Use the product of matrices (1, 2, -3; 3, 2, -2; 2, -1, 1) (0, 1, 2; -7, 7, -7; -7, 5, -4) to solve the following system of equations: x+2 y-3 z=6; 3…

$$\mathrm{A} B=\left(\begin{array}{ccc} 1 & 2 & -3 \\ 3 & 2 & -2 \\ 2 & -1 & 1 \end{array}\right)\left(\begin{array}{ccc} 0 & 1 & 2 \\ -7 & 7 & -7 \\ -7 & 5 & -4 \end{array}\right)=\left(\begin{array}{lll} 7 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 7 \end{array}\right)=7 I $$ Thus, $\displaystyle \mathrm{A}^{-1}=\frac{1}{7} B=\frac{1}{7}\left(\begin{array}{ccc} 0 & 1 & 2 \\ -7 & 7 & -7 \\ -7 & 5 & -4 \end{array}\right)$ so, Given equation can be written into a matrix equation as $$\left(\begin{array}{ccc} 1 & 2 & -3 \\ 3 & 2 & -2 \\ 2 & -1 & 1 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{l} 6 \\ 3 \\ \end{array}\right) \Rightarrow \mathrm{X}=\mathrm{A}^{-1} \cdot \mathrm{C} $$ $\displaystyle \mathrm{A} \quad \mathrm{X}=\mathrm{C}$ $$\begin{aligned} & \left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\frac{1}{7}\left(\begin{array}{ccc} 0 & 1 & 2 \\ -7 & 7 & -7 \\ -7 & 5 & -4 \end{array}\right)\left(\begin{array}{l} 6 \\ 3 \\ \end{array}\right)=\frac{1}{7}\left(\begin{array}{c} 7 \\ -35 \\ -35 \end{array}\right)=\left(\begin{array}{c} 1 \\ -5 \\ -5 \end{array}\right) \\ & \therefore x=1, y=-5, z=-5 \end{aligned} $$