
Basis/Spans
Let $\displaystyle v_1=\begin{bmatrix}1 \\ 1 \\ a\end{bmatrix},v_2=\begin{bmatrix}1 \\ a \\ 1\end{bmatrix},$ and $\displaystyle v_3=\begin{bmatrix}a \\ 1 \\ 1\end{bmatrix}$
For which values of the parameter a is $\displaystyle v_1$,$\displaystyle v_2$ and $\displaystyle v_3$ a basis for $\displaystyle R^3$?

Hint : evaluate the determinant of the matrix whose columns are those three vectors. Solve for the values of $\displaystyle a$ for which this determinant is nonzero. (Do you see why?)

Bruno, I found the determinant of the matrix to be $\displaystyle a^33a+2$ and the values of a that make it nonzero are $\displaystyle a=/2$ and $\displaystyle a=/1$
Is this right? And could you explain why this works (I am used to row reducing from class)? Thanks

How can you possibly get $\displaystyle a^3$ in there? You should get a second degree polynomial in $\displaystyle a$.
This works, because for the three vectors to be a basis of $\displaystyle \mathbb{R}^3$ they should form the sides of a parallelepiped having nonzero volume, and the determinant is that volume. Or, more abstractly put, if $\displaystyle \beta=\{v_1,v_2,v_3\}$ is a basis, then the change of basis transformation from $\displaystyle \beta$ to the standard basis is invertible; its matrix representation is the one whose determinant you evaluated (with a small mistake).

Determinant is easy method. But when dimension is high, it's difficult to solve equation.
I suggest processing by rref
$\displaystyle rref(A)=rref[\vec{v}_1\ \vec{v}_2\ \vec{v}_3]=\begin{bmatrix} 1&1&a\\0&a1&1a\\0&0&2aa^2 \end{bmatrix}$
if three vectors span a basis, then rank(A)=3, $\displaystyle a1\neq 0 \& 2aa^2\neq 0$

Span