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2: Systems of Equations and Matrices - Mathematics


2: Systems of Equations and Matrices - Mathematics

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Systems of Linear Equations

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JOEL LEWIS: Hi. Welcome back to recitation. You've been learning in lecture about matrices and their various applications, and one of them is to solving systems of linear equations. So I have here a system of three linear equations for you. 2x plus c*z equals 4, x minus y plus 2z equals pi, and x minus 2y plus 2z equals minus 12. So what I'd like you to do is the following.

Find the value of c-- or all values of c-- for which, first of all, there's a unique solution to this system. Second of all, for which the corresponding homogeneous system has a unique solution. So remember that the corresponding homogeneous system is the system where you just replace these constants on the right by 0. So it's a very similar-looking system. The left-hand sides are all the same, but the right-hand sides are replaced with 0. So you want to find the value of c for which this system has a unique solution, the value of c for which the corresponding homogeneous system has a unique solution, and also the values of c for which the corresponding homogeneous system has infinitely many solutions.

Note that I'm not asking you to solve this system of equations, although you're welcome to do so if you like. Although, of course, whether you can or not might depend on the value of c. So why don't you pause the video, take a little while to work out the solutions to these three questions, come back, and we can work it out together.

So hopefully you have some luck working out these problems. Let's start working through them together. So I'm actually going to take parts a and b together at the same time.

And the reason that I'm going to do that is that one thing you've learned is that a system has a unique solution for, on the right-hand side-- sorry-- a system has a unique solution, like this, a square system of linear equations has a unique solution if and only if it has a unique solution regardless of what the right-hand side is. So in particular, the answer to a and the answer to b are exactly the same.

So values of c for which this system has a unique solution are exactly the same as values of c for which the homogeneous system has a unique solution. Now the solutions will be different, of course. But the value of c-- or the values of c-- that make it solvable uniquely, make it solvable uniquely for all right-hand sides.

And so which values of c are those? Well, those are the values of c for which the coefficient matrix on the left-hand side is invertible. So if the coefficient matrix on the left-hand side is invertible, then we can solve this system and we get a unique solution. If it's not invertible, then either we can't solve this system-- like, there are no solutions-- or we can solve this system, but there are infinitely many solutions.

So in both questions a and b, we're asking for the value of c for which the coefficient matrix of the left-hand side is invertible, and that will be when we have a unique solution. So how do we know when a matrix is invertible? Well, let's write down what the matrix is first of all.

So this matrix M that we're after is equal to the matrix 2, 0, c 1, minus 1, 2 1, minus 2, 2. So this is the coefficient matrix M of that system, and we want to know for which values of c is it invertible.

Well, when is a matrix invertible? A matrix is invertible-- square matrix is invertible-- precisely when it has non-zero determinant. So we just need to look at the determinant of this matrix. So you've learned how to compute determinants of matrices, I think.

So let's, in this case, we have the det M. So it's a sum or difference of six different terms, and you could get it, for example, by the Laplace expansion if you wanted to. So I'm just going to write out what the six terms are. So it's 2 times minus 1 times 2, plus 0 times 2 times 1, plus c times 1 times minus 2, minus c times minus 1 times 1, minus 2 times minus 2 times 2, minus 0 times 1 times 2. So this is the determinant of this matrix.

You can get it either just by remembering which terms are which and which get a plus sign and which get a minus sign, or by doing the Laplace expansion, or by whatever other tricks you might happen to know. So now we need to know whether or not this determinant is 0. So let's work out what this is.

So this is-- let me start simplifying it. So this is minus 4 plus 0 minus 2c-- this is minus minus c, so plus c-- this is minus minus 8, so plus 8, which is equal to 4 minus c. So the determinant-- right, two of those terms are 0, and so I just get to leave them out. So the determinant of this matrix is 4 minus c. And what we're interested in is when this determinant is non-zero.

So in particular, for c not equal to 0-- sorry, for c not equal to 4-- when c is not 4, the determinant of M is not 0. So when c is not 4, determinant of M is not 0, so both systems-- both the original system and the corresponding homogeneous system-- have a unique solution. So when c is not 4-- so for most values of c-- the determinant is not 0, and the system has a unique solution.

