I want to solve this, i need help. Justify your answer please.

Question

Consider the linear system:

x - 2y + 3z = 4

2x - 3y + az = 5

3x - 4y + 5z = b

 

For which values of a and b does the linear system have

a) no solution,

b) a unique solution,

c) infinitely many solutions.

Answer 1

x-2y+3z = 4

2x-3y+az = 5

3x-4y+5z = b

 

Multiply the first equation by -2 and add to the second equation.  Multiply the first equation by -3 and add to the third equation.

 

x - 2y + 3z = 4

      y + (a-6)z = -3

     2y - 4z = (b-12)

 

Multiply the second equation by -2 and add to the third equation:

 

x - 2y + 3z = 4

      y + (a-6)z = -3

           (-2a+8)z = (b-6)

 

Infinitely many solutions when -2a+8 = 0 and b-6 = 0

 

                                                  So, a = 4 and b = 6

 

No solution if -2a+8 = 0 and b-6 ≠ 0

 

                 So, a = 4 and b ≠ 6

 

Unique solution if a ≠ 4 and b = any real number