1. Tardigrade
  2. Question
  3. Mathematics
  4. A cubic equation x3 + rx - p = 0 has roots a,b and c . A square matrix M = [mij], i, j = 0, 1 and 2 , of size 3 × 3 is made such that m00= a, m11 = b and m22 = c . All other elements of M are 1 . What should be the least value of p so that |M| is an odd prime?

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Q. 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?

J & K CETJ & K CET 2017Determinants

Solution:

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, $M=\left[\begin{matrix}a&1&1\\ 1&b&1\\ 1&1&c\end{matrix}\right]$
$\therefore \left|M\right|=a\left(bc-1\right)-1\left(c-1\right)+1\left(1-b\right)$
$=abc-a-c+1+1-b=2+abc-\left(a+b+c\right)$
$=2+p-0=2+p$
Since, $\left|M\right|$ is an odd prime so least value of p is 1