Question

The value of the determinant of $n$th order, being given by $\left|\begin{array}{cccc}x& 1& 1& ...\\ 1& x& 1& ...\\ 1& 1& x& ...\\ ...& ...& ...& ...\end{array}\right|$
is

• A. ${(x-1)}^{n-1}(x+n-1)$
• B. ${(x-1)}^{n-1}(x+n-1)$
• C. ${(1-x)}^{n-1}(x+n-1)$
• D. none of these

Answer 1

Answer: (A)

${(x-1)}^{n-1}(x+n-1)$

we have

$\left|\begin{array}{cccc}x& 1& 1& ...\\ 1& x& 1& ...\\ 1& 1& x& ...\\ ...& ...& ...& ...\end{array}\right|=\left|\begin{array}{cccc}x& 1& 1& ...\\ 1-x& x-1& 0& ...\\ 1-x& 0& x-1& ...\\ ...& ...& ...& ...\end{array}\right|$

(applying $R_2\to R_2-R_1,R_3\to R_3-R_1,....R_n\to R_n-R_1$)

$=x{(x-1)}^{n-1}+{(x-1)}^{n-1}+{(x-1)}^{n-1}+....+{(x-1)}^{n-1}(n-1)times$

expanding along $R_i$

$=x{(x-1)}^{n-1}+(n-1){(x-1)}^{n-1}={(x-1)}^{n-1}(x+n-1)$