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

Answer 1

The correct option is A ${(x-1)}^{n-1}(x+n-1)$

$\left|\begin{array}{cccc}x& 1& 1& ...\\ 1& x& 1& ...\\ 1& 1& x& ...\\ ...& ...& ...& ...\end{array}\right|$
Using $C_1\to C_1+C_2+C_3+\dots$ in the above matrix, it reduces to,
$\left|\begin{array}{cccc}x+n-1& 1& 1& ...\\ x+n-1& x& 1& ...\\ x+n-1& 1& x& ...\\ ...& ...& ...& ...\end{array}\right|$
Take $x+n-1$ out of the determinant,
$(x+n-1)\ast \left|\begin{array}{cccc}1& 1& 1& ...\\ 1& x& 1& ...\\ 1& 1& x& ...\\ ...& ...& ...& ...\end{array}\right|$
Now apply, $R_a\to R_a-R_{a-1}$ where $a\ge 2$
$(x+n-1)\ast \left|\begin{array}{cccc}1& 1& 1& ...\\ 0& x-1& 0& ...\\ 0& 0& x-1& ...\\ ...& ...& ...& ...\end{array}\right|$, which results in $(x+n-1)(x-1{)}^{n-1}$
Hence, Option A.