Help with solving for variables in a small linear equation system. I came across this question in my math book, I would please like some help with how to approach this question. 
Determine for all values on $a, b_1, b_2, b_3$ if there are solution to the equation system:
\begin{array} {lcl} x - y+ (a+1)z & = & b_1 \\ 2x - y+ az & = & b_2\\ x + (a+2)y + (2a+3)z & = & b_3 \end{array}
How would you approch this question and where would you start?
Thank you kindly for your help!
 A: Hints: what must you have for a solution to exist? Think determinant.
The determinant of the LHS is $a^2+6a+8$.
What values of $a$ would be problematic?
A: One very good method is of azmoti, if you want extensive conditions, follow this.
You have tagged Linear Algebra so you should have knowledge of Gaussian Elimination. Follow the general method to solve the system of linear equations, i.e. write coefficients in Matrix form. The condition, when the system has $a$ solution is when it can be reduced to identity matrix. It has infinitely many solution if one or more rows are reduced to complete zeros. Of course, these will depend upon values of $a,b_1,b_2,b_3$ So, work with matrix and try to reduce to row echelon form, you will get conditions.


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*Determinant zero may mean, it still has solution, but infinitely many


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A: You may first find the determinant of the coefficient matrix $A$ for the left hand side of the system of equations. You should find that it is $(a+2)(a+4)$. In other words when $a\neq-2,-4$, the system always has a unique solution $(x,y,z)$, regardless of the values of $b_1,b_2,b_3$.
When $a=-2$ or $-4$, the coefficient matrix is singular. So, the system can be inconsistent (i.e. insolvable) if $b_1,b_2,b_3$ have not right values. For $a=-2$, you can apply row operations to bring the system of equations into a row reduced echelon form. You will find that the bottom row of the LHS will become zero, while the RHS becomes a nonzero multiple of $b_2-b_1-b_3$. Hence the system is solvable if and only if $b_2=b_1+b_3$.
Alternatively, you may use column operations to show that the column space is the span of $(0,1,1)^T$ and $(1,1,0)^T$. Hence the system is solvable iff $(b_1,b_2,b_3)=p(0,1,1)+q(1,1,0)$ for some $p$ and $q$, i.e. iff $b_2=b_1+b_3$. The case $a=-4$ can be handled in a similar manner.
