Let

$q(x) =x_1^2+x_2^2+3x_3^2-2x_1x_3+2x_2x_3+x_1+x_3+5 =x^TQx+b^Tx+5$

$\begin{matrix} x=( x_1 & x_2 & x_3 )^T\\ \end{matrix}$

the first question was find a symmetric matrix Q and a vector b. so I found that matrix by using way to solve quadratic form

$$Q= \begin{matrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ -1 & 1 & 3 \\ \end{matrix}$$

# and $$b =\begin{matrix} (1 & 0 & 1)^T\\ \end{matrix}$$

The second question was 'find, in terms of Q and b and find the minimum of q(x)'

If $\partial q(x)/∂x= \begin{matrix} (∂q/∂x_1& ∂q/∂x_2&∂q/∂x_3)^T\\ \end{matrix}$

how can I solve second question? how can I approach in good way?

I just wanna share my try,

I guess we can find the minimum of this function by differentiation. Let's find our critical points, that is $$\exists x_{0} \in R^3, / \bigtriangledown q(x_{0})=0?$$ note that $q(x): R^3 \rightarrow R$ $$\bigtriangledown q(x)=(2x-2z+1,2y+2z,6z-2x+2y+1)$$ Now we should solve the following system

$2x-2z+1=0$

$2y+2z=0$

$6z-2x+2y+1=0$

it's east to find $x_{o}=(-3/2 , 1,-1)$ as its solution. Thus we just need to check if the Hessian is a positive matrix (cause it implies this point is a minimum). If you derivative again, you'll find:

$Hq(x_{o})= \begin{matrix} 2 & 0 & -2 \\ 0 & 2 & 2 \\ -2 & 2 & 6 \\ \end{matrix}$

As far as I remember there is a theorem which relate Hessian with the quadratic form, check your notes cause it will help. However, in our case $Hq(x_{0})=2Q$, then we just need to check if Q is a positive matrix. $A_{1} >0, A_{2}>0, A_{3}>0$ so Q is a positive matrix thus $x_{0}$ is the minimum of q(x) $\forall b$

• it was really good help to understand thank you. – matthew Sep 24 '16 at 14:20
• It's better to use correct English: "wanna" $\rightarrow$ "want to" (I am not native English speaker) – Jean Marie Sep 24 '16 at 19:29