Question

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

• A. $(x-1{)}^{n-1}(x+n-1)$
• B. $(x-1{)}^n(x+n-1)$
• C. $(x-1{)}^{-1}(x+n-1)$
• D. $(x-1{)}^{-1}(x-n+1)$

Answer 1

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

$\left|\begin{array}{cccc}x& 1& 1& \dots \\ 1& x& 1& \dots \\ 1& 1& x& \dots \\ \cdot & \cdot & \cdot & \cdots \end{array}\right|$

Using $C_1\to C_1+C_2+C_3+\dots$ in the above determinant, it reduces to, $\left|\begin{array}{cccc}x+n-1& 1& 1& \dots \\ x+n-1& x& 1& \dots \\ x+n-1& 1& x& \dots \\ \cdot & \cdot & \cdot & \cdots \end{array}\right|$

Take $x+n-1$ out of the determinant, $(x+n-1)\left|\begin{array}{cccc}1& 1& 1& \dots \\ 1& x& 1& \dots \\ 1& 1& x& \dots \\ \cdot & \cdot & \cdot & \cdots \end{array}\right|$

Now apply, $R_a\to R_a-R_{a-1}$ where $a\ge 2$

$(x+n-1)\left|\begin{array}{cccc}1& 1& 1& \dots \\ 0& x-1& 0& \dots \\ 0& 0& x-1& \dots \\ \cdot & \cdot & \cdot & \cdots \end{array}\right|$
which results in $(x+n-1)(x-1{)}^{n-1}$
(Applying concept of upper triangular determinant).