A furniture workshop produces three types of furniture - chairs, tables and beds each day. On a particular day the total number of furniture pieces…

Let the numbers of chairs, tables and beds produced be $\displaystyle \mathrm{x}, \mathrm{y}$ and z respectively. \[\begin{aligned} & \therefore \mathrm{x}+\mathrm{y}+\mathrm{z}=45 ; \quad-\mathrm{x}+0 . \mathrm{y}+\mathrm{z}=8 ; \quad \mathrm{x}-2 \mathrm{y}+\mathrm{z}=0 \\ & \text { Let } A=\left[\begin{array}{ccc} 1 & 1 & 1 \\ -1 & 0 & 1 \\ 1 & -2 & 1 \end{array}\right], X=\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right], B=\left[\begin{array}{c} 45 \\ 8 \\ 0 \end{array}\right] \\ & |A|=1(0+2)-1(-1-1)+1(2-0)=6 \neq 0 \\ & \therefore A^{-1} \text { exists } \\ & \quad A X=B \Rightarrow X=A^{-1} B \\ & \operatorname{adj}(A)=\left[\begin{array}{ccc} 2 & -3 & 1 \\ 2 & 0 & -2 \\ 2 & 3 & 1 \end{array}\right] \\ & A^{-1}=\frac{1}{6}\left[\begin{array}{ccc} 2 & -3 & 1 \\ 2 & 0 & -2 \\ 2 & 3 & 1 \end{array}\right] \\ & \therefore\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\frac{1}{6}\left[\begin{array}{ccc} 2 & -3 & 1 \\ 2 & 0 & -2 \\ 2 & 3 & 1 \end{array}\right]\left[\begin{array}{l} 45 \\ 8 \\ 0 \end{array}\right]=\left[\begin{array}{l} 11 \\ 15 \\ 19 \end{array}\right] \end{aligned} \] So, $\displaystyle \mathrm{x}=11, \mathrm{y}=15, \mathrm{z}=19$ Hence the numbers of chairs, tables and beds produced are $\displaystyle 11$, $\displaystyle 15$ and $\displaystyle 19$ respectively.