So when c is equal to 4, what happens? Well, when c is equal to 4, we're in the bottom case. We're in the case where the homogeneous system has infinitely many solutions. OK? So let me write that over here.

When c equals 4-- I'm going to abbreviate again-- the homogeneous system has-- I'm going to use this symbol-- this sort of sideways eight symbol means infinity, so I'm going to use it for infinitely many solutions. So when c is 4, the homogeneous system has infinitely many solutions. And you might be curious-- well, so let me say one more thing about that. We know when the coefficient matrix isn't invertible that the system either has zero or infinitely many solutions. But the homogeneous system always has a solution. It always has the solution where everything is all 0. Right? So that's why we know that it's infinitely many here.

And one thing you might ask is can you find any others? Can you find any solutions that aren't just [0, 0, 0]? And the answer is yes. So this is now going beyond when I asked you to do, but I think it's, you know, an interesting thing to see. So if you wanted to find another solution, what do you know? Well, let's go back to the equations that we had.

So when we're dealing with a homogeneous system, the right-hand sides are 0. So I'm just going to cross out these right-hand sides and replace them with 0 so we don't get confused. So this is 0, 0, and 0. So we're dealing with this system: 2x plus c*z equals 0, x minus y plus 2z equals 0, and x minus 2y plus 2z equals 0.

OK, so if you want a solution [x, y, z] to this system, what do you know? Well, from the second equation, you know that the vector [x, y, z] is orthogonal to the vector 1, minus 1, 2. How do you know that? Because this left-hand side, x minus y plus 2z, is equal to [x, y, z] dot 1, minus 1, 2.

And similarly from the third equation, you know that the vector [x, y, z] is orthogonal to the vector 1, minus 2, 2, because this left-hand side is equal to [x, y, z] dot 1, minus 2, 2. Yeah? And that's equal to 0. So from the second and third equations, you know that you're looking for a vector that's orthogonal to both x-- or sorry-- both 1, minus 1, 2, and 1, minus 2, 2.

How do you get a vector perpendicular to two known vectors? Well, you just take their cross product. So let's go back over here. So to find one, you take a cross product of two rows of the coefficient matrix. So in this case, for example, we can take these rows, 1, minus 1, 2 and 1, minus 2, 2. So, for example, the vector 1, minus 1, 2-- OK-- cross the vector 1, minus 2, 2.

Now I've kind of run out of board space, so I'm not going to work out precisely what this vector is for you. But if you like, you can certainly check. You can compute this cross product out with our nice formula for the cross product. It will give you some vector, and then you can check that that vector is indeed a solution of the homogeneous system. So that will give us a second solution of the homogeneous system. Nontrivial we say, because it's not just the 0 solution.

So to quickly recap, we had a system of linear equations. I've now crossed out what the original right-hand side was. We had a system of linear equations, and we were looking for a choice of c for which that system had a unique solution and for which the corresponding homogeneous system had a unique solution. And the values of c that make that work are precisely the values of c such that the coefficient matrix has a non-zero determinant. So that's true for both parts a and b.

And for part c, when we were looking for what values of c give the homogeneous system infinitely many solutions, the answer is any other value of c. Any value of c for which the coefficient matrix does have 0 determinant will give you infinitely many solutions in the homogeneous case, and in non-homogeneous cases will either give you 0 solutions or infinitely many solutions.

And then we also at the end, we briefly discussed one way to find nontrivial solutions in the homogeneous case when there are infinitely many solutions. So I'll end there.


Systems of Linear Equations and Matrices

Disclaimer: The concepts to know on the sheet is not necessarily all inclusive, but it should be a guide of
main concepts to know and kinds of problems to be able to solve as you study for the test.

Section 2.1: Systems of Linear Equations
• Be able to solve systems of linear equations (2 equations and 2 variables)
algebraically using Method of Elimination and Substitution Method .
• Be able to solve system of linear equations graphically (remember you need to
write down the equations you plugged into your calculator, sketch the graph and
label the solution)
• Understand the different types of solutions for a system of 2 equations and 2
variables that can be obtained algebraically (one solution, no solution, infinitely
many solutions) and how it correlates to the graphically solution (lines intercept at
one point, lines are parallel, or same line twice)
• Know what it means when a system of linear equations is said to be consistent,
inconsistent, independent or dependent.

