The value of left| {begin{array}{*{20}{c}}1&{ | vedclass

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

Question

(B) Let $\Delta = \left| {\begin{array}{*{20}{c}}1&{\cos (\beta - \alpha )}&{\cos (\gamma - \alpha )}\{\cos (\alpha - \beta )}&1&{\cos (\gamma - \bet

Vedclass · Accepted Answer

$\left| {\begin{array}{*{20}{c}}{\sin \alpha }&{\cos \alpha }&0\{\sin \beta }&{\cos \beta }&0\{\sin \gamma }&{\cos \gamma }&0\end{array}} ight|^2$

Vedclass · Answer

(B) Let $\Delta = \left| {\begin{array}{*{20}{c}}1&{\cos (\beta - \alpha )}&{\cos (\gamma - \alpha )}\{\cos (\alpha - \beta )}&1&{\cos (\gamma - \beta )}\{\cos (\alpha - \gamma )}&{\cos (\beta - \gamma )}&1\end{array}} ight|$.Using the identity $\cos(x-y) = \cos x \cos y + \sin x \sin y$,we can write each element as a dot product of two vectors:$\Delta = \left| {\begin{array}{*{20}{c}}{\cos^2 \alpha + \sin^2 \alpha}&{\cos \alpha \cos \beta + \sin \alpha \sin \beta}&{\cos \alpha \cos \gamma + \sin \alpha \sin \gamma}\{\cos \beta \cos \alpha + \sin \beta \sin \alpha}&{\cos^2 \beta + \sin^2 \beta}&{\cos \beta \cos \gamma + \sin \beta \sin \gamma}\{\cos \gamma \cos \alpha + \sin \gamma \sin \alpha}&{\cos \gamma \cos \beta + \sin \gamma \sin \beta}&{\cos^2 \gamma + \sin^2 \gamma}\end{array}} ight|$.This determinant can be expressed as the product of two matrices:$\Delta = \left| {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }&0\{\cos \beta }&{\sin \beta }&0\{\cos \gamma }&{\sin \gamma }&0\end{array}} ight| 	imes \left| {\begin{array}{*{20}{c}}{\cos \alpha }&{\cos \beta }&{\cos \gamma}\{\sin \alpha }&{\sin \beta }&{\sin \gamma}\{0}&{0}&{0}\end{array}} ight|$.Since the second determinant is the transpose of the first,we have:$\Delta = \left| {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }&0\{\cos \beta }&{\sin \beta }&0\{\cos \gamma }&{\sin \gamma }&0\end{array}} ight|^2 = \left| {\begin{array}{*{20}{c}}{\sin \alpha }&{\cos \alpha }&0\{\sin \beta }&{\cos \beta }&0\{\sin \gamma }&{\cos \gamma }&0\end{array}} ight|^2$ (by swapping columns).Thus,the correct option is $B$.

The value of $\left| {\begin{array}{*{20}{c}}1&{\cos (\beta - \alpha )}&{\cos (\gamma - \alpha )}\\{\cos (\alpha - \beta )}&1&{\cos (\gamma - \beta )}\\{\cos (\alpha - \gamma )}&{\cos (\beta - \gamma )}&1\end{array}} \right|$ is

Similar Questions

Let $A$ and $B$ be real matrices of the form $\begin{bmatrix} \alpha & 0 \\ 0 & \beta \end{bmatrix}$ and $\begin{bmatrix} 0 & \gamma \\ \delta & 0 \end{bmatrix}$,respectively.
Statement $1$: $AB - BA$ is always an invertible matrix.
Statement $2$: $AB - BA$ is never an identity matrix.

For some $\alpha, \beta \in R$, let $A = \begin{bmatrix} \alpha & 2 \\ 1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 1 \\ 1 & \beta \end{bmatrix}$ be such that $A^{2} - 4A + 2I = B^{2} - 3B + I = O$. Then $(\text{det}(\text{adj}(A^{3} - B^{3})))^{2}$ is equal to ....

Let $A$ be the set of all $3 \times 3$ determinants with entries $0$ or $1$ only and $B$ be the subset of $A$ consisting of all determinants with value $1$. If $C$ is the subset of $A$ consisting of all determinants with value $-1$, then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module