6 3 22 3 22 18 22 18 154975 a0 a1 a2 335 184 1368SolutionTh

[6 3 2.2 3 2.2 1.8 2.2 1.8 1.54975] {a_0 a_1 a_2} = [3.35 1.84 1.368]

Solution

The Augmented matrix of the given linear system of equations , after converting decimals into fractions, is B =

6

3

22/10

335/100

3

22/10

18/10

184/100

22/10

18/10

154975/100000

1368/1000

To solve the given equations, we will reduce B to its RREF as under:

Multiply the 1st row by 1/6

Add -3 times the 1st row to the 2nd row

Add -11/5 times the 1st row to the 3rd row

Multiply the 2nd row by 10/7

Add -7/10 times the 2nd row to the 3rd row

Multiply the 3rd row by 300/464473

Add -1 times the 3rd row to the 2nd row

Add -11/30 times the 3rd row to the 1st row

Add -1/2 times the 2nd row to the 1st row

Then the RREF of A is

1

0

0

-3828511151/ 325131100

0

1

0

1604544533/ 65026220

0

0

1

-25688 / 2322365

Therefore, a₀ =-3828511151/ 325131100, a₁ =1604544533/ 65026220 and a₂ =-25688 / 2322365 .

6	3	22/10	335/100
3	22/10	18/10	184/100
22/10	18/10	154975/100000	1368/1000