Determining if a vector is in the column space of a matrix Hi there I'm having some trouble with the following problem:
I have a $3\times3$ symmetric matrix
$$
A=\pmatrix{1+t&1&1\\ 1&1+t&1\\ 1&1&1+t}.
$$
I am trying to determine the values of $t$ for which the vector $b = (1,t,t^2)^\top$ (this is a column vector) is in the column space of $A$.
I think I'm fairly aware of how to go about it, forming the augmented matrix $[A|b]$ and basically using row ops to find a solution with which I could solve for the value(s) of $t$.  But I've been trying this and have no luck.  May I be missing something?
Thank you

Gaussian elimination is not difficult in this case:
\begin{align}
\left[\begin{array}{ccc|c}
1+t & 1 & 1 & 1 \\
1 & 1+t & 1 & t \\
1 & 1 & 1+t & t^2
\end{array}\right]
&\to
\left[\begin{array}{ccc|c}
1 & 1 & 1+t & t^2 \\
1 & 1+t & 1 & t \\
1+t & 1 & 1 & 1 
\end{array}\right]
\\
&\to
\left[\begin{array}{ccc|c}
1 & 1 & 1+t & t^2 \\
0 & t & -t & t-t^2 \\
0 & -t & -2t-t^2 & 1-t^2(1+t) 
\end{array}\right]
\end{align}
If $t\ne0$ we can go on:
\begin{align}
\left[\begin{array}{ccc|c}
1 & 1 & 1+t & t^2 \\
0 & t & -t & t-t^2 \\
0 & -t & -2t-t^2 & 1-t^2-t^3 
\end{array}\right]
&\to
\left[\begin{array}{ccc|c}
1 & 1 & 1+t & t^2 \\
0 & 1 & -1 & 1-t \\
0 & 0 & -3t-t^2 & 1+t-2t^2-t^3
\end{array}\right]
\end{align}
If $t=-3$, the last row becomes
$$
\begin{array}{ccc|c}
0 & 0 & 0 & 7
\end{array}
$$
so the system has no solution.
If $t\ne-3$, the system has a solution.
If $t=0$, we get, from the place we stopped at,
$$
\left[\begin{array}{ccc|c}
1 & 1 & 1 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1
\end{array}\right]
$$
and the system has no solution.