Section 2.2: Using Matrices to Solve systems of linear equations.
• Understand what is a matrix and be able to determine the dimension (i.e. size) of a
matrix
• Be able to write a system of linear equations as an augmented matrix
• Understand what it means for a matrix to be in RREF ( Reduced Row Echelon
Form) and be able to determine if a matrix is in RREF or not.
• Be able to solve a system of linear equations using Gaussian Elimination . You
need to be able to solve a system algebraically showing all Row Operations and
get the matrix in RREF to obtain the solution.
• Understand how to interpret a matrix in RREF to determine the solution of the
system (remember a system can have one solution, no solutions, or infinitely
many solutions). If you have an infinite number of solutions for the system, be
able to give the solutions in terms of one of the variables and list some solutions
to the system.

Section 2.3: Applications involving Systems of Linear Equations
• Use calculator to get an augmented matrix in RREF.
• Given an application problem, be able to set up a system of linear equations and
solve the system to get the solution. Remember to identify your variables
completely (i.e. let x =number of tickets in Upper Level region etc) and use units
for the answers.

Section 3.1: Matrix Addition and Scalar Multiplication
• Be able to add and subtract matrices (and know when matrix addition or
subtraction is undefined)
• Be able to perform scalar multiplication on a matrix

Section 3.2: Matrix multiplication and Inverses
• Be able to perform matrix multiplication (and know when matrix multiplication is
undefined).
• Understand what is an Identity Matrix.
• Understand what an inverse matrix is and given 2 matrices, be able to show one
matrix is the inverse matrix of the other.
• For a 2x2 matrix, be able to find the inverse matrix using a formula. Note: On a
quiz or exam you will not be asked to find the inverse matrix of a 2x2 matrix
using the formula from this section. However you may want to know the formula
as another method for finding the inverse matrix of a 2x2 matrix if you choose so.
• Understand what kinds of matrices have an inverse matrix.
• Understand what it means for a matrix to be Singular and Invertible.

Section 3.3: Solving Matrix Equations (Using the Inverse Matrix to Solve Systems of
Linear Equations)

• Be able to find the inverse matrix algebraically (or determine algebraically if an
inverse matrix does not exist) of any square size matrix using the row operations
(i.e. set up augmented matrix and use row operations to get matrix in RREF)
• Be able to use the graphing calculator to find an inverse matrix (or determine if an
inverse matrix does not exist). Give inverse matrix using exact values and not
approximations (convert repeating decimals to fraction form )
• Be able to write a system of linear equations (those with the same number of
equations as variables) as a product of matrices and write as a matrix equation
AX=B. Using matrix “A”, be able to solve a system of equations using the inverse
matrix of A, (i.e. A 𕒵 ) and solve the system of equations using the inverse matrix
and matrix multiplication.
• Understand the restrictions of using the inverse matrix to solve systems of linear
equations (only works on systems that are independent (i.e. have a unique
solution). You will need to use other algebraic methods to determine if the
system is dependent (infinitely many solutions) or inconsistent (no solutions)

On-Line Section: Cramer’s Rule:
• Be able to calculate the determinant of a 2x2 matrix
• Be able to use Cramer’s rule to solve a system of 2 equations and 2 variables.
• Understand the restrictions of using Cramer’s rule (get determinant of zero in the
denominator ) to solve a system of linear equations (only can solve systems that
are independent - i.e. have a unique solution. Need to use other algebraic methods
to determine if the system is dependent (infinitely many solutions) or inconsistent
(no solutions).

Section 4.1: Graphing Linear Inequalities
• Given a linear inequality or a system of linear inequalities, be able to graph the
solution region (make sure you can find the x and y intercepts (i.e. horizontal and
vertical intercepts algebraically)
• Be able to determine all corner points of the solution region (be able to
algebraically find the corner points that are not intercepts)

Section 4.2: Solving Linear Programming Problems Graphically.
• Given a linear programming problem, be able to find the optimal solution and
optimal value. You will need to be able to graph the constraints and determine
the feasible region, find all corner point to the feasible region and use the
objective function to find the optimal solution (i.e. the maximum or minimum
value). Understand cases where an optimal solution may not exist (If the feasible
region is unbounded). You will need to be able to find the x and y intercepts
(horizontal and vertical intercepts) algebraically and find corner points that are
not intercepts algebraically.
• Given an application problem be able to identify the variables, and come up with
the objective function and all the constraints, and solve the linear programming
problem.

