Question

How do you solve `a+3b+2c=3` , `2a-b-3c= -8`, `5a+2b+c=9` using matrices?

Answer 1

`a=2`, `b=-3` and `c=5`

Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

`A=((1,3,2,|,3),(2,-1,-3,|,-8),(5,2,1,|,9))`

I have written the equations not in the sequence as in the question in order to get `1` as pivot.

Perform the folowing operations on the rows of the matrix

`R2larrR2-2R1`; `R3larrR3-5R1`

`A=((1,3,2,|,3),(0,-7,-7,|,-14),(0,-13,-9,|,-6))`

`R2larr(R2)/(-7)`

`A=((1,3,2,|,3),(0,1,1,|,2),(0,-13,-9,|,-6))`

`R3larrR3+13R2`

`A=((1,3,2,|,3),(0,1,1,|,2),(0,0,4,|,20))`

`R3larr(R3)/4`

`A=((1,3,2,|,3),(0,1,1,|,2),(0,0,1,|,5))`

`R1larrR1-2R3`; `R2larrR2-R3`

`A=((1,3,0,|,-7),(0,1,0,|,-3),(0,0,1,|,5))`

`R1larrR1-3R2`

`A=((1,0,0,|,2),(0,1,0,|,-3),(0,0,1,|,5))`

Thus `a=2`, `b=-3` and `c=5`