1. is Negative Definite Matrix ?
`[[-3,0,0],[0,-2,0],[0,0,-1]]`

Solution:

`A`

=

`-3` `0` `0` `0` `-2` `0` `0` `0` `-1`

Test method 1: Existence of all negative Pivots.
First apply Gaussian Elimination method to find Pivots

`A`

=

`-3` `0` `0` `0` `-2` `0` `0` `0` `-1`

`:.` Pivots are `-3,-2,-1`

Here all pivots are negative, so matrix is negative definite.

---

Test method 2: Determinants of all upper-left sub-matrices.

`A`

=

`-3` `0` `0` `0` `-2` `0` `0` `0` `-1`

`-3`

`=-3`

`-3`   `0`   `0`   `-2`

`=6`

`-3`   `0`   `0`   `0`   `-2`   `0`   `0`   `0`   `-1`

`=-6`

Determinants are `-3,6,-6`

Here all odd determinants `D_1,D_3<0` and even determinants `D_2>0`, so matrix is negative definite.

---

Test method 3: All negative eigen values.
`|A-lamdaI|=0`

`(-3-lamda)`   `0`   `0`   `0`   `(-2-lamda)`   `0`   `0`   `0`   `(-1-lamda)`

= 0

`:.(-3-lamda)((-2-lamda) × (-1-lamda) - 0 × 0)-0(0 × (-1-lamda) - 0 × 0)+0(0 × 0 - (-2-lamda) × 0)=0`

`:.(-3-lamda)((2+3lamda+lamda^2)-0)-0(0-0)+0(0-0)=0`

`:.(-3-lamda)(2+3lamda+lamda^2)-0(0)+0(0)=0`

`:. (-6-11lamda-6lamda^2-lamda^3)-0+0=0`

`:.(-lamda^3-6lamda^2-11lamda-6)=0`

`:.-(lamda+1)(lamda+2)(lamda+3)=0`

`:.(lamda+1)=0 or(lamda+2)=0 or(lamda+3)=0 `

`:.` The eigenvalues of the matrix `A` are given by `lamda=-3,-2,-1`,

Here all determinants are negative, so matrix is negative definite.