System of equations with parameters

Question

Given the equations
$x+3y+z+au=1$
$x+2y+az+u=1$
$2x+y+2z+3u=b$
$x+y+az+2u=a$
Determine the parameters $a$ and $b$ such that system of equations has a solution.
As far as I know for those system to have a solution determinant has to be non-zero, so I found values of $a$ for which determinant is 0, so a has to be a value different from those numbers. But what about parameter $b$ do I have to take that in account? Since it doesn't appear in coefficients.

Answer 1

You gave the correct answer for $a$, concerning the determinant of the LHS Matrix ( let's call this matrix $\mathbf M$). When this determinant is nonzero, the system is invertible and you get for the vector of variables $\mathbf v$:

$$ \mathbf v = \mathbf M^{-1} (1, 1, b, a)^T $$

So there will always be a solution, regardless of $b$.

Answer 2

for the variable $x$ i have found $$-a^3+2 a^2 x+14 a^2-18 a-2 x+7=\left(a^2+3 a-3\right) b$$ can you proceed?