Summary of methods for solving system of linear equations:
Given any system of 2 equations and 2 variables you should be able to solve using the
following methods:
• Graphically
• Method of Elimination
• Substitution Method
• Gaussian Elimination (algebraically and using calculator – rref feature)
• Inverse Matrix method (be able to find inverse matrix algebraically and by
calculator)
• Cramer’s Rule

Given any system of 3 equations and 3 variables you should be able to solve using
the following methods:
• Gaussian Elimination (algebraically and using calculator – rref feature)
• Inverse Matrix method (be able to find inverse matrix algebraically and by
calculator


Solving Linear Systems

By now we have seen how a system of linear equations can be transformed into a matrix equation, making the system easier to solve.

can be written in the following way:

Now, by augmenting the matrix with the vector on the right and using row operations, this equation can easily be solved by hand. However, if our system did not have nice integer entries, solving it by hand using row reduction could become very difficult. MATLAB provides us with an easier way to get an answer.

A system of this type has the form Ax = b, so we can enter these numbers into MATLAB using the following commands:

Notice that for the column vector b , we include semicolons after each entry to ensure that the entries are on different rows. If instead we had typed

we would have gotten a row vector, which is not the same thing. Now that we've defined A and b , the command

will find the solution to our equation Ax = b if it exists. In this case, MATLAB tells us

Please take care in entering the command A . It has a backslash ( ), not a forward slash ( / ).

Consider the system of equations

Convert this system of equations into a matrix equation of the form Cx = d . Solve it by hand, and record your solution in your document.

Enter the matrix C and the column vector d into MATLAB, and use the command

We would expect to get the column vector d in MATLAB if we ran the command C*x , right? In other words, C*x-d should be zero. Enter this expression into MATLAB:

The discrepancy in the last part of the above exercise is simply due to rounding error. You'll notice that the error is a vector multiplied by a very small number, one on the order of 10 -15 . But why is there any error at all? After all, solving by row reduction gave very nice numbers, right? The answer is in the way MATLAB stores numbers. In this calculation MATLAB represents numbers in "floating point form," which means it represents them in scientific notation to an accuracy of about 10 -14 . Thus when you see 10 -14 in calculations during this course, it is usually equivalent to zero.

There are drawbacks to using the command x = Cd , unfortunately. Let's explore them now.

Consider the system of equations

As you did in the previous exercise, enter the corresponding matrix C and column vector d into MATLAB. Then type in

Note the strange output. Include it in your write-up. Now go ahead and solve this system by hand. How many free variables do you have in your solution? Based on your answer, can you explain why you got the error message when trying to use the command x = Cd ?

To deal with the case of inconsistent systems or systems with infinitely many solutions, it may sometimes be better to simply use MATLAB to row-reduce your matrix and then read off the solutions yourself. Luckily, MATLAB has a command that performs Gaussian elimination for you.

Consider the following homogeneous system of equations:

Enter the corresponding matrix C and column vector d into MATLAB. Now we want to perform row reduction on the augmented matrix [C | d]. The command that performs row reduction in MATLAB is rref (the name stands for "reduced row echelon form"). Type in

Recall from class that we can use this row-reduced form to see that x1 = 3.5x3 and that x2 = -12x3 . Here, x3 is the free variable, and we can choose whatever value we want for it once we do that, the other two variables are fixed. For instance, if we pick x3 = 2 , then x1 = 7 and x2 = -24 . There are infinitely many solutions because there are infinitely many choices for the value of x3 , but they all follow this pattern.

Consider the homogeneous system of equations

By using the rref command, write down the general solution to this system of equations. How many free variables are required?

Now that we've seen MATLAB's basic tools for solving linear systems, let's turn to some applications.


Subsection 2.3.1 The Matrix Equation

In this section we introduce a very concise way of writing a system of linear equations:

are vectors (generally of different sizes), so first we must explain how to multiply a matrix by a vector.

Remark

In this book, we do not reserve the letters

for the numbers of rows and columns of a matrix. If we write “

Definition

is the linear combination

Example

to make sense, the number of entries of

has to be the same as the number of columns of

we are using the entries of

as the coefficients of the columns of

in a linear combination. The resulting vector has the same number of entries as the number of rows of

has that number of entries.

