<s>[INST] Finding $p$ and $t$ such that the system $px+y+z=1$, $x+2y+4z=t$, $x+4y+10z=t^2$ has one, infinitely-many, or no solutions 
Consider the following equations:
$$\begin{align}
px+\phantom{2}y+\phantom{10}z &=1 \\
x+2y+\phantom{1}4z &=t \\
x+4y+10z &=t^2
\end{align}$$
Now find the values of $p$ and $t$ for which
(a) there is a unique solution
(b) there are infinitely many solutions
(c) there is no solution

My attempt: For (a) we require the determinant $\begin{vmatrix}p&1&1\\1&2&4\\1&4&10\end{vmatrix}\neq 0$. Hence $p\neq 1$.
For (b) we need two of the equations to represent the same line - I tried subbing in the difference between the second and third equation into the first but got nowhere. Similarly for (c).
 [/INST] System of equations give the below matrix:
$\left(\begin{array}{rrr|r}
p & 1 & 1 & 1\\
1 & 2 & 4 & t\\
1 & 4 & 10 & t^2
\end{array}\right)$
We set third row as $(R_3 - 2 \cdot R_2)$ and second row as $(R_2 - 2 \cdot R_1)$.
$\left(\begin{array}{rrr|r}
p & 1 & 1 & 1\\
1-2p & 0 & 2 & t-2\\
-1 & 0 & 2 & t^2-2t
\end{array}\right)$
If $p = 1$, $t$ must be either $1$ or $2$ for row $2$ and $3$ to be consistent. Also note that it leads to only two equations in $3$ variables.
So we have no solutions, if $p = 1, t \notin (1,2)$
If $p = 1, t \in (1,2)$, we have infinite solutions
If $p \ne 1$, we have unique solution.
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