Solve the equation for x,y,z and t if 2\(\begin{bmatrix}x&y\\y&t\end{bmatrix}\)+3\(\begin{bmatrix}1&-1\\0&2\end{bmatrix}\)=3\(\begin{bmatrix}3&5\\4&6\end{bmatrix}\)
2\(\begin{bmatrix}x&y\\y&t\end{bmatrix}\)+3\(\begin{bmatrix}1&-1\\0&2\end{bmatrix}\)=3\(\begin{bmatrix}3&5\\4&6\end{bmatrix}\)
\(\Rightarrow \) \(\begin{bmatrix}2x&2z\\2y&2t\end{bmatrix}\)+\(\begin{bmatrix}3&-3\\0&6\end{bmatrix}\)=\(\begin{bmatrix}9&15\\12&18\end{bmatrix}\)
\(\Rightarrow \begin{bmatrix}2x+3&2z-3\\2y&2t+6\end{bmatrix}\)=\(\begin{bmatrix}9&15\\12&18\end{bmatrix}\)
Comparing the corresponding elements of these two matrices, we get:
2x+3=9
\(\Rightarrow\) x=3
2y=12
\(\Rightarrow\) y=6
2z-3=15
\(\Rightarrow\) 2z=18
z=9
2t+6=18
2t=12
t=6.
\(\therefore\) x=3,y=6,z=9 and t=6
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 \(\begin{vmatrix} cos\alpha cos\beta &cos\alpha sin\beta &-sin\alpha \\ -sin\beta&cos\beta &0 \\ sin\alpha cos\beta&sin\alpha\sin\beta &cos\alpha \end{vmatrix}\)