86.1 Simultaneous Linear Equations Concept Overview

SIMULTANEOUS LINEAR EQUATIONS | CONCEPT OVERVIEW The TOPIC of SIMULTANEOUS LINEAR EQUATIONS is not one that is directly referenced in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. However, it’s a situation where knowing how to deal with SIMULTANEOUS LINEAR EQUATIONS can be a major contributor the success on any given problem. For this reason, we will break it down despite it being neglected from the NCEES Supplied Reference Handbook.

CONCEPT INTRO: Given any NUMBER of LINEAR EQUATIONS each having a unique set of UNKNOWN SHARED VARIABLES, such that 𝐴𝑥! + 𝐵𝑥! + 𝐶𝑥! + . . . +𝑍𝑥! = 𝑦! 𝐷𝑥! + 𝐸𝑥! + 𝐹𝑥! + . . . +𝑌𝑥! = 𝑦! 𝐺𝑥! + 𝐻𝑥! + 𝐼𝑥! + . . . +𝑋𝑥! = 𝑦! Where: • 𝑥! , 𝑥! , 𝑥! , … , 𝑥! are UNKNOWN SHARED VARIABLES • 𝐴, 𝐵, 𝐶, … , 𝑍 are KNOWN COEFFICIENTS and • 𝑦! , 𝑦! , 𝑎𝑛𝑑 𝑦! are KNOWN QUANTITIES



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We can solve this set of equations using one of a two approaches, either: 1. Solve them using the method of LINEAR ELIMINATION or LINEAR COMBINATION…this is a manual process done By hand, acceptable, but too time consuming for the more complex systems. 2. Solve using your NCEES APPROVED CALCULATOR…using MATRICES or the SYSTEM SOLVER, both a standard functionality of the calculator which will significantly decrease our time solving these problems. Let’s lay out each of these approaches. LINEAR ELIMINATION: In the LINEAR ELIMINATION method, we will take a set of equations, line them up in a VERTICAL manner and either ADD or SUBTRACT the equations with the goal of simplifying it down to one equation with one single VARIABLE. We can then take that defined VARIABLE and work our way back up our CHAIN of work to define the remaining UNKNOWN VARIABLES, until ultimately, all are defined giving us the whole SOLUTION to the set of SIMULTANEOUS LINEAR EQUATIONS. MATRIX FORMAT: It’s possible to take a set of SIMULATRENOUS LINEAR EQUATIONS and place them in to a specific MATRIX FORMAT, such that:



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𝐴𝑋 = 𝐵 Where each of these VARIABLES represent a UNIQUE MATRIX developed through the definition of each of our EQUATIONS, such that: • The matrix “A” is the COEFICIENT MATRIX of the system used to represent the coefficients of variables in an equation. These will be the numbers in front of the variables on the left side of the equal sign. • The matrix “X” is the MATRIX OF VARIABLES used to represent the variables in the equation, which are usually X, Y, and Z. These will be any variables on the left side of the equal sign. • The matrix “B” is the MATRIX OF CONSTANTS used to represent the constants that are on the other side of the equal signs for each equation. These are usually the numbers on the right side of the equal sign. Visually how this all plays out is pretty straight forward, let’s revisit our set of equations defined above, limiting it to three unknown variables, such that we have: 𝐴𝑥! + 𝐵𝑥! + 𝐶𝑥! = 𝑦! 𝐷𝑥! + 𝐸𝑥! + 𝐹𝑥! = 𝑦! 𝐺𝑥! + 𝐻𝑥! + 𝐼𝑥! = 𝑦! Laying this out in the MATRIX FORM: 𝐴𝑋 = 𝐵

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We get: 𝐴 𝐷 𝐺

𝐵 𝐸 𝐻

𝐶 𝐹 𝐼

𝑥! 𝑦! 𝑥! = 𝑦! 𝑥! 𝑦!

