Answer:
a) If [tex]2-a^{2}=0[/tex] and [tex]a-2\neq 0[/tex]. The system has no solution
b) If [tex]2-a^{2}\neq 0[/tex] and a different to 2, we have a unique solution.
c) If [tex]2-a^{2}=0[/tex] and [tex]a-2=0[/tex] we have infinitely solutions.
Step-by-step explanation:
We need to rewrite the third equation in terms of a and z to solve this.
Let's start solving the first equation:
[tex]x+2y+z=2[/tex] (1)
If we subtract z in both side of this equation we will have:
[tex]x+2y=2-z[/tex] (2)
Now we can put the equation 2 into the last equation:
[tex]x+2y-(a^{2}-3)z=a[/tex]
[tex]2-z-(a^{2}-3)z=a[/tex] (3)
Simplifying the equation 3 we have:
[tex]z(2-a^{2})=a-2[/tex] (4)
Now we can analyzes the equation 4 for each case:
a) If [tex]2-a^{2}=0[/tex] and [tex]a-2\neq 0[/tex]. The system has no solution
b) If [tex]2-a^{2}\neq 0[/tex] and a different to 2, we have a unique solution.
c) If [tex]2-a^{2}=0[/tex] and [tex]a-2=0[/tex] we have infinitely solutions.
I hope it helps you!
