For what values of $a$ does the set $\{(a,a,1), (1,a,1), (1,2,4)\}$ form a basis for $\Bbb R^3$ I've tried working with the matrix form to find values for $a$ that would work in a homogeneous solution set but the $a$ has made it very complicated and I don't feel that would be the way to solve this. I've also thought to use the determinant and find all values that don't make it zero though I have come into some issues with this too. Can anyone think of a better way or maybe how or why my other ones haven't or can work?
 A: It's a basis iff the matrix with $(a,a,1),(1,a,1),(1,2,4)$ as columns (or rows) has nonzero determinant.
In this case, its determinant will be a polynomial in the variable $a$, so it will be a basis as long as you choose values of $a$ which aren't zeros of the determinant.
A: I recommend pursuing the method of making the determinant non-zero. First calculate the determinant and then set your equation equal to zero. Whichever values of $a$ you get are the values that $a$ cannot take on in order for your three tuples to form a basis. 
Hint: You should end up with the polynomial $(a-1)(4a-2) = 0$.
Let $A = \left( \begin{array}{ccc}
a & a & 1\\
1 & a & 1 \\
1 & 2 & 4 \end{array} \right)$.
Then $\det(A) = det \left( \begin{array}{ccc}
a & 1\\
2 & 4 \\
 \end{array} \right) \cdot a \cdot (-1)^{1+1} + \left( \begin{array}{ccc}
a & 1\\
2 & 4 \\
 \end{array} \right)\cdot1\cdot(-1)^(1+2) + \left( \begin{array}{ccc}
a & 1\\
a & 1 \\
 \end{array} \right)\cdot1\cdot(-1)^{1+3}$
Can you take it from here?
A: Since the dimension of $\mathbb{R}^3$ is 3, any subset of 3 linearly independent vectors of  $\mathbb{R}^3$ will be a basis. Just choose $a$ so that the 3 vectors are linealry independent and you've got your answer. 