Properties of the Matrix-Vector Product
Definition

A matrix equation is an equation of the form

is a vector whose coefficients

In this book we will study two complementary questions about a matrix equation

    Given a specific choice of

what are all of the solutions to

The first question is more like the questions you might be used to from your earlier courses in algebra you have a lot of practice solving equations like

The second question is perhaps a new concept for you. The rank theorem in Section 2.9, which is the culmination of this chapter, tells us that the two questions are intimately related.

Matrix Equations and Vector Equations

Consider the vector equation

This is equivalent to the matrix equation

is equivalent to the vector equation

Example
Four Ways of Writing a Linear System

We now have four equivalent ways of writing (and thinking about) a system of linear equations:

In particular, all four have the same solution set.

We will move back and forth freely between the four ways of writing a linear system, over and over again, for the rest of the book.

Another Way to Compute

The above definition is a useful way of defining the product of a matrix with a vector when it comes to understanding the relationship between matrix equations and vector equations. Here we give a definition that is better-adapted to computations by hand.

Definition

A row vector is a matrix with one row. The product of a row vector of length


System of Two Linear Equations in Matrix Form

In this tutorial we shall convert equations of straight lines into matrix form. First we will discuss one linear equation in matrix form.

One Linear Equation:
Consider the equation of a straight line is given as:
[ax + by + c = 0,,,,< ext< – – – >>left( < ext> ight)]

Equation (i) is a linear equation and the two variables, $x$ and $y$, can be written in matrix form as follows:

Equation (i) becomes
[egin Rightarrow ax + by = – c Rightarrow left[ ight] = left[ < – c> ight] end ]

It can be further written as
[egin Rightarrow left[ <egin<*<20>> a&b end> ight]left[ <egin<*<20>> x y end> ight] = left[ < – c> ight] AX = C end ] Where $A = left[ <egin<*<20>> a&b end> ight]$ is the coefficient matrix, $X = left[ <egin<*<20>> x y end> ight]$ is variable matrix and $C = left[ < – c> ight]$ is the constant matrix.

Now we shall discuss the system of two equations in matrix form.

A System of Two Linear Equations
Consider the system of two equations of straight lines is given as:
[egin x + y + = 0,,,,< ext< – – – >>left( < ext> ight) x + y + = 0,,,,< ext< – – – >>left( << ext>> ight) end ]

Equation (i) and (ii) are linear equations, and two variables, $x$ and $y$, can be written in matrix form as follows:


System of Equations and Matrices

Okay, so I need some help. I have these word problems I have to do for homework, like this:

The Arcadium ardcade in Lynchburg, Tennessee uses 3 different colored tokens for their game machines. For 20$ you can purchase any of the following mixtures of tokens: 14 gold, 20 silver, and 24 bronze OR, 20 gold, 15 silver, and 19 bronze OR, 30 gold, 5 silver, and 13 bronze.

These problems tell me to write a system of equations (and I did!), which is:

14x+20y+24z=20
20x+15y+19z=20
30x+5y+13z+20

x represents the value of gold
y represents the value of silver
z represents the value of bronze

I then had to represent the system as a matrix, which is:

14 20 24 x 20
20 15 19 y 20
30 5 13 z 20

All I want to know is how do I figure out the monetary value of each token? Could someone explain to me how I do that?

Subhotosh Khan

Super Moderator

Okay, so I need some help. I have these word problems I have to do for homework, like this:

The Arcadium ardcade in Lynchburg, Tennessee uses 3 different colored tokens for their game machines. For 20$ you can purchase any of the following mixtures of tokens: 14 gold, 20 silver, and 24 bronze OR, 20 gold, 15 silver, and 19 bronze OR, 30 gold, 5 silver, and 13 bronze.

These problems tell me to write a system of equations (and I did!), which is:

14x+20y+24z=20
20x+15y+19z=20
30x+5y+13z+20

x represents the value of gold
y represents the value of silver
z represents the value of bronze

I then had to represent the system as a matrix, which is:

14 20 24 x 20
20 15 19 y 20
30 5 13 z 20

All I want to know is how do I figure out the monetary value of each token? Could someone explain to me how I do that?

