Gaussian Elimination Method for Solving Linear Equations: Step-by-Step Solution, Study notes of Engineering

A step-by-step solution to solve a system of linear equations using the gaussian elimination method. The augmented coefficient matrix, the row operations, and the calculation of the solution vector.

Typology: Study notes

Pre 2010

Uploaded on 08/07/2009

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Solution of Linear Equation by Gaussian Elimination

1 3 -2 1 5 = Augmented Coefficient Matrix 2 -1 6 2 26 A(1) = 3 2 -5 1 - 2 6 1 -4 1

1 3 -2 1 5 Line same as A(1) m(2,1) 2 0 -7 10 0 16 m(3,1) 3 A(2) = 0 -7 1 -2 - m(4,1) 2 0 0 5 -6 -

1 3 -2 1 5 Line same as A(2) 0 -7 10 0 16 Line same as A(2) m(3,2) 1 A(3) = 0 0 -9 -2 - m(3,4) 0 0 0 5 -6 -

1 3 -2 1 5 Line same as A(3) 0 -7 10 0 16 Line same as A(3) A(4) = 0 0 -9 -2 -35 Line same as A(3) m(4,4) -0.556 0 0 0 -7.11 -28.

Solution: x(1) = 1 x(2) = 2 x(3) = 3 x(4) = 4

(^3)

(^1)

5

(^6)

(^2)

26

(^2)

(^1)

(^6)

(^1)

1

(^1)

(^3)

(^1)

5

m(2,1)

(^) =A4/A

=A4-$D$9*A

=B4-$D$9*B

=C4-$D$9*C

=D4-$D$9*D

=E4-$D$9*E

m(3,1)

(^) =A5/A

A(2) =

=A5-$D$10*A

=B5-$D$10*B

=C5-$D$10*C

=D5-$D$10*D

=E5-$D$10*E

m(4,1)

(^) =A6/A

=A6-$D$11*A

=B6-$D$11*B

=C6-$D$11*C

=D6-$D$11*D

=E6-$D$11*E

(^1)

(^3)

(^1)

5

(^0)

10

(^0)

16

m(3,2)

(^) =H10/H

A(3) =

=G10-$D$16*G

=H10-$D$16*H

=I10-$D$16*I

=J10-$D$16*J

=K10-$D$16*K

m(3,4)

(^) =H11/H

=G11-$D$17*G

=H11-$D$17*H

=I11-$D$17*I

=J11-$D$17*J

=K11-$D$17*K

(^1)

(^3)

(^1)

(^5)

x(1) =

=(K20-(J20N23+I20N22+H20*N21))/G

(^0)

10

(^0)

16

x(2) =

=(K21-J21N23-I21N22)/H

A(4) =

(^0)

(^0)

x(3) =

=(K22-J22*N23)/I

m(4,4)

(^) =I17/I

=G17-$D$23*G

=H17-$D$23*H

=I17-$D$23*I

=J17-$D$23*J

=K17-$D$23*K

x(4) =

=K23/J