Let a, b, c be such that b(a+c) neq 0. If lef | vedclass

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

Question

(D) Let the given equation be $D_1 + D_2 = 0$. We have $D_1 = \left|\begin{array}{ccc}a & a+1 & a-1 \ -b & b+1 & b-1 \ c & c-1 & c+1\end{array}igh

Vedclass · Accepted Answer

any odd integer

Vedclass · Answer

(D) Let the given equation be $D_1 + D_2 = 0$. We have $D_1 = \left|\begin{array}{ccc}a & a+1 & a-1 \ -b & b+1 & b-1 \ c & c-1 & c+1\end{array}ight|$. Taking the transpose of $D_2$, we get $D_2 = \left|\begin{array}{ccc}a+1 & a-1 & (-1)^{n+2} a \ b+1 & b-1 & (-1)^{n+1} b \ c-1 & c+1 & (-1)^{n} c\end{array}ight|$. Now, swapping the first and third columns of $D_2$ twice (or rearranging columns) to match the structure of $D_1$: $D_2 = \left|\begin{array}{ccc}(-1)^{n+2} a & a+1 & a-1 \ (-1)^{n+1} b & b+1 & b-1 \ (-1)^{n} c & c-1 & c+1\end{array}ight|$. Adding $D_1$ and $D_2$, we get: $\left|\begin{array}{ccc}a(1 + (-1)^{n+2}) & a+1 & a-1 \ b(-1 + (-1)^{n+1}) & b+1 & b-1 \ c(1 + (-1)^{n}) & c-1 & c+1\end{array}ight| = 0$. For the determinant to be zero for arbitrary $a, b, c$, the first column must be zero. $1 + (-1)^{n+2} = 0 \Rightarrow (-1)^{n+2} = -1$, which implies $n+2$ is odd, so $n$ is odd. $-1 + (-1)^{n+1} = 0 \Rightarrow (-1)^{n+1} = 1$, which implies $n+1$ is even, so $n$ is odd. $1 + (-1)^{n} = 0 \Rightarrow (-1)^{n} = -1$, which implies $n$ is odd. Thus, $n$ is any odd integer.

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

Similar Questions

If $A$ is a square matrix of order $n$ and $A = kB$,where $k$ is a scalar,then $|A|=$

Prove that $\left|\begin{array}{ccc}a & a+b & a+b+c \\ 2a & 3a+2b & 4a+3b+2c \\ 3a & 6a+3b & 10a+6b+3c\end{array}\right|=a^{3}$

Let $A$ be a $3 \times 3$ matrix with $\operatorname{det}(A) = 4$. Let $R_{i}$ denote the $i^{\text{th}}$ row of $A$. If a matrix $B$ is obtained by performing the operation $R_{2} \rightarrow 2R_{2} + 5R_{3}$ on $2A$,then $\operatorname{det}(B)$ is equal to:

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 $k$.

The parameter on which the value of the determinant $\left| \begin{array}{ccc} 1 & a & a^2 \\ \cos(p-d)x & \cos px & \cos(p+d)x \\ \sin(p-d)x & \sin px & \sin(p+d)x \end{array} \right|$ does not depend is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module