For the matrix A=\(\begin{bmatrix}1&5\\6&7\end{bmatrix}\),verify that
I. (A+A') is a symmetric matrix
II. (A-A') is a skew symmetric matrix
A'= \(\begin{bmatrix}1&6\\5&7\end{bmatrix}\)
(i)A+A'=\(\begin{bmatrix}1&5\\6&7\end{bmatrix}\)+\(\begin{bmatrix}1&6\\5&7\end{bmatrix}\)
=\(\begin{bmatrix}2&11\\11&14\end{bmatrix}\)
therefore (A+A')'= \(\begin{bmatrix}2&11\\11&14\end{bmatrix}\)=A+A'
Hence,(A+A') is a symmetric matrix.
(ii)A-A'= \(\begin{bmatrix}1&5\\6&7\end{bmatrix}\)-\(\begin{bmatrix}1&6\\5&7\end{bmatrix}\)
=\(\begin{bmatrix}0&-1\\1&0\end{bmatrix}\)
(A-A')'=\(\begin{bmatrix}0&1\\-1&0\end{bmatrix}\)=-\(\begin{bmatrix}0&-1\\1&0\end{bmatrix}\)=-(A-A')
Hence,(A-A') is a skew-symmetric matrix.
Let
\( A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha \end{bmatrix} \)
and \(|2A|^3 = 2^{21}\) where \(\alpha, \beta \in \mathbb{Z}\). Then a value of \(\alpha\) is:
What is the Planning Process?
Evaluate \(\begin{vmatrix} cos\alpha cos\beta &cos\alpha sin\beta &-sin\alpha \\ -sin\beta&cos\beta &0 \\ sin\alpha cos\beta&sin\alpha\sin\beta &cos\alpha \end{vmatrix}\)