Let A = begin{bmatrix} 0 & 2q & r p & q & -r | vedclass

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

Question

(C) Given that $AA^T = I_3$,$A$ is an orthogonal matrix. Thus,the product $AA^T$ is: $\begin{bmatrix} 0 & 2q & r \ p & q & -r \ p & -q & r \end{bmat

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}}$

Vedclass · Answer

(C) Given that $AA^T = I_3$,$A$ is an orthogonal matrix. Thus,the product $AA^T$ is: $\begin{bmatrix} 0 & 2q & r \ p & q & -r \ p & -q & r \end{bmatrix} \begin{bmatrix} 0 & p & p \ 2q & q & -q \ r & -r & r \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$ Performing matrix multiplication: Row $1$ $\cdot$ Col $1$: $0^2 + (2q)^2 + r^2 = 4q^2 + r^2 = 1$ (Eq. $1$) Row $2$ $\cdot$ Col $2$: $p^2 + q^2 + (-r)^2 = p^2 + q^2 + r^2 = 1$ (Eq. $2$) Row $2$ $\cdot$ Col $3$: $p^2 - q^2 - r^2 = 0$ (Eq. $3$) Adding (Eq. $2$) and (Eq. $3$): $2p^2 = 1 \implies p^2 = \frac{1}{2} \implies |p| = \frac{1}{\sqrt{2}}$.

Let $A = \begin{bmatrix} 0 & 2q & r \\ p & q & -r \\ p & -q & r \end{bmatrix}$. If $AA^T = I_3$,then $|p|$ is

Similar Questions

If $A = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$ such that $XA = B^T$ and $A^{-1}Y = B$, then find the value of $XY$.

If $A = \begin{bmatrix} 3 & 4 \\ 5 & 7 \end{bmatrix}$,then $A(adj A) = $

If possible,using elementary row transformations,find the inverse of the following matrix:
$\left[\begin{array}{ccc}2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3\end{array}\right]$

If $A$ and $B$ are invertible matrices of the same order, then which of the following is not correct?

If $A = \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ \alpha & 6 & -5 \\ \beta & -2 & 2 \end{bmatrix}$,then the values of $\alpha$ and $\beta$ are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module