We have \(\begin{bmatrix} x &-5&-1 \end{bmatrix}\)\(\begin{bmatrix} 1 & 0 & 2\\ 0 & 2 & 1 \\2&1&3 \end{bmatrix}\)\(\begin{bmatrix} x \\ 4\\1 \end{bmatrix}=O\)
⇒ \(\begin{bmatrix} x+0-2 &0-10+-0 &2x-5-3\end{bmatrix}\) \(\begin{bmatrix} x \\ 4\\1 \end{bmatrix}=O\)
⇒ \(\begin{bmatrix} x-2 &-10&2x-8 \end{bmatrix}\) \(\begin{bmatrix} x \\ 4\\1 \end{bmatrix}=O\)
⇒ \(\begin{bmatrix} x(x-2) -40+2x-8 \end{bmatrix}=0\)
⇒ \(\begin{bmatrix} x^2 -2x-40+2x-8 \end{bmatrix}=[0]\)
⇒ \(\begin{bmatrix} x^2 -48 \end{bmatrix}=[0]\)
⇒\(\therefore x^2=48\)
⇒ \(x= 4 \sqrt3\)
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 \(\begin{vmatrix} cos\alpha cos\beta &cos\alpha sin\beta &-sin\alpha \\ -sin\beta&cos\beta &0 \\ sin\alpha cos\beta&sin\alpha\sin\beta &cos\alpha \end{vmatrix}\)