Find all values for t such that $\{(t,3),(7,t)\}$ is not a basis for $\mathbb{R}^2$ Find all values of $t$ such that 
$$
        \begin{pmatrix}
        t & 7 \\
        3 & t \\
        \end{pmatrix}
$$
is not a basis for $\mathbb{R}^2$
$$
        \begin{pmatrix}
        t & 0 \\
        1 & t \\
        \end{pmatrix}
$$
is a basis for $\mathbb{R}^2$
For first question I know i have to show that its either linear dependent or it's span is not $\mathbb{R}^2$ but I'm not sure how I would go about showing that. I'm new to linear algebra please help me as much as possible sorry.
 A: I wonder what the determinant of those matrices is for various values of $t$...
Why this matters:

 If the determinant is zero, the vectors are linearly dependent.  If two vectors are linearly dependent, they do not span a 2-dimensional space.

A: You know any two vectors$\begin{pmatrix}
        p  \\
        q  \\
        \end{pmatrix}$,$\begin{pmatrix}
        p'  \\
        q'  \\
        \end{pmatrix}$ in $\mathbb{R^2}$ are linearly independent if $$x\begin{pmatrix}
        p  \\
        q  \\
        \end{pmatrix}+y\begin{pmatrix}
        p'  \\
        q'  \\
        \end{pmatrix}=\begin{pmatrix}
        0  \\
        0  \\
        \end{pmatrix} \,\,\,\Rightarrow x=y=0 \,\,\,\,\, x,y\in\mathbb{R}$$
Here in this case, we have $$x\begin{pmatrix}
        t  \\
        3  \\
        \end{pmatrix}+y\begin{pmatrix}
        7  \\
        t  \\
        \end{pmatrix}=\begin{pmatrix}
        0  \\
        0  \\
        \end{pmatrix}$$  which can be written as $$\begin{pmatrix}
        t & 7 \\
        3 & t  \\
        \end{pmatrix}\begin{pmatrix}
        x  \\
        y  \\
        \end{pmatrix}=\begin{pmatrix}
        0  \\
        0  \\
        \end{pmatrix}$$ We want to find the values of $t$ in such a way the solution is non-trivial( then only the basis would be linearly dependent). Now we know only trivial solution exists only if the determinant of the matrix $\begin{pmatrix}
        t & 7 \\
        3 & t \\
        \end{pmatrix}$ is non-zero. Hence for a non trivial solution we need $t= \sqrt{21},-\sqrt{21}$
A: Calculate for which $t$
\begin{align}
\begin{vmatrix}
t & 7\\
3 & t
\end{vmatrix}= 0\qquad \text{and}\qquad\quad\begin{vmatrix}
t & 0\\
1 & t
\end{vmatrix}\neq 0\qquad \text{for the second part}
\end{align}
That is $t=\pm \sqrt{21}$ and $t\in\mathbb{R}\backslash\lbrace 0\rbrace$ for the second part.
A: $\begin{pmatrix}t&7\\ 3&t\end{pmatrix}$ is not a basis for $\mathbb R^2$ if $\begin{pmatrix}t\\3\end{pmatrix}$ and $\begin{pmatrix}7\\t\end{pmatrix}$ are linearly dependent.
You can let $\begin{pmatrix}t\\3\end{pmatrix}=a\begin{pmatrix}7\\t\end{pmatrix}$ and solve for $a$ and $t$.
Solving yields $a=\sqrt{\frac 37}$ and $t=\sqrt {21}$
Edit: another solution is $a=-\sqrt{\frac 37}$ and $t=-\sqrt{21}$