Where: • The COEFICIENT MATRIX, A, would represent the coefficients of variables in the equations, which in this case, are the variables A,B,C etc, all on the left side of the equation. • The MATRIX OF VARIABLES, X, would represent the variables in the equation, which are x1, x2, and x3. • The MATRIX OF CONSTANTS, B, would represent the constants on the right side of the equation, D, H, and L. Once we have written the SYSTEM OF LINEAR EQUATIONS in the form AX=B, we can now SOLVE for the MATRIX OF VARIABLES, X, by MULTIPLYING each side by the INVERSE OF MATRIX A. This will isolate the matrix of variables on the left side, making the matrix operation: 𝐴!! 𝐴𝑋 = 𝐴!! 𝐵



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It is important to remember that a matrix multiplied by the inverse of itself will produce the identity matrix, which gives us: 𝑋 = 𝐴!! 𝐵 This may be a simple definition problem on the exam, so remember that. Now that we know how to write an equation to solve for the MATRIX OF VARIABLES, we need to figure out how to calculate the INVERSE OF MATRIX A and then MULTIPLY it by MATRIX B. The inverse of any matrix is defined as:

𝐴!! =

𝑎𝑑𝑗(𝐴) 𝐴

Where: • Adj(A) is the ADJOINT of MATRIX A obtaining by replacing 𝐴! elements with their cofactors • |A| is the determinant of the matrix [A] We aren’t going to worry about learning how to calculate the MATRIX INVERSE, the ADJOIINT or the DETERMINANT of a MATRIX, it’s confusing and the calculator can do it super easily!



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If we can HACK a problem, then we are going to HACK IT, that’s the way we roll. So that’s the basic workflow, to recap: Express the set of SIMULATANEOUS LINEAR EQUATIONS compactly in the matrix form: 𝐴𝑋 = 𝐵 Pre-multiply both sides by the INVERSE MATRIX such that: 𝐴!! 𝐴𝑋 = 𝐴!! 𝐵 Since 𝐴!! 𝐴𝑥 = 𝐼𝑥 = 𝑥, the expression can equivalently be written as: 𝑋 = 𝐴!! 𝐵 So as long as 𝐴!! exists, the MATRIX OF VARIABLES, X, can be fully solved. If the inverse does not exist, then the set of SIMULATANEOUS LINEAR EQUATIONS does not have a unique solution. We will work this process by hand as we make our way through our practice problems, but will reiterate over and over the importance of knowing it can be HACKED and learning that HACK in it’s most intimate form.



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SIMULTANEOUS LINEAR EQUATIONS | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material. Solve the following set of linear equations: −𝑥 + 5𝑦 = 4 2𝑥 + 5𝑦 = −2 A. 𝑥 = 5 and 𝑦 = 6 B. 𝑥 = 0 and 𝑦 = 3 C. 𝑥 = 3 and 𝑦 = 3 D. 𝑥 = −2 and 𝑦 =

! !

SOLUTION: The TOPIC of SIMULTANEOUS LINEAR EQUATIONS is not one that is directly referenced in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. However, it’s a situation where knowing how to deal with SIMULTANEOUS LINEAR EQUATIONS can be a major contributor the success on any given problem. For this reason, we will break it down despite it being neglected from the NCEES Supplied Reference Handbook.



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Given any NUMBER of LINEAR EQUATIONS each having a unique set of UNKNOWN SHARED VARIABLES, such that 𝐴𝑥! + 𝐵𝑥! + 𝐶𝑥! + . . . +𝑍𝑥! = 𝑦! 𝐷𝑥! + 𝐸𝑥! + 𝐹𝑥! + . . . +𝑌𝑥! = 𝑦! 𝐺𝑥! + 𝐻𝑥! + 𝐼𝑥! + . . . +𝑋𝑥! = 𝑦! Where: • 𝑥! , 𝑥! , 𝑥! , … , 𝑥! are UNKNOWN SHARED VARIABLES • 𝐴, 𝐵, 𝐶, … , 𝑍 are KNOWN COEFFICIENTS and • 𝑦! , 𝑦! , 𝑎𝑛𝑑 𝑦! are KNOWN QUANTITIES We can solve this set of equations using one of a two approaches, either: 1. Solve them using the method of LINEAR ELIMINATION or LINEAR COMBINATION…this is a manual process done By hand, acceptable, but too time consuming for the more complex systems. 2. Solve using your NCEES APPROVED CALCULATOR…using MATRICES or the SYSTEM SOLVER, both a standard functionality of the calculator which will significantly decrease our time solving these problems. In this problem, we will deploy our knowledge of MATRICES, solving it by hand, but illustrating how it is set up so you can HACK it using your CALCULATOR…which is always our go to approach.

