    Case of consistent dependent in two unknown.

    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 consistent dependent in two unknown.

    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.

    Solved problem for the consistent dependent . of the system of linear equation.

    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.

    page 9- matrix 6.-case of inconsistent of the system of linear equation.

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