Explanation:
(A) : Given, \(x+2 y+3 z=6 \quad \ldots\) (i)
\(3 x-2 y+z=2 \tag{ii}\)
\(4 x+2 y+z=7 \tag{iii}\)
\(A=\left[\begin{array}{ccc}1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1\end{array}\right], B=\left[\begin{array}{l}6 \\ 2 \\ 7\end{array}\right]\) and \(X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]\)
\(\begin{array}{ll}\therefore & |\mathrm{A}|=\left|\begin{array}{ccc}1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1\end{array}\right| \\ =1(-2-2)-2(3-4)+3(6+8)\end{array}\)
\(=-4+2+42=40\)
Now, adj \(A=\left[\begin{array}{ccc}-4 & 4 & 8 \\ 1 & -11 & 8 \\ 14 & 6 & -8\end{array}\right]\)
\(\therefore A^{-1}=\frac{1}{|A|}\) adj \(A=\frac{1}{40}\left[\begin{array}{ccc}-4 & 4 & 8 \\ 1 & -11 & 8 \\ 14 & 6 & -8\end{array}\right]\)
Now, \(X=A^{-1} B=\frac{1}{40}\left[\begin{array}{ccc}-4 & 4 & 8 \\ 1 & -11 & 8 \\ 14 & 6 & -8\end{array}\right]\left[\begin{array}{l}6 \\ 2 \\ 7\end{array}\right]\)
\(\frac{1}{40}\left[\begin{array}{c}-24+8+56 \\ 6-22+56 \\ 84+12-56\end{array}\right]=\frac{1}{40}\left[\begin{array}{l}40 \\ 40 \\ 40\end{array}\right]=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]\)
\(\Rightarrow x=1, y=1, z=1\)
\(\Rightarrow \mathrm{x}=1, \mathrm{y}=1, \mathrm{z}=1\)
