#GATE 2014 IN

Question

Answer 1

I am considering that b and B are same.

 

$f(x) =x^{T}Ax + B^{T}x +c$

A n*n symmetric real matrix $A$ is said to be positive definite if $x^{T}Ax >0$ for all non zero $x$ in $R^{n}$

$B$ and $x$ are vectors of dimensions n *1.

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Let $n =2$ ,$A_{2*2}$ = $\begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix}$, $B_{2*1}$ = $\begin{bmatrix} m\\n \end{bmatrix}$

Now since $x^{T}Ax >0$ and $c$ is a constant value so we should try to make $b^{T}x$ as negative so that we could get a minimum value for f(x).

So either option b or option c should be the answer.

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Option B

$x =-(A^{T}A)^{-1}B$

$\Rightarrow x = -\left ( \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix}\right )^{-1}$ $\begin{bmatrix} m\\n \end{bmatrix}$

$\Rightarrow x = -\left ( \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix}\right )^{-1}$ $\begin{bmatrix} m\\n \end{bmatrix}$

$\Rightarrow x =- \frac{\left ( \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix}\right )}{1}$ $\begin{bmatrix} m\\n \end{bmatrix}$

$\Rightarrow x = \begin{bmatrix} -m\\-n \end{bmatrix}$

Also, $x^{T} =\begin{bmatrix} -m & -n \end{bmatrix}$

$x^{T}Ax$ = $\begin{bmatrix} -m & -n \end{bmatrix}\begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} -m\\-n \end{bmatrix}$=$m^{2}+n^{2}$

$B^{T}x$ = $\begin{bmatrix} m & n \end{bmatrix}\begin{bmatrix} -m\\-n \end{bmatrix}$=$-m^{2}-n^{2}$

$\therefore$ $f(x)$ =$x^{T}Ax + B^{T}x +c$ = $m^{2}+n^{2}-m^{2}-n^{2}+c$= $c$

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Option C

$x$ =$-\left ( \frac{A^{-1}B}{2}\right )$

$\Rightarrow x$ = $-\left ( \frac{\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}^{-1}\begin{bmatrix} m\\n \end{bmatrix} }{2}\right )$=$-\left ( \frac{\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\begin{bmatrix} m\\n \end{bmatrix} }{2}\right )$=$ -\frac{\begin{bmatrix} m\\n \end{bmatrix} }{2}$=$\begin{bmatrix} -\frac{m}{2}\\-\frac{n}{2} \end{bmatrix}$

Also, $x^{T} =\begin{bmatrix} -\frac{m}{2} & -\frac{n}{2} \end{bmatrix}$

$x^{T}Ax$ = $\begin{bmatrix} -\frac{m}{2} & -\frac{n}{2} \end{bmatrix}\begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} -\frac{m}{2}\\-\frac{n}{2} \end{bmatrix}$=$\frac{m^{2}}{4}+\frac{n^{2}}{4}$

$B^{T}x$ = $\begin{bmatrix} m & n \end{bmatrix}\begin{bmatrix} -\frac{m}{2}\\-\frac{n}{2} \end{bmatrix}$=$-\frac{m^{2}}{2}-\frac{n^{2}}{2}$

$\therefore$ $f(x)$ =$x^{T}Ax + B^{T}x +c$ = $\frac{m^{2}}{4}+\frac{n^{2}}{4}-\frac{m^{2}}{2}-\frac{n^{2}}{2}+c$= $-\frac{m^{2}}{2}-\frac{n^{2}}{2}+c$

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$\because$ $-\frac{m^{2}}{2}-\frac{n^{2}}{2}+c < c$

$\therefore$ Option C should be the right answer.

Comments

Thank you very much for the explanation @Satbir