If A=\(\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\) and I=\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\),find k so that A2=kA-2I
A2=A.A
\(\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\)\(\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\)
=\(\begin{bmatrix}3(3)+(-2)(4)&3(-2)+(-2)(-2)\\4(3)+(-2)(4)&4(-2)+(-2)(-2)\end{bmatrix}\)=\(\begin{bmatrix}1&-2\\4&-4\end{bmatrix}\)
Now A2=kA-2I
\(\Rightarrow \begin{bmatrix}1&-2\\4&-4\end{bmatrix}\)=\(\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\)-2\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)
\(\Rightarrow\) \(\begin{bmatrix}1&-2\\4&-4\end{bmatrix}\)=\(\begin{bmatrix}3k&-2k\\4k&-2k\end{bmatrix}\)-2\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)
\(\Rightarrow\) \(\begin{bmatrix}1&-2\\4&-4\end{bmatrix}\)=\(\begin{bmatrix}3k-2&-2k\\4k-2k&-2\end{bmatrix}\)
Comparing the corresponding elements, we have:
3k-2=1
3k=2
\(\Rightarrow\) k=1
Thus, the value of k is 1.
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 \(\begin{vmatrix} cos\alpha cos\beta &cos\alpha sin\beta &-sin\alpha \\ -sin\beta&cos\beta &0 \\ sin\alpha cos\beta&sin\alpha\sin\beta &cos\alpha \end{vmatrix}\)