We have \(\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}\)\(\begin{bmatrix} 1 & 2 & 0\\ 2 & 0 & 1 \\1&0&2 \end{bmatrix}\)\(\begin{bmatrix} 0 \\2\\x\end{bmatrix}\)\(=O\)
⇒ \(\begin{bmatrix} 1+4+1 & 2+0+0 & 0+2+2 \end{bmatrix}\) \(\begin{bmatrix} 0 \\2\\x\end{bmatrix}\)\(=O\)
⇒ \(\begin{bmatrix} 6 & 2 & 4 \end{bmatrix}\) \(\begin{bmatrix} 0 \\2\\x\end{bmatrix}\)\(=O\)
⇒\(\begin{bmatrix} 6(0) +2(2)+4(x) \end{bmatrix}=O\)
⇒[4+4x]=[0]
\(\therefore 4+4x=0\)
⇒x=-1
Thus, the required value of x is −1
What is the Planning Process?
Evaluate \(\begin{vmatrix} cos\alpha cos\beta &cos\alpha sin\beta &-sin\alpha \\ -sin\beta&cos\beta &0 \\ sin\alpha cos\beta&sin\alpha\sin\beta &cos\alpha \end{vmatrix}\)