Let a, b, c be such that b + c ne 0. If left| | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the given equation be $D_1 + D_2 = 0$. First,we observe that the second determinant $D_2$ can be transposed without changing its value. $D_2 =

Vedclass · Accepted Answer

any odd integer

Vedclass · Answer

(C) Let the given equation be $D_1 + D_2 = 0$. First,we observe that the second determinant $D_2$ can be transposed without changing its value. $D_2 = \left| \begin{array}{ccc} a+1 & a-1 & (-1)^{n+2}a \ b+1 & b-1 & (-1)^{n+1}b \ c-1 & c+1 & (-1)^n c \end{array} ight|$. By adding the two determinants,we find that for the sum to be zero for arbitrary $a, b, c$,the columns must cancel out. Specifically,if $n$ is an odd integer,then $(-1)^{n+2} = -1$,$(-1)^{n+1} = 1$,and $(-1)^n = -1$. Substituting these into the sum of determinants,the terms cancel out to zero. Thus,$n$ must be any odd integer.

Similar Questions

If $\begin{vmatrix} x^k & x^{k+2} & x^{k+3} \\ y^k & y^{k+2} & y^{k+3} \\ z^k & z^{k+2} & z^{k+3} \end{vmatrix} = (x-y)(y-z)(z-x)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$, then find the value of $k$.

If $D = \begin{vmatrix} a^2 + 1 & ab & ac \\ ba & b^2 + 1 & bc \\ ca & cb & c^2 + 1 \end{vmatrix}$,then $D =$

$\left| {\begin{array}{ccc} 1/a & a^2 & bc \\ 1/b & b^2 & ca \\ 1/c & c^2 & ab \end{array}} \right| = $

The determinant $\left| {\begin{array}{*{20}{c}}{1 + a + x}&{a + y}&{a + z}\\{b + x}&{1 + b + y}&{b + z}\\{c + x}&{c + y}&{1 + c + z}\end{array}} \right|$ is equal to:

$\left| \begin{array}{ccc} a - b & b - c & c - a \\ x - y & y - z & z - x \\ p - q & q - r & r - p \end{array} \right| = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module