Evaluate the determinant: left| begin{array}{ | vedclass

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

Question

(A) Let the given determinant be $\Delta$. Using the identity $(p+q)^2 - (p-q)^2 = 4pq$,we apply the column operation $C_1 	o C_1 - C_2$: $(a^x + a^{

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Let the given determinant be $\Delta$. Using the identity $(p+q)^2 - (p-q)^2 = 4pq$,we apply the column operation $C_1 	o C_1 - C_2$: $(a^x + a^{-x})^2 - (a^x - a^{-x})^2 = 4(a^x)(a^{-x}) = 4(a^0) = 4$. Similarly,for the second and third rows,we get $4$ and $4$. Thus,the determinant becomes: $\Delta = \left| \begin{array}{ccc} 4 & (a^x - a^{-x})^2 & 1 \ 4 & (b^x - b^{-x})^2 & 1 \ 4 & (c^x - c^{-x})^2 & 1 \end{array} ight|$. Since column $C_1$ and column $C_3$ are proportional (specifically,$C_1 = 4C_3$),the value of the determinant is $0$.

Evaluate the determinant: $\left| \begin{array}{ccc} (a^x + a^{-x})^2 & (a^x - a^{-x})^2 & 1 \\ (b^x + b^{-x})^2 & (b^x - b^{-x})^2 & 1 \\ (c^x + c^{-x})^2 & (c^x - c^{-x})^2 & 1 \end{array} \right|$

Similar Questions

The determinant of a skew-symmetric matrix of order $3$ is always:

$A$ value of $\theta$ lying between $0$ and $\frac{\pi}{2}$ and satisfying $\left|\begin{array}{ccc} 1+\sin^2 \theta & \cos^2 \theta & 4\sin 4\theta \\ \sin^2 \theta & 1+\cos^2 \theta & 4\sin 4\theta \\ \sin^2 \theta & \cos^2 \theta & 1+4\sin 4\theta \end{array}\right| = 0$ is:

$\left|\begin{array}{ccc}1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3\end{array}\right|=$

The total number of distinct $x \in \mathbb{R}$ for which $\left|\begin{array}{ccc}x & x^2 & 1+x^3 \\ 2x & 4x^2 & 1+8x^3 \\ 3x & 9x^2 & 1+27x^3\end{array}\right|=10$ is

$\left| \begin{array}{ccc} 11 & 12 & 13 \\ 12 & 13 & 14 \\ 13 & 14 & 15 \end{array} \right| = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module