Unique soln for system of linear eqns: ax - y + bz = 1, 2y + az = 0, 2ax - y = 0

[bouhrassa]

What condition the parameter a and b must satisfy for this system of linear equations have a unique solution

ax - y + bz = 1
2y + az = 0
2ax - y = 0

So I guess the determinant must not be equal to zero so what I got is :

-a^2 - 4ab

My choices are

A) a^2b+b^2 = 0
B) 4ab+a^2 ≠0
C) ab + 2= 0
D) a^2b - 2b^2 ≠0
E) a^2 - 2ab = 0
F) None of these answers

Thank you
bnone of these answersnone of these answers

 

[Deleted member 4993]

What condition the parameter a and b must satisfy for this system of linear equations have a unique solution

ax - y + bz = 1
2y + az = 0
2ax - y = 0

So I guess the determinant must not be equal to zero so what I got is :

(2a)*(-a-2b) - -1(a^2)

= -2a^2 - 2ab + a^2

= -a^2 - 2ab

My choices are

A) a^2b+b^2 = 0
B) 4ab+a^2 ≠0
C) ab + 2= 0
D) a^2b - 2b^2 ≠0
E) a^2 - 2ab = 0
F) None of these answers

Thank you
bnone of these answersnone of these answers

Your calculation of determinant does not seem to be correct. What is the matrix that you use?

 

[bouhrassa]

Just edited it ... but I don't really know what to do with the determinent from there...
Can I multiply -1 on each side of -a^2 - 4ab ≠ 0 so the answer would be B)

 

[Deleted member 4993]

Just edited it ... but I don't really know what to do with the determinent from there...
Can I multiply -1 on each side of -a^2 - 4ab ≠ 0 so the answer would be B)

Yes...