given three vectors in $\mathbb R^3$, they span all of $\mathbb R^3$ if and only if what does not equal zero?

Question

Given three column vectors, $a_1=(a,-8,0)^t$, $a_2=(7,2,6)^t$, and $a_3=(-7,10,-8)^t$, what can not equal zero (in terms of $a$) in order for these three vectors to span $\mathbb R^3$?enter image description here

Answer 1

Note that $\lambda_1(a,-8,0)+\lambda_2(7,2,6)+\lambda_3(-7,10,-8)=0$ can be written as the linear system $$\begin{pmatrix}a&7&-7\\ -8&2&10\\ 0&6&-8\end{pmatrix}\Lambda=0.$$ Note that this system has only the zero solution if the determinant of the matrix is not zero. Developing w.r.t. the last row we find that $\det(A)=-6(10a-56)-8(2a+56)=-76a-112$. Hence if $-76a-112\neq 0$, the three vectors are linearly independent. Three linearly independent vectors in $\mathbb{R}^3$ must be generating as well.

Long story short, the determinant of these vectors put in rows (or columns) cannot be zero.

Answer 2

We have $\{a_1,a_2,a_3\}$ spans $\mathbb{R}^3$ if and only if $\det(a_1,a_2,a_3)\neq 0$. Since $$\det(a_1,a_2,a_3)=-4(19a+28)$$ if follows that the vectors $a_1,a_2,a_3$ spans $\mathbb{R}^3$ if and only if $19a+28\neq 0$.