If P(theta) = begin{bmatrix} 1 & cot theta - | vedclass

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

Question

(B) Given $P(	heta) = \begin{bmatrix} 1 & \cot 	heta \ -\cot 	heta & 1 \end{bmatrix}$.First,we find the determinant of $P(	heta)$:$|P(	heta)| =

Vedclass · Accepted Answer

$P(-	heta)$

Vedclass · Answer

(B) Given $P(	heta) = \begin{bmatrix} 1 & \cot 	heta \ -\cot 	heta & 1 \end{bmatrix}$.First,we find the determinant of $P(	heta)$:$|P(	heta)| = (1)(1) - (\cot 	heta)(-\cot 	heta) = 1 + \cot^2 	heta = \csc^2 	heta$.Since $PQ = I$,$Q$ is the inverse of $P$,i.e.,$Q = P^{-1} = \frac{1}{|P(	heta)|} 	ext{adj}(P(	heta))$.The adjoint of $P(	heta)$ is $\begin{bmatrix} 1 & -\cot 	heta \ \cot 	heta & 1 \end{bmatrix}$.Thus,$Q = \frac{1}{\csc^2 	heta} \begin{bmatrix} 1 & -\cot 	heta \ \cot 	heta & 1 \end{bmatrix}$.We need to find $(\csc^2 	heta)Q$:$(\csc^2 	heta)Q = \csc^2 	heta \cdot \frac{1}{\csc^2 	heta} \begin{bmatrix} 1 & -\cot 	heta \ \cot 	heta & 1 \end{bmatrix} = \begin{bmatrix} 1 & -\cot 	heta \ \cot 	heta & 1 \end{bmatrix}$.Now,observe $P(-	heta) = \begin{bmatrix} 1 & \cot(-	heta) \ -\cot(-	heta) & 1 \end{bmatrix} = \begin{bmatrix} 1 & -\cot 	heta \ \cot 	heta & 1 \end{bmatrix}$.Therefore,$(\csc^2 	heta)Q = P(-	heta)$.

If $P(\theta) = \begin{bmatrix} 1 & \cot \theta \\ -\cot \theta & 1 \end{bmatrix}$ and $PQ = I$,then find $(\csc^2 \theta)Q$,where $I$ is an identity matrix of order $2 \times 2$.

Similar Questions

If $A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}$,then

If $A = \begin{bmatrix} k & 2 \\ -2 & -k \end{bmatrix}$,then $A^{-1}$ does not exist if $k =$

If $A = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}$,then $A^{-1} = $

The inverse of the matrix $ \begin{bmatrix} 2 & 5 & 0 \\ 0 & 1 & 1 \\ -1 & 0 & 3 \end{bmatrix} $ is

Assertion $(A)$: If $B$ is a $3 \times 3$ matrix and $|B|=6$,then $|\operatorname{Adj}(B)|=36$.
Reason $(R)$: If $B$ is a square matrix of order $n$,then $|\operatorname{Adj}(B)|=|B|^{n}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module