(a) If A = [1, 2, -3; 2, 0, -3; 1, 2, 0], then find A^-1 and hence solve the following system of equations: x+2 y-3 z=1; 2 x-3 z=2; x+2 y=3

$\displaystyle |\mathrm{A}|=1(6)-2(3)-3(4)=-12 \neq 0, \therefore \mathrm{A}^{-1}$ exist

$$\begin{aligned} \operatorname{adj} \mathrm{A} & =\left[\begin{array}{ccc} 6 & -6 & -6 -3 & 3 & -3 4 & 0 & -4 \end{array}\right] \therefore \mathrm{A}^{-$\displaystyle 1$} & =-\frac{1}{12}\left[\begin{array}{ccc} 6 & -6 & -6 -3 & 3 & -3 4 & 0 & -4 \end{array}\right] \end{aligned}

$$The given system of equations can be written as $\displaystyle \mathbf{A X}=\mathbf{B}, \mathbf{X}=\left[\begin{array}{l}\mathbf{x} \\ \mathbf{y} \\ \mathbf{z}\end{array}\right], \mathbf{B}=\left[\begin{array}{l}1 \\ 3\end{array}\right]$

$$X=\mathrm{A}^{-$\displaystyle 1$} B \Rightarrow\left[\begin{array}{l} x y z \end{array}\right]=-\frac{1}{12}\left[\begin{array}{ccc} 6 & -6 & -6 -3 & 3 & -3 4 & 0 & -4 \end{array}\right]\left[\begin{array}{l} 1 2 \end{array}\right]=\left[\begin{array}{c} 2 1 / 2 2 / 3 \end{array}\right]

$$∴ The solution of the given system of equations is: $\displaystyle x=2, y=\frac{1}{2}, z=\frac{2}{3}$

Or $$ \text { (b) } \begin{aligned} & {\left[\begin{array}{ccc} 1 & 2 & -3 2 & 3 & 2 3 & -3 & -4 \end{array}\right]\left[\begin{array}{ccc} -6 & 17 & 13 14 & 5 & -8 -15 & 9 & -1 \end{array}\right]=\left[\begin{array}{ccc} 67 & 0 & 0 0 & 67 & 0 0 & 0 & 67 \end{array}\right] } \Rightarrow & {\left[\begin{array}{ccc} 1 & 2 & -3 2 & 3 & 2 3 & -3 & -4 \end{array}\right]^{-$\displaystyle 1$}=\frac{1}{67}\left[\begin{array}{ccc} -6 & 17 & 13 14 & 5 & -8 -15 & 9 & -1 \end{array}\right] } \end{aligned}

$$Solution of the system of equations is given by:

$$\begin{gathered} {\left[\begin{array}{l} x y z \end{array}\right]=\left[\begin{array}{ccc} 1 & 2 & -3 2 & 3 & 2 3 & -3 & -4 \end{array}\right]^{-$\displaystyle 1$}\left[\begin{array}{c} -4 2 11 \end{array}\right]=\frac{1}{67}\left[\begin{array}{ccc} -6 & 17 & 13 14 & 5 & -8 -15 & 9 & -1 \end{array}\right]\left[\begin{array}{c} -4 2 11 \end{array}\right]=\left[\begin{array}{c} 3 -2 \end{array}\right],} \therefore x=$\displaystyle 3$, y=-$\displaystyle 2$, z=$\displaystyle 1$ \end{gathered}