Question

Using the matrix rankings, find the value of the parameters a,b so that the system does not have solution, the values so that it has a solution and in this case show how many solutions there are?

`3x_1+3x_2+a=6`
`x_1+x_2+x_3-x_4=-2`
`2x_1-x_2-x_3+2x_4=b`

Answer 1

There will be infinitely many solutions for all values of `a` and `b`

Explanation:

Starting with the augmented matrix `(A|b}` we carry out row operations to reduce the system

`( (3,3,0,0,|, 6-a),(1,1,1,-1,|, -2),(2,-1,-1,2, | ,b))`

`R_1 to R_1/3`

`( (1,1,0,0,|, 2-a/3),(1,1,1,-1,|, -2),(2,-1,-1,2, | ,b))`

`R_2 to R_2-R_1,R_3 to R_3-2R_1`

`( (1,1,0,0,|, 2-a/3),(0,0,1,-1,|, -4+a/3),(0,-3,-1,2, | ,b+{2a}/3-4))`
Interchange `R_2` and `R_3`

`( (1,1,0,0,|, 2-a/3),(0,-3,-1,2, | ,b+{2a}/3-4),(0,0,1,-1,|, -4+a/3))`
`R_2 to R_2/(-3)`
`( (1,1,0,0,|, 2-a/3),(0,1,1/3,-2/3, | ,-b/3-{2a}/9+4/3),(0,0,1,-1,|, -4+a/3))`

It is easy to see that the augmented matrix always has the same rank (namely 3) as the coefficient matrix `A` - so there will always be a solution - and there will be an infinite number of them (as the rank is less than the number of unknowns.