For a quick review - please go to:

HallsofIvy

Elite Member

Why did you write it as a matrix? That is a perfectly good method but the fact that you mention "matrices" makes me think you must know something about them!

You have (displaystyle egin14 & 20 & 24 20 & 15 & 19 30 & 5 & 13endeginx y zend= egin20 20 20end)
Writing it like that, the obvious thing to do is to find the inverse matrix of the coefficient matrix, then multiply both sides by that. That is, you solve Ax= b by multiplying both sides by (displaystyle A^<-1>): (displaystyle A^<-1>Ax= x= A^<-1>x),

Another way to solve a systme of equations like that is to write the "augmented matrix":
(displaystyle egin14 & 20 & 24 & 20 20 & 15 & 19 & 20 30 & 5 & 13 & 20 end)
and "row reduce" so that the first three columns are (displaystyle egin 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1end) and the last column will give x, y, and z.

But I will admit that I, personally would not use "matrices" at all. From the equation (displaystyle 14x+ 20y+ 24z= 20) we can divide by 2 to get (displaystyle 7x+ 10y+ 12z= 10). The second equation is (displaystyle 20x+ 15y+ 19z= 20). If we multiply that equation by 2, the first equation by 3, and subtract the second from the first, we get (displaystyle (40x+ 30y+38z)- (21x+ 30y+ 36z)= 60-20) or (displaystyle 19x+ 2z= 40), eliminating y. The subtract (displaystyle 7x+ 10y+ 12x= 10) from, say, twice the third equation, (displaystyle 60x+ 10y+ 26z= 40), to get (displaystyle 53x+ 14z= 30). Now we have two equations in two unknowns. Manipulate those equations to eliminate one of those.


Introduction to Matrices and Systems of Equations


A matrix is a rectangular array of numbers arranged in rows and columns. We call each number in this array an element of the matrix. When we write a matrix, we typically enclose the array in brackets. Matrices (plural) come in many sizes, determined by the number of rows and the number of columns. If a matrix has n rows and m columns, then we say that the size of the matrix is m x n, read "m by n". The following are examples of matrices of various sizes.

When we enumerate rows and columns of a given matrix, we count rows from top to bottom and count columns from left to right.Since a matrix is an array of numbers, we often see matrices used to record information, especially if rows and columns of a matrix can be understood to represent categories. As such, we can certainly use matrices to record pertinent information about a system of linear equations - coefficients of the variables as well as constants on the right-hand side of equations in the system.

We adopt a convention here of using subscripted variables rather than individual letter variables to avoid possible difficulties in the number of available letters. We construct a 2 x 3 matrix, called the augmented matrix for the system, where each row represents information for a particular equation and each column represents either coefficients of a variable or the constants on the right-hand side of the equations.

We write this matrix as follows.

Notice the correspondence between the rows of this matrix and the equations in the system as well as the correspondence between the columns of the matrix and the coefficients and constant terms in the equations. The vertical line has no real purpose except to serve as a visual reminder of the location of the equal signs in the system and hence a separation between the coefficients of the variables and the constants on the right-hand side of the equations. Before delving into the use of these augmented matrices representing systems, we will pause to put forth some terminology and notation. Recall that in the method of elimination we had three operations that we could use to produce equivalent systems of linear equations. We have a similar collection of row operations that we perform on matrices. We say two matrices are row-equivalent if one is obtained from the other by some sequence of row operations. These operations are as follows:

  • Interchange any two rows.
  • Multiply (all elements in) a row by any nonzero constant and replace that row with the result.
  • Multiply (all elements in) a row by any constant and add (corresponding elements) to any other row, replacing the second row in this sum with the result.

Performing any sequence of these operations results in a row-equivalent matrix.

Notice the similarity between these operations and the operations employed in the method of elimination. We use a similar shorthand notation to indicate performing a particular row operation as well.

Notation for Row Operations


In the case where a matrix is the augmented matrix representing a system of linear equations, performing a row operation on the matrix is equivalent to performing the corresponding operation on a system of equations. So, row-equivalent matrices represent equivalent systems of linear equations. To demonstrate how to use augmented matrices to find solutions to systems of linear equations, we will show parallel operations in the method of elimination and the corresponding row operations.