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But for edification purposes, we will solidify an understanding of the process behind the scenes, so let’s get in to it. It’s possible to take any set of SIMULATRENOUS LINEAR EQUATIONS and place them in to a specific MATRIX FORMAT, such that: 𝐴𝑋 = 𝐵 Where each of these VARIABLES represent a UNIQUE MATRIX developed through the definition of each of our EQUATIONS, such that: • The matrix “A” is the COEFICIENT MATRIX of the system used to represent the coefficients of variables in an equation. These will be the numbers in front of the variables on the left side of the equal sign. • The matrix “X” is the MATRIX OF VARIABLES used to represent the variables in the equation, which are usually X, Y, and Z. These will be any variables on the left side of the equal sign. • The matrix “B” is the MATRIX OF CONSTANTS used to represent the constants that are on the other side of the equal signs for each equation. These are usually the numbers on the right side of the equal sign. In this problem, we are given: −𝑥 + 5𝑦 = 4 2𝑥 + 5𝑦 = −2

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Our first step would be to always ensure that the equations are in STANDARD FORM, meaning all the VARIABLES are on the left side and all the CONSTANTS are on the right side of the equation. We are good, lets move forward. Laying this out in the MATRIX FORM: 𝐴𝑋 = 𝐵 We get: −1 2

5 𝑥 4 = 𝑦 −2 5

Where: • The COEFICIENT MATRIX, A, would represent the coefficients of variables in the equations, which in this case, are the variables -1, 5, 2, and 5…all on the left side of the equation. • The MATRIX OF VARIABLES, X, would represent the variables in the equation, which are x and y. • The MATRIX OF CONSTANTS, B, would represent the constants on the right side of the equation, which are 4 and -2.



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Once we have written the SYSTEM OF LINEAR EQUATIONS in the form AX=B, we can now SOLVE for the MATRIX OF VARIABLES, X, by MULTIPLYING each side by the INVERSE OF MATRIX A. This will isolate the MATRIX OF VARIABLES on the left side, making the matrix operation: 𝐴!! 𝐴𝑋 = 𝐴!! 𝐵 It is important to remember that a matrix multiplied by the inverse of itself will produce the identity matrix, which gives us: 𝑋 = 𝐴!! 𝐵 This may be a simple definition problem on the exam, so remember that. Now that we know how to write an equation to solve for the MATRIX OF VARIABLES, we need to figure out how to calculate the INVERSE OF MATRIX A and then MULTIPLY it by MATRIX B. The inverse of any matrix is defined as:

𝐴!! =



𝑎𝑑𝑗(𝐴) 𝐴

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Where: • Adj(A) is the ADJOINT of MATRIX A obtaining by replacing 𝐴! elements with their cofactors • |A| is the determinant of the matrix [A] The ADJOINT of a MATRIX is a complex term made extremely simple when viewing the GENERAL FORMULA expressed as: 𝑎𝑑𝑗(𝐴) = (𝐴! )! Where: • 𝐴! represents the COEFFICIENT MATRIX of MATRIX A • (𝐴! )! represents the TRANSPOSE of the COEFFICIENT MATRIX A To find the ELEMENTS of a COEFFICIENT MATRIX, you will go ELEMENT by ELEMENT of the ORIGINAL MATRIX, replacing the ELEMENT with the new ELEMENT represented in the COEFFICIENT MATRIX. Let’s illustrate this with our current problem.



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We are given the MATRIX: −1 2

5 5

The ADJOINT of this MATRIX will then be is found by first swapping the leading diagonal of the 2 𝑥 2 matrix and switching the signs of the other two elements such that: 5 −2

−5 −1

Now to define the DETERMINATE of “𝐴”, we proceed as we would with any COFACTOR, such that: 𝐴 = 5 −1 − −5 −2 Or: 𝐴 = −15 This gives us what we need to determine the INVERSE MATRIX, which is:

𝐴!! = −



1 5 15 −2

−5 −1

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Or:

𝐴!!

1 3 = 2 15 −

1 3 1 15

With everything defined to carry out our OPERATION, we can determine the SOLUTIONS to this set of SIMULTANEOUS LINEAR EQUATIONS but plugging in what we have up to this point, which gives us: 1 3 𝑋= 2 15 −

1 4 3 1 −2 15

Carrying out our standard MATRIX OPERATIONS, we get: −2 𝑥= 2 5 The SOLUTION to our set of SIMULTANEOUS LINEAR EQUATIONS is: 𝑥 = −2 𝑦=



2 5

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We can do a quick confirmation of these results by plugging them in to the original equations, such that:

− −2 + 5

2 =2+2=4 5

2 −2 + 5

2 = −4 + 2 = −2 5 𝟐

The correct answer choice is D. 𝒙 = −𝟐 𝒂𝒏𝒅 𝒚 = 𝟓



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