How to show this particular $4 \times 4$ matrix is positive definite? I am preparing for an exam in Numerical Analysis, and I am solving some practice problems. The question is to show that the following matrix $A$ is positive definite. I would only have about 10 minutes to work on this problem, so I am trying to solve this as fast as possible.
We are also told that all the eigenvalues of $A$ are distinct (but this information may not be useful here; there is a part b to this question which might make use of this fact)
$$A = \begin{bmatrix}
1 & -1 & 2 & 0\\
-1 & 4 & -1 & 1\\
2 & -1 & 6 & -2\\
0 & 1 & -2 & 4
\end{bmatrix}$$
My Attempts: My first tought is to use Greshgorin's Circle Theorem to show that all the eigenvalues are positive. However, this does not work because  the first Greshgorin disk contains negative reals.
My second tought is to use Sylvester's Criterion. This is perhaps doable in under 10 minutes, but it is prone to mistakes (especially when going fast). I am also not sure if Sylvester's Criterion was taught in the class that this problem comes from.
 A: @2rd_7's suggestion to calculate the pivots seems quick and straightforward. You just need to keep subtracting out multiples of the remaining rows (i.e., add the first row to the second, subtract 2 times the first row from the third, etc.) until you get an upper triangluar matrix:
$U = \begin{bmatrix} 1 & -1 & 2 & 0\\ 0 & 3 & 1 &1\\ 0 & 0 & 3 & -1\\ 0 & 0 & 0 & 26/9\end{bmatrix}$.
The pivots are now on the diagonal and are all positive, so the matrix is positive definite.
A: HINT:
The rows from $2$ to $4$ are diagonally dominant. The first one is not. But we can make it so by multiplying on both sides by the matrix $\operatorname{diag}(t, 1, 1, 1)$, where $t$ is large.
$\bf{Added:}$ Looks plausible, but the other rows are affected. Indeed, need $t> 3$, and then the second row is not dominant anymore. So the solution is not good. In fact, one can check that there is no way to transform our matrix with a diagonal matrix to make it diagonally dominant.
Maybe just showing directly that the determinant is positive. Since the principal minor $(2,3,4)$ is positive definite, being dominant, this would be good enough.
Using WolframAlpha, I got the Cholesky decomposition of the matrix
$$\left[\begin{matrix} 1 & -1 & 2 & 0  \\ -1& 4 & -1& 1\\ 2& -1& 6&-2\\ 0&1 &-2&4\end{matrix} \right]=\\=\left[\begin{matrix} 1 & 0 & 0 & 0  \\ -1& 1 & 0& 0\\ 2& 1/3 & 1&0\\ 0&1/3 &-7/5&1\end{matrix} \right]\left[\begin{matrix} 1 & 0& 0 & 0  \\ 0& 3 & 0& 0\\ 0& 0& 5/3&0\\ 0&0 &0&2/5\end{matrix} \right]\left[\begin{matrix} 1& -1& 2& 0\\ 0& 1& 1/3& 1/3\\ 0& 0& 1& -7/5\\0& 0& 0& 1\end{matrix} \right]$$
The diagonal part has moderate eigenvalues. Now, $A$ has a small eigenvalue, $\approx 0.02$, and this is possible since the upper diagonal part has a small singular value.
