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Use the method of Gaussian elimination to solve the following systems of linear equations:?

0 votes

Use the method of Gaussian elimination to solve the following systems of linear equations: 

Thanks guys i don't know how to do it with the extra variable

asked Nov 3, 2014 in PRECALCULUS by anonymous

2 Answers

0 votes

(a)

The given system of equations are

  x1 -   x2 -  7x3  +  7x4 = 5

 -x1 +  x2 +  8x3  -   5x4 = -7

3x1 - 2x2 - 17x3 + 13x4 = 14

2x1 -   x2 - 11x3 +   8x4 = 7

We can write above equations are in matrix form as

image

R indicates Rows

C indicates Columns

R2 = R2+R1

R3 = R3-3R1

R4 = R4-2R1

image

Interchange R3 with R2

image

R4 = R4-R2

image

image

R4 = R4+R3

image

image

R4 = R4/4

image

image

We can write above matrix is in equations form as

x1 -   x2 -  7x3  +  7x4 = 5      -----------------(1)

        x2 +  4x3  -  8x4 = -1     -----------------(2)

                  x3  + 2x4 = -2     -----------------(3)

                            x4 = -1     -----------------(4)

Substitute x4 = -1 in equation (3)

x3  + 2(-1) = -2

x3  = -2 + 2

x3  = 0

Substitute x4 = -1 and x3  = 0  in equation (2)

x2 +  4(0)  - 8 (-1) = -1

x2 = -1 - 8

x2 = -9

Substitute x2 = -9 , x4 = -1 and x3  = 0  in equation (1)

x1 - (-9) -  7(0) +  7(-1) = 5

x1 = 5 - 9 + 7

x1 = 3

Solution :

x1 = 3

x2 = -9

x3  = 0

x4 = -1

answered Nov 3, 2014 by lilly Expert
0 votes

(b)

The given system of equations are

  x1 -   x2 -  7x3  +  7x4 = 5

 -x1 +  x2 +  8x3  -   5x4 = -7

3x1 - 2x2 - 17x3 + 13x4 = 14

0 x1 - 0 x2 - 0  x3 + 0  x4 = 0

We can write above equations are in matrix form as

image

R indicates Rows

C indicates Columns

R2 = R2+R1

R3 = R3-3R1

image

image

Interchange R3 with R2

image

We can write above matrix is in equations form as

x1 -   x2 -  7x3  +  7x4 = 5           ----------------------------(1)

        x2 +  4x3  -  8x4 = -1          ---------------------------(2)

                  x3  + 2x4 = -2          ---------------------------(3)

Let x4 = k

Substitute x4 = k in equation (3)

x3  + 2(k) = -2

x3  = -2 - 2k

Substitute x4 = k and x3  = -2 - 2k  in equation (2)

x2 +  4(-2 - 2k)  - 8(k) = -1

x2 = -1 + 8k + 8k + 8

x2 = 7 + 16k

Substitute x2 = 7+16k , x4 = k and x3  = -2 + 2k in equation (1)

x1 - (7+16k) -  7(-2 + 2k) +  7(k) = 5

x1 = 5 + 7 +16k + 14 - 7k + 14k

x1 = 26 + 23k

Solution :

x1 = 26+23k

x2 = 7+16k

x3  = -2 - 2k

x4 = k

answered Nov 3, 2014 by lilly Expert

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