SOLVING SYSTEMS OF LINEAR EQUATIONS BY GAUSSIAN ELIMINATION

Solve the following systems of linear equations by Gaussian elimination method.

Problem 1 :

2x - 2y + 3z = 2, x + 2y - z = 3 and 3x - y + 2z = 1

Solution :

x + 2y - z = 3 -----(1)

-6y + 5z = -4 ------(2)

-5z = -20 -----(3)

z = 4

Applying the value of z in (2), we get

-6y + 5(4) = -4

-6y + 20 = -4

-6y = -4 - 20

-6y = -24

y = 4

Applying the value of z = 4 and y = 4 in (1), we get

x + 2(4) - 4 = 3

x + 8 - 4 = 3

x + 4 = 3

x = 3 - 4

x = -1

So, the solution is (-1, 4, 4)

Problem 2 :

2x + 4y + 6z = 22, 3z + 8y + 5z = 27, -x + y + 2z = 2

Solution :

-x + y + 2z = 2 ----(1)

11y + 11z = 33 ----(2)

44z = 88 -----(3)

z = 88/44

z = 2

Applying the z = 2 in (2), we get

11y + 11(2) = 33

11y = 33 - 22

11y = 11

y = 1

Applying y = 1 and z = 2 in (1), we get

-x + 1 + 2(2) = 2

-x + 5 = 2

x = 3

So, the solution is (3, 1, 2).

Problem 3 :

If ax²+bx+c is divided by x + 3, x - 5 and x - 1, the remainders are 21, 61 and 9. Find a, b and c.(Use Gaussian elimination method)

Solution :

Let p(x) = ax²+bx+c

x + 3 = 0

x = -3

p(-3) = a(-3)²+b(-3)+c

21 = 9a - 3b + c

9a - 3b + c = 21 -----(1)

x - 5 = 0

x = 5

p(5) = a(5)²+b(5)+c

61 = 25a + 5b + c

25a + 5b + c = 61 -----(2)

x - 1 = 0

x = 1

p(1) = a(1)²+b(1)+c

9 = a + b + c

a + b + c = 9 -----(3)

x + y + z = 9 -----(1)

-12y - 8z = -60 -----(2)

64z = 384  -----(3)

z = 384/64

z = 6

Applying z = 6 in (2), we get

-12y - 8(6) = -60

-12y - 48 = -60

-12y = -60 + 48

-12y = -12

y = 1

Applying y = 1 and z = 6 in (1)

x + 1 + 6 = 9

x + 7 = 9

x = 9 - 7

x = 2

So, the solution is (2, 1, 6).

Problem 4 :

An amount of 65000 is invested in three bonds at the rates of 6%, 8% and 9% per annum respectively. The total income is 4800. The income from the third bond is 600 more than that from the second bond. Determine the price of each bond.

Solution :

Let x, y and z be amounts in 3 bonds.

x + y + z = 65000-----(1)

(6/100)x + (8/100)y + (9/100)z = 4800

6x + 8y + 9z = 480000-----(2)

(9/100)z = (8/100)y + 600

-8y + 9z = 60000 ----(3)

x + y + z = 65000 -------(1)

2y + 3z = 90000 ------(2)

21z = 420000 ------(3)

z = 420000/21

z = 20000

Applying the value of z in (2), we get

2y + 3(20000) = 90000

2y = 90000-60000

2y = 30000

y = 15000

Applying the value of y and z in (1)

x + 15000 + 20000 = 65000

x = 65000 - 35000

x = 30000

Problem 5 :

A boy is walking along the path y = ax² + bx + c through the points (-6, 8), (-2, -12) and (3, 8). He wants to meet his friend at P (7, 60) . Will he meet his friend? (Use Gaussian elimination method.)

Solution :

y = ax² + bx + c

The curve passes through the point (-6, 8).

8 = a(6)² + b(-6) + c

36a - 6b + c = 8 ----------(1)

The curve passes through the point (-2, -12).

12 = a(-2)² + b(-2) + c

4a - 2b + c = -12 ----------(2)

The curve passes through the point (3, 8).

8 = a(3)² + b(3) + c

9a + 3b + c = 8 ----------(3)

36a - 6b + c = 8 -----(1)

-12b + 8c = -116 -----(2)

30c = -300

c = -300/30

c = -10

Applying the value of c in (2), we get

-12b + 8(-10) = -116

-12b - 80 = -116

-12b = -116 + 80

-12b = -36

b = 3

Applying the value of b = 3 and c = -10 in (1)

36a - 6(3) + (-10) = 8

36a - 18 - 10 = 8

36a - 28 = 8

36a = 8 + 28

a = 1

Height he will meet at at x = 7

y = ax² + bx + c

y = (1)x² + 3x -10

When x = 7

y = 7² + 3(7) -10

y = 49 + 21 - 10

y = 60

So, he will meet his friend.