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(C) Let $D_1 = \begin{vmatrix} a & a+1 & a-1 \ -b & b+1 & b-1 \ c & c-1 & c+1 \end{vmatrix}$ and $D_2 = \begin{vmatrix} a+1 & b+1 & c-1 \ a-1 & b-1

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Any odd integer

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(C) Let $D_1 = \begin{vmatrix} a & a+1 & a-1 \ -b & b+1 & b-1 \ c & c-1 & c+1 \end{vmatrix}$ and $D_2 = \begin{vmatrix} a+1 & b+1 & c-1 \ a-1 & b-1 & c+1 \ (-1)^{n+2} a & (-1)^{n-1} b & (-1)^n c \end{vmatrix}$.We are given $D_1 + D_2 = 0$.In $D_2$,take $(-1)^n$ common from the third row: $D_2 = (-1)^n \begin{vmatrix} a+1 & b+1 & c-1 \ a-1 & b-1 & c+1 \ a & -b & c \end{vmatrix}$.Now,perform column operations on $D_2$: Swap $C_1$ and $C_2$,then swap $C_2$ and $C_3$. Two swaps result in no change in sign.$D_2 = (-1)^n \begin{vmatrix} a & a+1 & a-1 \ -b & b+1 & b-1 \ c & c-1 & c+1 \end{vmatrix} = (-1)^n D_1$.Thus,$D_1 + (-1)^n D_1 = 0 \Rightarrow (1 + (-1)^n) D_1 = 0$.For this to hold for arbitrary $a, b, c$,we must have $1 + (-1)^n = 0$,which implies $(-1)^n = -1$.This is true when $n$ is an odd integer.

Let $a, b, c$ be such that $(b+c) \neq 0$ and $\left|\begin{array}{ccc} a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1 \end{array}\right|+\left|\begin{array}{ccc} a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2} a & (-1)^{n-1} b & (-1)^n c \end{array}\right|=0$. Then the value of $n$ is

Similar Questions

Using properties of determinants,prove that:
$\left|\begin{array}{ccc}3 a & -a+b & -a+c \\ -b+a & 3 b & -b+c \\ -c+a & -c+b & 3 c\end{array}\right|=3(a+b+c)(a b+b c+c a)$

Let $D_1 = \begin{vmatrix} a & b & a+b \\ c & d & c+d \\ a & b & a-b \end{vmatrix}$ and $D_2 = \begin{vmatrix} a & c & a+c \\ b & d & b+d \\ a & c & a+b+c \end{vmatrix}$. Then the value of $\frac{D_1}{D_2}$,where $b \neq 0$ and $ad \neq bc$,is:

$\left|\begin{array}{ccc} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{array}\right| = $

The determinant $\left| {\begin{array}{*{20}{c}}{{a^2} + {x^2}}&{ab}&{ca}\\{ab}&{{b^2} + {x^2}}&{bc}\\{ca}&{bc}&{{c^2} + {x^2}}\end{array}} \right|$ is a divisor of:

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