The system of equations:
$\begin{cases}-px+5y+3z=3\\2x-4y-z=p\\x+3py+pz=p\\\end{cases}$
This is how I learned to solve, using the Kronecker – Capelli Theorem:
Step1: Find the ranks of the coefficient matrix and the augmented matrix. Both the ranks should be equal to each other for the system to be consistent, if the ranks of the coefficient matrix and the augmented matrix are not equal, then the system is inconsistent and the system does not have a solution.
Let $A$ and $A|B$ be -subsequently- the coefficient and the augmented matrix, with:
$A=\begin{bmatrix}-p&5&3\\2&-4&-1\\1&3p&p\end{bmatrix} \text{and}\hspace{7px}A|B=\left[\begin{matrix}-p&5&3\\2&-4&-1\\1&3p&p\end{matrix}\left|\,\begin{matrix}3\\p\\p\\\end{matrix}\right.\right]$
We first find $rn(A)$ using determinants (at least, that is how I learned):
$\left|A\right|=\begin{vmatrix}-p&5&3\\2&-4&-1\\1&3p&p\end{vmatrix}=p^2+8p+7 ,\hspace{7px}\begin{array}\\\text{if}\hspace{7px}\det(A)=0,\hspace{7px} rn(A)\in\left[0,3\right)\\\text{if}\hspace{7px}\det(A)\neq0,\hspace{7px}rn(A)=3\end{array}$
Then we find $rn(A|B)$, also using determinants. I won't write it all down but I will explain here: Because $A|B$ is a $3\text{x}4$ matrix, I calculate the determinants of its $3\text{x}3$ minors (or go down further if all the minors of order 3 have $det=0$). There are only four different minors of order 3, thus I obtain four different expressions (with one being the same as the $\det(A)$).
$\begin{cases}D_1=p^2+8p+7\\D_2=3p^3+4p^2+13p+12\\D_3=p^3+p^2+3p+3\\D_4=4p^2+4p\end{cases}$
This is my problem here, I don't know how to go further with this.
Step2: After finding the range, we check if $rn(A)=rn(A|B)=n$
$\left(n-\text{the number of the linear equations, in this case:}\hspace{7px}n=3 \right)$
$\begin{cases} \text{if}\hspace{7px}rn(A)=rn(A|B)=n, \text{then the solution is unique and the system is consistent.}\\ \text{if}\hspace{7px}rn(A)=rn(A|B)<n, \text{then this system has infinitely many solutions, but it is still consistent.} \end{cases}$
Step3: After determining the equivalence with $n$, we solve the system via. the Cramer's Rule or Gaussian elimination method. If the ranks are not equal to the $n$, then we remove a row from the coefficient matrix (or an equation from the system). The rows left are the linearly independent ones. And then we solve the system (my lecturer in his examples, treats one of the coefficients as a parameter and find others in terms of that parameter).
I feel very confused with this solution. I don't know how should I apply these cubic and quadratic expressions to the solutions. There seem to be many answers to this, especially since I have many expressions (from the determinants of submatrices of $A|B$) in my hand. I would appreciate it if you would explain what kind of a way I should follow.
Best Answer
When you get that $\det(A)=p^2+8p+7=(p+7)(p+1)$ you can argue that:
If $p\ne -1,-7$ then $rn(A)=3$, so $rn(A|B)=3$ (since $A$ is a 3x3 minor) and you have unique solution.
There are only 2 values of $p$ for which yo do NOT know $rn(A)$.
So, you can study these 2 particular cases (namely $p=-7$ and $p=-1$). There is no need of study the range of $A|B$ with the parameter $p$.
Observe that in $A$ you allways have a 2x2 nonzero minor, so $rn(A)=2$ if $p=-7,-1$.