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
If \[ A = \begin{bmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{bmatrix} \] then find \( A^{-1} \). Hence, solve the system of linear equations: \[ x - 2y = 10, \] \[ 2x - y - z = 8, \] \[ -2y + z = 7. \]