find values of b for which the linear system has infinitely many solutions with 1 arbitrary parameter?

86 Views Asked by At

$$\left[\begin{array}{ccc|c} b(b-1) & 0 & b & b^2 \\ 0 & 1-b^2 & b^2 & -b \\ 0 & 0 & b(b+1) & 0 \\ 0 & 0 & 0 & b(2-b)\end{array}\right]$$ Edit:Would the same conditions hold for row equivalent matrices like A? $$A = \left[\begin{array}{ccc|c} b^2-b & 0 & b & b^2 \\ 0 & 1-b^2 & b^2 & -b \\ 0 & b^2-1 & b & b \\ b^2-b & 0 & b & 2b\end{array}\right]$$

Question: Determine the values of b so that the system has

(1) No solution: b $\ne 0 $ and b $\ne 2 $

Consistent System : b = $ 0 $ or b = $ 2 $

(2)unique solution: b = 2

(3) Infinity many solns, one parameter : no values of b found

(4) Infinitely many solns, 2 parameters: b = $0$

I am not sure how to find values of b for (3)? I have tried -1 and 1,but both values of b give no solutions.

Please correct me if I am wrong about the conditions for b, thank you!