IIT JEE 1983 Maths - 'Adapted' - FIB to Single-Digit Integer Q6

consider the matrix

$A=\left[\begin{matrix} 0&3&7\\ 2&0&2\\ 7&6&0\\ \end {matrix}\right]$

the charactestic equation of the matrix is given by

$\begin{aligned} \det(A-\lambda I) &=0\\ \rightarrow \begin{vmatrix} -\lambda&3&7\\ 2&-\lambda&2\\ 7&6&-\lambda\\ \end {vmatrix}&=0\end{aligned}$

we can get the required equation by using the transformation $\lambda=-x$

also we have sum of roots of characterstic equation $=trace(A)=0$

the sum of roots of given equation will be $\sum_{i=0}^{3} -\lambda_i=-\sum_{i=0}^{3} \lambda_i=0$

Thus $\alpha +\beta=9$

$5+\left\lfloor\dfrac{\alpha+\beta}{3}\right\rfloor=5+3=\boxed{8}$

The above determinant computes to:

$x^3 - 67x + 126 = (x + 9)(x - \alpha)(x - \beta) = x^3 - (\alpha + \beta - 9)x^2 + (\alpha\beta - 9\alpha - 9\beta)x + 9\alpha\beta = 0$

By Vieta's Formulae, we obtain:

$\alpha + \beta - 9 = 0; \alpha\beta - 9\alpha - 9\beta = -67; 9\alpha\beta = 126 \Rightarrow \beta = \frac{14}{\alpha}$

and now taking $\alpha + \frac{14}{\alpha} - 9 = 0 \Rightarrow \alpha^2 - 9\alpha + 14 = (\alpha - 2)(\alpha - 7) = 0 \Rightarrow \alpha = 2, 7.$

This gives the points $(\alpha, \beta) = (2,7); (7,2)$ as solutions, which makes $5+\left\lfloor\dfrac{\alpha+\beta}3\right\rfloor = \boxed{8}.$

I mean like is there a shorter way of reducing this determinant like subtracting the rows/columns method.Because otherwise we have to directly use the co-factor sum ,and expanding along row $R_1$ we get cubic $x^3$ $-$ $67 x$ $+$ $126$ $=$ $0$ .And it can be factored to $(x+9)(x^2 -9x +14)$ $=$ $0$ clearly the sum of the other 2 roots is $-b/a$ $=$ $9$ .Hence answer is $5 + 3 = 8$ .