Given a matrix $A$, show that it is positive. 
Show that
  $$A := \begin{bmatrix}7 & 2 & -4\\2 & 4 & -2\\-4 & -2 & 7 \end{bmatrix}$$
  is positive definite.

Could this be proven by showing that each of the vectors of the standard basis gives a positive result, e.g.:
$$\begin{bmatrix}1 & 0 & 0\end{bmatrix} \begin{bmatrix}7 & 2 & -4\\2 & 4 & -2\\-4 & -2  & 7 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \\ 0\end{bmatrix} > 0.$$
The second part of the question asks me to diagonalize the matrix using an orthogonal matrix, which as I understand, is to use elementary matrices on the rows and columns of the matrix to get it to a diagonal form. Would it make a difference if Ifirstly only dealt with the rows and only afterward used the exact matrices only on the columns?
Thanks for your time.

I'm assuming we're thinking of $\Bbb R^3$.
If you have $v=\sum \alpha_ib_i$ written in terms of a basis of column vectors $b_i$, then 
$$
v^\top Av=(\sum_i \alpha_ib_i^\top)A(\sum_j \alpha_jb_j)=\sum_{i,j}\alpha_i\alpha_jb_i^\top Ab_j
$$
Clearly the diagonal terms ($\alpha_i^2 b_i^\top A b_i\geq 0$) but you see there is some potential for negatives among the terms with $i\neq j$.
The two main tools are Sylvester's criterion or checking eigenvalues. Sylvester's criterion clears this one up rather quickly.
