Question

Let $A=a_{ij}$ be a matrix of order $3,$ where
$a_{ij}=\{\begin{array}{cc}x;& \text{if}i=j,x\in R\\ 1;& \text{if}|i-j|=1\\ 0;& \text{otherwise}\end{array},$
then which of the following hold(s) good:

• A. for $x=2,A$ is a diagonal matrix
• B. $A$ is a symmetric matrix
• C. Let $f\left(x\right)=detA$, then the function$f\left(x\right)$ has only maxima
• D. Let $f\left(x\right)=detA$, then the function$f\left(x\right)$ has both the maxima and minima

Answer 1

Answer: (B), (D)

Let $f\left(x\right)=detA$, then the function$f\left(x\right)$ has both the maxima and minima

$a_{ij}=\{\begin{array}{cc}x;& \text{if}i=j,x\in R\\ 1;& \text{if}|i-j|=1\\ 0;& \text{otherwise}\end{array}$

$\Rightarrow A=\left[\begin{array}{ccc}x& 1& 0\\ 1& x& 1\\ 0& 1& x\end{array}\right]$
$\Rightarrow |A|=x^3-2x$

If $f\left(x\right)=x^3-2x$
$\Rightarrow f^{\prime }\left(x\right)=3x^2-2$
$=(\sqrt{3}x-\sqrt{2})(\sqrt{3}x+\sqrt{2})$

$\Rightarrow f^{\prime \prime }\left(x\right)=6x$

So, $x=\frac{\sqrt{2}}{\sqrt{3}}$ is a point of minima and $x=-\frac{\sqrt{2}}{\sqrt{3}}$ is a point of maxima.

Hence, $f\left(x\right)$ has both the maxima and minima