Question

A cubic equation $x^3+rx-p=0$ has roots $a,b$ and $c$ . A square matrix $M=[m_{ij}],i,j=0,1$ and $2$ , of size $3\times 3$ is made such that $m_{00}=a,m_{11}=b$ and $m_{22}=c$ . All other elements of $M$ are $1$ . What should be the least value of $p$ so that $|M|$ is an odd prime?

• A. $0$
• B. $1$
• C. $-1$
• D. $-2$

Answer 1

Answer: (B)

We have , $x^3+rx-p=0$ has roots $a,b$ and $c$
$\therefore a+b+c=0,ab+bc+ca=r,abc=p$
Now, $M=\left[\begin{array}{ccc}a& 1& 1\\ 1& b& 1\\ 1& 1& c\end{array}\right]$
$\therefore \left|M\right|=a\left(bc-1\right)-1\left(c-1\right)+1\left(1-b\right)$
$=abc-a-c+1+1-b=2+abc-\left(a+b+c\right)$
$=2+p-0=2+p$
Since, $\left|M\right|$ is an odd prime so least value of p is 1