<s>[INST] For which a the matrix is positive-definite We have $A=\begin{pmatrix}a&2\\ 2&a+1\end{pmatrix}$
And I want to know for which a the matrix is positive-definite.
So our requirement is $det(A)\ge0$
$(a*(a+1))-4\ge0$
$a^2+a-4\ge0$
I understand that in the points $\frac{-1+\sqrt{17}}{2},\frac{-1-\sqrt{17}}{2}$ the function are equal to zero.
Is it true to say that because the function is equal to zero in the points so $ ∀a>\frac{-1+\sqrt{17}}{2},\frac{-1-\sqrt{17}}{2}$
$a^2+a-4\ge0$?
 [/INST] You can prove a matrix is positive definite by checking if the determinant of all of its leading minors are greater then 0.
$$\begin{vmatrix} a\end{vmatrix}>0 \iff a>0 \space\space\space(1)$$
$$\begin{vmatrix} 
a & 2 \\
2 & a+1
\end{vmatrix} = a^2+a-4$$
$$a^2+a-4>0 \iff a>\frac{-1+\sqrt{17}}{2} \space\space\space OR \space\space\space\space a<\frac{-1-\sqrt{17}}{2}\space\space\space\space\space(2)$$
From $(1)$ and $(2)$ we can get that A is positive definite for any $a>\frac{-1+\sqrt{17}}{2}$
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