Intro to LaTeXNatalya JacksonSeptember 12th, 2010AbstractGaussian elimination is used to place an augmented matrix in row echelon form to solve asystem of equations using back substitution.IntroductionIn this paper I will solve a system of three equations in three unknown. I will represent the systemusing an augmented matrix, use the method of Gaussian elimination to place the augmented matrixin row echelon form, and solve the new system of equations using back substitution.The System of EquationsHere is the original system of equations I will solve, which will hereafter be referred to as system (1).x + 2y + 3z = 62x − y + 4z = 8−x + 8y + 2z = 12(1)The goal is to solve this system of three equations in three unknowns to find values of x, y, and zwhich make all three statements true.Elimination Using Augmented MatricesI have set up an augmented matrix to represent this system of three equations, shown below:1 2 3 62 −1 4 8−1 8 2 12(2)I will use the Gaussian elimination method to place (2) in row echelon form. First I will multiplyrow 1 by 2, subtract the result of this multiplication from row 2, and replace row 2 with the resultof this subtraction. The result of this operation is shown in (3), displayed below:1 2 3 60 −5 −2 −4−1 8 2 12(3)1Next I will multiply row 1 by −1 and subtract the result from row 3, and replace row 3 with theresult of this subtraction. The result of this operation is shown in (4):1 2 3 60 −5 −2 −40 10 5 18(4)Next I will divide row 2 by −5 to obtain a one in the second position of row 2, as shown in (5):1 2 3 60 125450 10 5 18(5)Next I will multiply row 2 by 10, subtract the result from row 3, and replace row 3 with the resultof this subtraction. The resulting augmented matrix is in row echelon form, as shown in (6):1 2 3 60 125450 0 1 10(6)Back SubstitutionNow that the augmented matrix is in row echelon form, shown above in (6), I can use back substi-tution to solve the system of equations represented by this new matrix, displayed below in (7):x + 2y + 3z = 6y +25z =45z = 10(7)As shown above in (7), I already have a value of z = 10, so I can substitute this value into thesecond equation of (7) and solve for y, as shown in (8):y +25(10) =45y + 4 =45y = −165(8)Now that I have values for y and z, I can substitute −16/5 and 10 into the first equation of (7) andsolve for x, as shown in (9):x + 2−165+ 3(10) = 6x −325+ 30 = 6x −325= −24x = −885(9)2SolutionAs shown in (9), there is a single solution for the system of equations depicted in (1). Shown belowin (10) are the values for x, y, and z:x = −885y = −165z = 10(10)ConclusionI started with the original system of equations shown in (1). I first formulated an augmented matrix,shown in (2), to represent the system. I then used the Gaussian elimination method found at [2]and refined by [3] to place this matrix in row echelon form. The row echelon form of this matrix isshown in (6). From this matrix I derived a new system of equations, (7), which I solved to get thesolutions for x, y, and z. The final unique solution of this non-singlular system is shown in (10).References[1] David Arnold Writing Scientific Papers in LaTeX, Revised 2008.[2] John Hawkins, Ph.D. Gaussian Elimination Row Echelon Form Algorithm, http://personal.georgiasouthern.edu/~jbhawkin/Handouts/GaussianInstr.htm[3] Gilbert Strang Introduction to Linear Algebra, Fourth Edition, Wesley-Cambridge,
View Full Document
Unlocking...