Let A be the following symmetric matrix
A =
[4 2 -2 2
2 2 1 1
-2 1 14 -1
2 1 -1 x]
find the values of x , such that A is positive definite.
a symmetric matrix is a square matrix, A, that is equal to its transpose and An n × n real symmetric matrix M is positive definite if zTMz > 0 for all non-zero vectors z with real entries (i.e. z ∈ Rn), where zT denotes the transpose of z.
There is a theorem telling that $\displaystyle A=(a_{ij})_{1\leq i,j\leq n}$ (here, $\displaystyle n=4$) is positive definite if, and only if, for $\displaystyle k=1,\ldots,n$ the determinant of $\displaystyle A_k=(a_{ij})_{1\leq i,j\leq k}$ is positive.
If you know this, then it is easy. The determinants of $\displaystyle A_1, A_2, A_3$ are positive, they do not depend on $\displaystyle x$, and the determinant of $\displaystyle A_4=A$ is found to be $\displaystyle 36(x-1)$, hence it is positive iff $\displaystyle x>1$. So $\displaystyle A$ is positive definite iff $\displaystyle x>1$.